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Grade 4 · Adding fractions with unlike denominators
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Why couldn't the two fractions get married?
They couldn't agree on a common denominator.
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Grade 4 · Adding fractions with unlike denominators
The board stays in the language your child is taught in. The script comes to you in yours.El tablero se queda en el idioma en que le enseñan a su hijo o hija. El guion le llega a usted en el suyo.The board stays in the language you are taught in. The script comes to you in the one you read best.El tablero se queda en el idioma en que te enseñan. El guion te llega en el que leas mejor.Bảng vẫn giữ nguyên ngôn ngữ mà con anh/chị được dạy. Kịch bản đến với anh/chị bằng ngôn ngữ của anh/chị.Bảng vẫn giữ nguyên ngôn ngữ em được học. Kịch bản đến với em bằng ngôn ngữ em đọc tốt nhất.O quadro continua no idioma em que seu filho ou filha estuda. O roteiro chega até você no seu.O quadro continua no idioma em que você estuda. O roteiro chega até você naquele que você lê melhor.
This is what a parent or coach reads to them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Coach Mia đọc to phần Nói — đúng những lời được in bên dưới.Mia đọc to phần Nói cho em nghe — đúng những lời được in bên dưới.A Coach Mia lê o bloco Diga em voz alta — as mesmas palavras impressas abaixo.A Mia lê o bloco Diga em voz alta para você — as mesmas palavras impressas abaixo.
ROUND 1Find Common Denominators
ON SCREEN — Board shows: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Four soccer balls sit above that sentence under the prompt “Tap the pieces to count!”, and four answer choices sit below it — A is 2/4, B is 3/4, C is 2/6, D is 1/6.
“Read it with me. She made 1/2 of her kicks in the first half. She made 1/4 in the second half. We want the whole game, so we add those two. Now look at the bottom numbers. One is 2. The other is 4. They are different, and that is the whole problem — we cannot add pieces that are not the same size. So we make them the same size. Count by twos: 2, 4. Count by fours: 4, 8. Both lists land on 4, so 4 is the size we will use. Change 1/2 into fourths. Cut each half in two and you get 2 pieces out of 4, so 1/2 is the same amount as 2/4. That is the same as multiplying the top and the bottom by 2: 1 times 2 is 2, and 2 times 2 is 4. Now the pieces match. Add the tops: 2 plus 1 is 3. Keep the bottom at 4, because the pieces did not change size. She made 3/4 of her penalty kicks.”
Point at the two bottom numbers on the screen — the 2 and the 4 — and ask which number both of them fit into. Then stop talking and wait. Do not fill it in for them. When they say 4, ask them to say out loud what 1/2 turns into before they touch a single answer choice. If they want to tap the four soccer balls to count the fourths, let them; that is what the balls are for.
The mistake to watch for is adding straight across — 1 plus 1 on top and 2 plus 4 on the bottom, which lands on 2/6. That is exactly why 2/6 is sitting there as choice C. Ask them: why can't we just add the 2 and the 4 on the bottom? If they can tell you the pieces are different sizes until we turn the halves into fourths, they have it. If they answer 2/4, they converted correctly and then forgot to add the second half.
3/4. First, change 1/2 to 2/4 so both fractions have the same denominator of 4. Then add: 2/4 + 1/4 = 3/4. The bottom stays 4 because making the halves into fourths changed how many pieces there are, not how big a piece is.
RONDA 1Buscar denominador común
EN PANTALLA — El tablero muestra, en inglés, tal como lo ve su hijo o hija: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Es decir: una jugadora anotó 1/2 de sus penales en el primer tiempo y 1/4 en el segundo. Arriba de esa frase hay cuatro balones que se pueden tocar, y debajo hay cuatro opciones: la A es 2/4, la B es 3/4, la C es 2/6 y la D es 1/6.
“Léelo conmigo. En el primer tiempo anotó 1/2 de sus penales. En el segundo anotó 1/4. Queremos el partido completo, así que hay que sumar los dos. Ahora fíjate en los números de abajo. Uno es 2 y el otro es 4. Son distintos, y ahí está toda la dificultad: no se pueden sumar pedazos de tamaños diferentes. Entonces los igualamos. Cuenta de dos en dos: 2, 4. Ahora de cuatro en cuatro: 4, 8. El 4 aparece en las dos listas, así que ese es el tamaño que vamos a usar. Convierte 1/2 a cuartos. Si partes cada mitad en dos, te quedan 2 pedazos de 4, así que 1/2 es la misma cantidad que 2/4. Es lo mismo que multiplicar arriba y abajo por 2: 1 por 2 es 2, y 2 por 2 es 4. Ya los pedazos son iguales. Suma los de arriba: 2 más 1 es 3. Abajo se queda el 4, porque los pedazos no cambiaron de tamaño. Anotó 3/4 de sus penales.”
Señale en la pantalla los dos números de abajo, el 2 y el 4, y pregunte en qué número caben los dos. Luego cállese y espere. No lo conteste usted. Cuando le diga 4, pídale que diga en voz alta en qué se convierte 1/2 antes de tocar cualquier opción. Si quiere tocar los cuatro balones para ir contando los cuartos, déjelo; para eso están los balones.
El error que hay que vigilar es sumar de frente: 1 más 1 arriba y 2 más 4 abajo, que da 2/6. Por eso la opción C es justamente 2/6. Pregúntele: ¿por qué no podemos sumar el 2 y el 4 de abajo? Si le contesta que los pedazos son de tamaños distintos hasta convertir las mitades en cuartos, ya lo entendió. Si contesta 2/4, hizo bien la conversión y se le olvidó sumar el segundo tiempo.
3/4. Primero convertimos 1/2 en 2/4 para que las dos fracciones tengan el mismo denominador, 4. Luego sumamos: 2/4 + 1/4 = 3/4. Abajo sigue el 4 porque convertir las mitades en cuartos cambió cuántos pedazos hay, no de qué tamaño es cada pedazo.
VÒNG 1Tìm mẫu số chung
TRÊN MÀN HÌNH — Bảng hiển thị bằng tiếng Anh, đúng như con của anh/chị nhìn thấy: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Nghĩa là: một nữ cầu thủ sút thành công 1/2 số quả phạt đền trong hiệp một và 1/4 trong hiệp hai. Phía trên câu đó có bốn quả bóng chạm được, và phía dưới có bốn lựa chọn: A là 2/4, B là 3/4, C là 2/6 và D là 1/6.
“Con đọc lại cùng nhé. Hiệp một bạn ấy sút vào 1/2 số quả. Hiệp hai bạn ấy sút vào 1/4. Mình muốn biết cả trận, nên phải cộng hai phân số đó lại. Bây giờ nhìn xuống mẫu số. Một bên là 2, bên kia là 4. Hai mẫu số khác nhau, và đó chính là chỗ khó: không cộng được những phần có kích thước khác nhau. Vậy mình làm cho chúng bằng nhau. Đếm bội của 2: 2, 4. Đếm bội của 4: 4, 8. Cả hai dãy đều có số 4, nên 4 là kích thước mình sẽ dùng. Đổi 1/2 thành phần tư. Cắt mỗi nửa ra làm đôi thì được 2 phần trong 4 phần, nên 1/2 bằng đúng 2/4. Cũng chính là nhân cả tử số và mẫu số với 2: 1 nhân 2 bằng 2, và 2 nhân 2 bằng 4. Giờ các phần đã bằng nhau rồi. Cộng tử số: 2 cộng 1 bằng 3. Giữ nguyên mẫu số là 4, vì các phần không đổi kích thước. Bạn ấy đã sút vào 3/4 số quả phạt đền.”
Anh/chị chỉ vào hai mẫu số trên màn hình — số 2 và số 4 — rồi hỏi con xem số nào chia hết cho cả hai số đó. Sau đó im lặng và chờ. Đừng trả lời thay con. Khi con nói 4, anh/chị hãy yêu cầu con nói thành tiếng 1/2 sẽ đổi thành gì, trước khi con chạm vào bất kỳ lựa chọn nào. Nếu con muốn chạm vào bốn quả bóng để đếm các phần tư thì cứ để con làm; bốn quả bóng ở đó là để dùng vào việc ấy.
Lỗi cần để ý là cộng thẳng hàng ngang — 1 cộng 1 ở tử số và 2 cộng 4 ở mẫu số, ra 2/6. Đó chính là lý do 2/6 nằm sẵn ở lựa chọn C. Anh/chị hãy hỏi con: vì sao mình không cộng luôn số 2 với số 4 ở dưới mẫu? Nếu con nói được rằng các phần có kích thước khác nhau cho tới khi mình đổi các nửa thành phần tư, thì con đã nắm được rồi. Nếu con trả lời 2/4 thì con quy đồng đúng nhưng quên cộng hiệp hai.
3/4. Trước hết đổi 1/2 thành 2/4 để hai phân số có cùng mẫu số là 4. Sau đó cộng: 2/4 + 1/4 = 3/4. Mẫu số vẫn là 4 vì việc đổi các nửa thành phần tư làm thay đổi số phần có bao nhiêu, chứ không làm thay đổi mỗi phần to bằng nào.
RODADA 1Encontrar o denominador comum
NA TELA — O quadro aparece em inglês, exatamente como seu filho ou filha vê: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Ou seja: uma jogadora acertou 1/2 das cobranças de pênalti no primeiro tempo e 1/4 no segundo. Acima dessa frase há quatro bolas em que dá para tocar, sob a chamada “Tap the pieces to count!”, que quer dizer: toque nas partes para contar. Embaixo há quatro alternativas: a A é 2/4, a B é 3/4, a C é 2/6 e a D é 1/6.
“Leia comigo. No primeiro tempo ela acertou 1/2 das cobranças. No segundo tempo ela acertou 1/4. A gente quer o jogo inteiro, então soma as duas frações. Agora olhe os números de baixo, os denominadores. Um é 2. O outro é 4. Eles são diferentes, e é aí que está toda a dificuldade: não dá para somar pedaços de tamanhos diferentes. Então a gente deixa os pedaços do mesmo tamanho. Conte de 2 em 2: 2, 4. Conte de 4 em 4: 4, 8. As duas contagens caem no 4, então 4 é o tamanho que a gente vai usar. Transforme 1/2 em quartos. Corte cada metade em duas e você fica com 2 pedaços de 4, então 1/2 é a mesma quantidade que 2/4. Dá no mesmo que multiplicar o de cima e o de baixo por 2: 1 vezes 2 é 2, e 2 vezes 2 é 4. Agora os pedaços são iguais. Some os de cima: 2 mais 1 é 3. Mantenha o 4 embaixo, porque os pedaços não mudaram de tamanho. Ela acertou 3/4 das cobranças de pênalti.”
Aponte na tela para os dois números de baixo, o 2 e o 4, e pergunte à criança em qual número os dois cabem. Depois fique quieto e espere. Não responda no lugar dela. Quando ela disser 4, peça que ela diga em voz alta no que 1/2 se transforma, antes de tocar em qualquer alternativa. Se ela quiser tocar nas quatro bolas para ir contando os quartos, deixe; as bolas estão ali para isso.
O erro que você precisa vigiar é somar tudo de frente: 1 mais 1 em cima e 2 mais 4 embaixo, o que dá 2/6. É exatamente por isso que 2/6 está ali na alternativa C. Pergunte à criança: por que a gente não pode simplesmente somar o 2 com o 4 lá embaixo? Se ela conseguir dizer que os pedaços têm tamanhos diferentes até a gente transformar as metades em quartos, ela entendeu. Se ela responder 2/4, ela fez a conversão certa e esqueceu de somar o segundo tempo.
3/4. Primeiro transforme 1/2 em 2/4, para que as duas frações fiquem com o mesmo denominador, 4. Depois some: 2/4 + 1/4 = 3/4. Embaixo continua o 4 porque transformar as metades em quartos mudou quantos pedaços existem, não o tamanho de cada pedaço.
ROUND 1Find Common Denominators
ON SCREEN — Your board says: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Four soccer balls sit above that sentence, under the words “Tap the pieces to count!” Four answers sit below it. A is 2/4. B is 3/4. C is 2/6. D is 1/6.
Read it out loud. She made 1/2 of her kicks in the first half. She made 1/4 in the second half. You want the whole game, so you add those two. Now look at the bottom numbers. One is 2. The other is 4. They are different, and that is the whole problem — you cannot add pieces that are not the same size. So you make them the same size. Count by twos: 2, 4. Count by fours: 4, 8. Both lists land on 4, so 4 is the size you will use. Change 1/2 into fourths. Cut each half in two and you get 2 pieces out of 4, so 1/2 is the same amount as 2/4. That is the same as multiplying the top and the bottom by 2: 1 times 2 is 2, and 2 times 2 is 4. Now the pieces match. Add the tops: 2 plus 1 is 3. Keep the bottom at 4, because the pieces did not change size. She made 3/4 of her penalty kicks.
Look at the two bottom numbers on the screen — the 2 and the 4 — which number do both of them fit into? Then stop and wait before you answer. Do not let anyone fill it in for you. When you get 4, say out loud what 1/2 turns into before you touch a single answer choice. If you want to tap the four soccer balls to count the fourths, go ahead; that is what the balls are for.
The mistake to watch for is adding straight across — 1 plus 1 on top and 2 plus 4 on the bottom, which lands on 2/6. That is exactly why 2/6 is sitting there as choice C. Ask yourself: why can't you just add the 2 and the 4 on the bottom? If you can say the pieces are different sizes until you turn the halves into fourths, you have it. If you answer 2/4, you converted correctly and then forgot to add the second half.
3/4. Her half was two quarters. One quarter more makes three quarters. The 4 on the bottom stays, because the pieces never changed size. You only counted more of them.
ROUND 2Add with Different Denominators
ON SCREEN — Your board says: “One team scores 1/3 of their penalty kicks on Monday and 1/6 on Tuesday. What fraction did they score in both days?” Six soccer balls sit above it. The four answers are A 1/9, B 1/2, C 2/9 and D 2/6.
Monday was 1/3. Tuesday was 1/6. You want the two days together. Look at the bottom numbers. A 3 and a 6. They are different again. Back to the pizza. Cut it into three big slices. That is what the 3 on the bottom means — the whole pizza is cut into three. Monday is one of those big slices. Now picture the same pizza cut into six small slices. That is what the 6 means. Tuesday is one of those small slices. A big slice and a small slice still don't have one name. So cut each big slice down the middle. Now the pizza is in six pieces. They are all the same size. Monday's big slice turned into two of those pieces. So one third is two sixths. Now everything is sixths. Two sixths plus one sixth is three sixths. Now look at your pizza. Three pieces out of six. That is half the pizza. So 3/6 and 1/2 are the same amount. Your answer list says 1/2. They scored 1/2 of their kicks.
Draw a circle and cut it into six pieces. Colour two pieces for Monday. Colour one piece for Tuesday. Now see how much of the circle you coloured.
The easy mistake is adding the bottoms too: 1 plus 1 on top, 3 plus 6 on the bottom, which gives 2/9. Check it. Is 2/9 bigger or smaller than the 1/3 they scored on Monday? Smaller. Tuesday added kicks, so the total cannot shrink, and that answer is wrong. Choice D, 2/6, is Monday on its own after you cut it. Tuesday is still missing from it. And if your paper says 3/6, don't hunt for 3/6 in the list. Your circle already showed you that three pieces out of six is half.
1/2. Monday's 1/3 is two of the six pieces. So Monday is 2/6. Two sixths plus one sixth is three sixths. Three pieces out of six is half the pizza. So 3/6 has a shorter name. That name is 1/2.
ROUND 3Three Fractions, One Goal
ON SCREEN — Your board says: “Alex makes 1/2 of penalty kicks in practice, 1/4 in warm-up, and 1/8 in the game. What fraction did Alex make total?” Eight soccer balls sit above it. The four answers are A 3/8, B 7/8, C 4/8 and D 3/14.
This time there are three numbers, not two. Practice was 1/2. Warm-up was 1/4. The game was 1/8. Look at the bottom numbers. A 2, a 4 and an 8. All three are different. Take the pizza again. Cut it into eight pieces. That is what the 8 on the bottom means — the whole pizza is cut into eight. The game is one of those eight pieces. Now find the other two on the same pizza. Half the pizza is four of those pieces. So 1/2 is 4/8. A quarter is two of those pieces. So 1/4 is 2/8. Now everything is eighths. Count what Alex made. Four pieces, then two more, then one more. That is seven pieces. Look at the pizza. One piece is left over. Alex made 7/8 of the kicks.
Draw a circle and cut it into eight pieces. Colour four pieces for practice. Colour two more for warm-up. Colour one more for the game. Now look at the piece you did not colour.
The easy mistake is adding straight across all three: 1 plus 1 plus 1 on top, 2 plus 4 plus 8 on the bottom, which gives 3/14. Check it against your picture. You coloured nearly the whole pizza. Is 3/14 nearly a whole pizza? No, it is tiny. So that answer is wrong. Choice C, 4/8, is only the practice half. Choice A, 3/8, is what you get if you add the tops before the pieces are the same size.
7/8. Half the pizza is 4/8. A quarter of it is 2/8. The last piece is already 1/8. Four and two and one is seven pieces. One piece out of eight is still uncoloured. That is exactly what 7/8 looks like.
ROUND 4Penalty Kicks Across Two Teams
ON SCREEN — Your board says: “Team A makes 2/3 of their penalty kicks and Team B makes 1/6 of theirs. What fraction of all kicks did both teams make together?” Six soccer balls sit above it. The four answers are A 3/9, B 3/6, C 2/6 and D 5/6.
Team A made 2/3. Team B made 1/6. You want both teams together. Look at the bottom numbers. A 3 and a 6. Different again. Cut the pizza into three big slices. The 3 on the bottom means the whole pizza is cut into three. The 2 on top means Team A has two of those slices. That is the part to slow down on. The bottom number is the size. The top number is how many. Now cut every big slice down the middle. The pizza is in six pieces now. Team A had two big slices. Each one became two small pieces. Two and two is four. So 2/3 is 4/6. Notice that the top number moved as well, from 2 up to 4. It has to. You cut Team A's slices, not just the empty ones. Now everything is sixths. Four sixths plus one sixth is five sixths. Together they made 5/6.
Draw a circle and cut it into three pieces. Colour two of them for Team A. Now cut every piece down the middle, coloured ones too. Count your coloured pieces. You should have four.
The easy mistake is changing the bottom number and leaving the top alone. That turns 2/3 into 2/6. Add Team B and you land on 3/6, which is choice B. Check it. Team A on its own made 2/3, and that is more than half. Is 3/6 more than half? No, it is exactly half. Your total cannot be smaller than what one team made alone, so that answer is wrong. Choice C, 2/6, is smaller still. Choice A, 3/9, is both tops and both bottoms added.
5/6. Cutting the pizza into six turned each of Team A's slices into two pieces. So 2/3 became 4/6. Four sixths plus one sixth is five sixths. One piece out of six is missing. That fits, because Team A alone had already made more than half.
RONDA 1Buscar denominador común
EN PANTALLA — Tu tablero dice, en inglés: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” O sea: una jugadora anotó 1/2 de sus penales en el primer tiempo y 1/4 en el segundo. Arriba de esa frase hay cuatro balones que puedes tocar, bajo las palabras “Tap the pieces to count!” Abajo hay cuatro opciones. La A es 2/4. La B es 3/4. La C es 2/6. La D es 1/6.
Léelo en voz alta. En el primer tiempo anotó 1/2 de sus penales. En el segundo anotó 1/4. Quieres el partido completo, así que tienes que sumar los dos. Ahora fíjate en los números de abajo. Uno es 2 y el otro es 4. Son distintos, y ahí está toda la dificultad: no se pueden sumar pedazos de tamaños diferentes. Entonces los igualas. Cuenta de dos en dos: 2, 4. Ahora de cuatro en cuatro: 4, 8. El 4 aparece en las dos listas, así que ese es el tamaño que vas a usar. Convierte 1/2 a cuartos. Si partes cada mitad en dos, te quedan 2 pedazos de 4, así que 1/2 es la misma cantidad que 2/4. Es lo mismo que multiplicar arriba y abajo por 2: 1 por 2 es 2, y 2 por 2 es 4. Ya los pedazos son iguales. Suma los de arriba: 2 más 1 es 3. Abajo se queda el 4, porque los pedazos no cambiaron de tamaño. Anotó 3/4 de sus penales.
Mira en la pantalla los dos números de abajo, el 2 y el 4, y pregúntate en qué número caben los dos. Luego párate y piensa antes de contestar. No dejes que nadie lo conteste por ti. Cuando llegues al 4, di en voz alta en qué se convierte 1/2 antes de tocar cualquier opción. Si quieres tocar los cuatro balones para ir contando los cuartos, hazlo; para eso están los balones.
El error que hay que vigilar es sumar de frente: 1 más 1 arriba y 2 más 4 abajo, que da 2/6. Por eso la opción C es justamente 2/6. Pregúntate: ¿por qué no puedes sumar el 2 y el 4 de abajo? Si puedes decir que los pedazos son de tamaños distintos hasta convertir las mitades en cuartos, ya lo entendiste. Si contestas 2/4, hiciste bien la conversión y se te olvidó sumar el segundo tiempo.
3/4. Su mitad eran dos cuartos. Un cuarto más son tres cuartos. El 4 de abajo se queda, porque los pedazos nunca cambiaron de tamaño. Solo contaste más pedazos.
RONDA 2Sumar con denominadores distintos
EN PANTALLA — Tu tablero dice, en inglés: “One team scores 1/3 of their penalty kicks on Monday and 1/6 on Tuesday. What fraction did they score in both days?” O sea: un equipo anotó 1/3 de sus penales el lunes y 1/6 el martes, y hay que decir cuánto anotó en los dos días juntos. Arriba hay seis balones. Las cuatro opciones son A 1/9, B 1/2, C 2/9 y D 2/6.
El lunes fue 1/3. El martes fue 1/6. Quieres los dos días juntos. Mira los números de abajo. Un 3 y un 6. Otra vez son distintos. Vuelve a la pizza. Córtala en tres pedazos grandes. Eso es lo que significa el 3 de abajo — que toda la pizza está cortada en tres. El lunes es uno de esos pedazos grandes. Ahora imagina la misma pizza cortada en seis pedazos chicos. Eso es lo que significa el 6. El martes es uno de esos pedazos chicos. Un pedazo grande y uno chico todavía no tienen un solo nombre. Entonces corta cada pedazo grande por el medio. Ahora la pizza está en seis pedazos. Todos miden lo mismo. El pedazo grande del lunes se volvió dos de esos pedazos. Entonces un tercio son dos sextos. Ahora todo son sextos. Dos sextos más un sexto son tres sextos. Ahora mira tu pizza. Tres pedazos de seis. Eso es media pizza. Entonces 3/6 y 1/2 son la misma cantidad. Tu lista de opciones dice 1/2. Metieron 1/2 de sus tiros.
Dibuja un círculo y córtalo en seis pedazos. Colorea dos pedazos por el lunes. Colorea un pedazo por el martes. Ahora fíjate cuánto del círculo coloreaste.
El error fácil es sumar también los de abajo: 1 más 1 arriba, 3 más 6 abajo, lo que da 2/9. Revísalo. ¿2/9 es más o menos que el 1/3 que metieron el lunes? Menos. El martes sumó tiros, así que el total no puede achicarse, y esa respuesta está mal. La opción D, 2/6, es solo el lunes ya cortado. Le falta el martes. Y si en tu papel dice 3/6, no busques 3/6 en la lista. Tu círculo ya te enseñó que tres pedazos de seis son la mitad.
1/2. El 1/3 del lunes son dos de los seis pedazos. Entonces el lunes es 2/6. Dos sextos más un sexto son tres sextos. Tres pedazos de seis son media pizza. Entonces 3/6 tiene un nombre más corto. Ese nombre es 1/2.
RONDA 3Tres fracciones, un solo gol
EN PANTALLA — Tu tablero dice, en inglés: “Alex makes 1/2 of penalty kicks in practice, 1/4 in warm-up, and 1/8 in the game. What fraction did Alex make total?” O sea: Alex anotó 1/2 en la práctica, 1/4 en el calentamiento y 1/8 en el partido, y hay que decir cuánto anotó en total. Arriba hay ocho balones. Las cuatro opciones son A 3/8, B 7/8, C 4/8 y D 3/14.
Esta vez son tres números, no dos. La práctica fue 1/2. El calentamiento fue 1/4. El partido fue 1/8. Mira los números de abajo. Un 2, un 4 y un 8. Los tres son distintos. Agarra otra vez la pizza. Córtala en ocho pedazos. Eso es lo que significa el 8 de abajo — que toda la pizza está cortada en ocho. El partido es uno de esos ocho pedazos. Ahora busca los otros dos en la misma pizza. Media pizza son cuatro de esos pedazos. Entonces 1/2 son 4/8. Un cuarto son dos de esos pedazos. Entonces 1/4 son 2/8. Ahora todo son octavos. Cuenta lo que metió Alex. Cuatro pedazos, luego dos más, luego uno más. Son siete pedazos. Mira la pizza. Sobra un pedazo. Alex metió 7/8 de sus tiros.
Dibuja un círculo y córtalo en ocho pedazos. Colorea cuatro pedazos por la práctica. Colorea dos más por el calentamiento. Colorea uno más por el partido. Ahora mira el pedazo que no coloreaste.
El error fácil es sumar de frente los tres: 1 más 1 más 1 arriba, 2 más 4 más 8 abajo, lo que da 3/14. Revísalo con tu dibujo. Coloreaste casi toda la pizza. ¿3/14 es casi una pizza entera? No, es chiquito. Entonces esa respuesta está mal. La opción C, 4/8, es solo la práctica, esa mitad. La opción A, 3/8, es lo que sale si sumas los de arriba antes de que los pedazos midan lo mismo.
7/8. Media pizza son 4/8. Un cuarto de ella son 2/8. El último pedazo ya es 1/8. Cuatro y dos y uno son siete pedazos. Queda un pedazo de ocho sin colorear. Así se ve justamente 7/8.
RONDA 4Penales de dos equipos
EN PANTALLA — Tu tablero dice, en inglés: “Team A makes 2/3 of their penalty kicks and Team B makes 1/6 of theirs. What fraction of all kicks did both teams make together?” O sea: el equipo A anotó 2/3 de sus penales y el equipo B anotó 1/6 de los suyos, y hay que decir cuánto anotaron juntos. Arriba hay seis balones. Las cuatro opciones son A 3/9, B 3/6, C 2/6 y D 5/6.
El equipo A metió 2/3. El equipo B metió 1/6. Quieres los dos equipos juntos. Mira los números de abajo. Un 3 y un 6. Otra vez distintos. Corta la pizza en tres pedazos grandes. El 3 de abajo significa que toda la pizza está cortada en tres. El 2 de arriba significa que el equipo A tiene dos de esos pedazos. Ahí hay que ir despacio. El número de abajo es el tamaño. El número de arriba es cuántos. Ahora corta cada pedazo grande por el medio. Ahora la pizza está en seis pedazos. El equipo A tenía dos pedazos grandes. Cada uno se volvió dos pedazos chicos. Dos y dos son cuatro. Entonces 2/3 son 4/6. Fíjate que el número de arriba también se movió, del 2 al 4. Tiene que moverse. Cortaste los pedazos del equipo A, no solo los vacíos. Ahora todo son sextos. Cuatro sextos más un sexto son cinco sextos. Juntos metieron 5/6.
Dibuja un círculo y córtalo en tres pedazos. Colorea dos por el equipo A. Ahora corta cada pedazo por el medio, también los coloreados. Cuenta tus pedazos coloreados. Te deben quedar cuatro.
El error fácil es cambiar el número de abajo y dejar el de arriba igual. Así 2/3 se vuelve 2/6. Le sumas el equipo B y caes en 3/6, que es la opción B. Revísalo. El equipo A por sí solo metió 2/3, y eso es más de la mitad. ¿3/6 es más de la mitad? No, es justo la mitad. Tu total no puede ser menor que lo que metió un solo equipo, así que esa respuesta está mal. La opción C, 2/6, es todavía menor. La opción A, 3/9, sale de sumar los de arriba y los de abajo.
5/6. Cortar la pizza en seis volvió cada pedazo del equipo A en dos pedazos. Entonces 2/3 se volvió 4/6. Cuatro sextos más un sexto son cinco sextos. Falta un pedazo de seis. Eso cuadra, porque el equipo A por sí solo ya había metido más de la mitad.
VÒNG 1Tìm mẫu số chung
TRÊN MÀN HÌNH — Bảng của em ghi bằng tiếng Anh: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Nghĩa là: một nữ cầu thủ sút vào 1/2 số quả phạt đền trong hiệp một và 1/4 trong hiệp hai. Phía trên câu đó có bốn quả bóng, dưới dòng chữ “Tap the pieces to count!” Phía dưới có bốn đáp án. A là 2/4. B là 3/4. C là 2/6. D là 1/6.
Em đọc lại thành tiếng nhé. Hiệp một bạn ấy sút vào 1/2 số quả. Hiệp hai bạn ấy sút vào 1/4. Em muốn biết cả trận, nên phải cộng hai phân số đó lại. Bây giờ nhìn xuống mẫu số. Một bên là 2, bên kia là 4. Hai mẫu số khác nhau, và đó chính là chỗ khó: không cộng được những phần có kích thước khác nhau. Vậy em làm cho chúng bằng nhau. Đếm bội của 2: 2, 4. Đếm bội của 4: 4, 8. Cả hai dãy đều có số 4, nên 4 là kích thước em sẽ dùng. Đổi 1/2 thành phần tư. Cắt mỗi nửa ra làm đôi thì được 2 phần trong 4 phần, nên 1/2 bằng đúng 2/4. Cũng chính là nhân cả tử số và mẫu số với 2: 1 nhân 2 bằng 2, và 2 nhân 2 bằng 4. Giờ các phần đã bằng nhau rồi. Cộng tử số: 2 cộng 1 bằng 3. Giữ nguyên mẫu số là 4, vì các phần không đổi kích thước. Bạn ấy đã sút vào 3/4 số quả phạt đền.
Em nhìn vào hai mẫu số trên màn hình — số 2 và số 4 — rồi tự hỏi xem số nào chia hết cho cả hai số đó. Sau đó dừng lại và nghĩ trước khi trả lời. Đừng để ai trả lời thay em. Khi em ra số 4, hãy nói thành tiếng 1/2 sẽ đổi thành gì, trước khi em chạm vào bất kỳ lựa chọn nào. Nếu em muốn chạm vào bốn quả bóng để đếm các phần tư thì cứ làm; bốn quả bóng ở đó là để dùng vào việc ấy.
Lỗi cần để ý là cộng thẳng hàng ngang — 1 cộng 1 ở tử số và 2 cộng 4 ở mẫu số, ra 2/6. Đó chính là lý do 2/6 nằm sẵn ở lựa chọn C. Em hãy tự hỏi: vì sao mình không cộng luôn số 2 với số 4 ở dưới mẫu? Nếu em nói được rằng các phần có kích thước khác nhau cho tới khi em đổi các nửa thành phần tư, thì em đã nắm được rồi. Nếu em trả lời 2/4 thì em quy đồng đúng nhưng quên cộng hiệp hai.
3/4. Một nửa của bạn ấy là hai phần tư. Thêm một phần tư nữa thành ba phần tư. Số 4 ở mẫu giữ nguyên, vì các miếng không hề đổi kích thước. Em chỉ đếm được nhiều miếng hơn thôi.
VÒNG 2Cộng phân số khác mẫu
TRÊN MÀN HÌNH — Bảng của em ghi: “One team scores 1/3 of their penalty kicks on Monday and 1/6 on Tuesday. What fraction did they score in both days?” Nghĩa là: một đội sút vào 1/3 số quả phạt đền hôm thứ Hai và 1/6 hôm thứ Ba; cả hai ngày họ sút vào bao nhiêu phần? Phía trên có sáu quả bóng. Bốn đáp án là A 1/9, B 1/2, C 2/9 và D 2/6.
Thứ Hai là 1/3. Thứ Ba là 1/6. Em muốn gộp hai ngày lại. Hãy nhìn xuống mẫu số. Một số 3 và một số 6. Lại khác nhau nữa. Quay lại cái pizza. Cắt nó thành ba miếng to. Số 3 ở mẫu có nghĩa là như vậy — cả cái pizza được cắt thành ba. Thứ Hai là một trong những miếng to đó. Bây giờ hãy hình dung cũng cái pizza ấy được cắt thành sáu miếng nhỏ. Số 6 có nghĩa là như vậy. Thứ Ba là một trong những miếng nhỏ đó. Một miếng to và một miếng nhỏ thì vẫn chưa có một tên gọi chung. Vậy hãy cắt đôi từng miếng to. Bây giờ cái pizza có sáu miếng. Sáu miếng đều bằng nhau. Miếng to của thứ Hai đã thành hai miếng trong số đó. Vậy một phần ba là hai phần sáu. Bây giờ tất cả đều là phần sáu. Hai phần sáu cộng một phần sáu là ba phần sáu. Bây giờ hãy nhìn cái pizza của em. Ba miếng trong sáu miếng. Đó là một nửa cái pizza. Vậy 3/6 và 1/2 là cùng một lượng. Danh sách đáp án ghi 1/2. Họ đã sút vào 1/2 số quả.
Vẽ một hình tròn và chia thành sáu phần. Tô hai phần cho thứ Hai. Tô một phần cho thứ Ba. Bây giờ xem em đã tô bao nhiêu phần hình tròn.
Lỗi dễ mắc là cộng cả mẫu số: 1 cộng 1 ở trên, 3 cộng 6 ở dưới, ra 2/9. Hãy kiểm tra lại. 2/9 lớn hơn hay nhỏ hơn 1/3 mà họ đã sút được hôm thứ Hai? Nhỏ hơn. Thứ Ba có thêm quả vào, nên tổng không thể nhỏ đi, vậy đáp án đó sai. Lựa chọn D, 2/6, chỉ là riêng thứ Hai sau khi em cắt. Trong đó vẫn còn thiếu thứ Ba. Và nếu trên giấy của em là 3/6 thì đừng đi tìm 3/6 trong danh sách. Hình tròn của em đã cho thấy ba miếng trong sáu miếng là một nửa.
1/2. 1/3 của thứ Hai là hai miếng trong sáu miếng. Vậy thứ Hai là 2/6. Hai phần sáu cộng một phần sáu là ba phần sáu. Ba miếng trong sáu miếng là một nửa cái pizza. Vậy 3/6 có một tên gọi ngắn hơn. Tên đó là 1/2.
VÒNG 3Ba phân số, một mục tiêu
TRÊN MÀN HÌNH — Bảng của em ghi: “Alex makes 1/2 of penalty kicks in practice, 1/4 in warm-up, and 1/8 in the game. What fraction did Alex make total?” Nghĩa là: Alex sút vào 1/2 số quả phạt đền trong buổi tập, 1/4 khi khởi động và 1/8 trong trận đấu; tổng cộng Alex sút vào bao nhiêu phần? Phía trên có tám quả bóng. Bốn đáp án là A 3/8, B 7/8, C 4/8 và D 3/14.
Lần này có ba số chứ không phải hai. Buổi tập là 1/2. Khởi động là 1/4. Trận đấu là 1/8. Hãy nhìn xuống mẫu số. Một số 2, một số 4 và một số 8. Cả ba đều khác nhau. Lại lấy cái pizza ra. Cắt nó thành tám miếng. Số 8 ở mẫu có nghĩa là như vậy — cả cái pizza được cắt thành tám. Trận đấu là một trong tám miếng đó. Bây giờ tìm hai phần còn lại trên chính cái pizza ấy. Một nửa cái pizza là bốn miếng trong số đó. Vậy 1/2 là 4/8. Một phần tư là hai miếng trong số đó. Vậy 1/4 là 2/8. Bây giờ tất cả đều là phần tám. Hãy đếm những gì Alex sút được. Bốn miếng, rồi thêm hai miếng, rồi thêm một miếng nữa. Tất cả là bảy miếng. Hãy nhìn cái pizza. Còn thừa lại một miếng. Alex đã sút vào 7/8 số quả.
Vẽ một hình tròn và chia thành tám phần. Tô bốn phần cho buổi tập. Tô thêm hai phần cho khởi động. Tô thêm một phần cho trận đấu. Bây giờ nhìn vào phần em chưa tô.
Lỗi dễ mắc là cộng thẳng hàng ngang cả ba: 1 cộng 1 cộng 1 ở trên, 2 cộng 4 cộng 8 ở dưới, ra 3/14. Hãy đối chiếu với hình em vẽ. Em đã tô gần hết cái pizza. 3/14 có gần bằng cả cái pizza không? Không, nó rất bé. Vậy đáp án đó sai. Lựa chọn C, 4/8, chỉ là một nửa của buổi tập. Lựa chọn A, 3/8, là kết quả khi em cộng các tử số trước lúc các miếng bằng nhau.
7/8. Một nửa cái pizza là 4/8. Một phần tư của nó là 2/8. Miếng cuối cùng vốn đã là 1/8. Bốn với hai với một là bảy miếng. Vẫn còn một miếng trong tám miếng chưa được tô. 7/8 trông đúng như vậy.
VÒNG 4Phạt đền của hai đội
TRÊN MÀN HÌNH — Bảng của em ghi: “Team A makes 2/3 of their penalty kicks and Team B makes 1/6 of theirs. What fraction of all kicks did both teams make together?” Nghĩa là: đội A sút vào 2/3 số quả phạt đền của họ và đội B sút vào 1/6 của đội mình; cả hai đội cùng sút vào bao nhiêu phần trong tổng số quả? Phía trên có sáu quả bóng. Bốn đáp án là A 3/9, B 3/6, C 2/6 và D 5/6.
Đội A sút vào 2/3. Đội B sút vào 1/6. Em muốn gộp cả hai đội lại. Hãy nhìn xuống mẫu số. Một số 3 và một số 6. Lại khác nhau. Cắt cái pizza thành ba miếng to. Số 3 ở mẫu nghĩa là cả cái pizza được cắt thành ba. Số 2 ở trên nghĩa là đội A có hai miếng trong số đó. Đây là chỗ cần đi chậm lại. Số ở dưới là kích thước. Số ở trên là bao nhiêu miếng. Bây giờ cắt đôi từng miếng to. Bây giờ cái pizza có sáu miếng. Đội A có hai miếng to. Mỗi miếng thành hai miếng nhỏ. Hai với hai là bốn. Vậy 2/3 là 4/6. Hãy để ý số ở trên cũng đổi theo, từ 2 lên 4. Nhất định phải như vậy. Em đã cắt cả những miếng của đội A, chứ không chỉ cắt những miếng trống. Bây giờ tất cả đều là phần sáu. Bốn phần sáu cộng một phần sáu là năm phần sáu. Cả hai đội cùng sút vào 5/6.
Vẽ một hình tròn và chia thành ba phần. Tô hai phần cho đội A. Bây giờ cắt đôi từng phần, cả những phần đã tô. Đếm những phần đã tô của em. Em sẽ có bốn phần.
Lỗi dễ mắc là đổi số ở dưới mà để nguyên số ở trên. Như vậy 2/3 thành 2/6. Cộng thêm đội B thì ra 3/6, chính là lựa chọn B. Hãy kiểm tra lại. Riêng đội A đã sút vào 2/3, mà 2/3 thì nhiều hơn một nửa. 3/6 có nhiều hơn một nửa không? Không, nó đúng bằng một nửa. Tổng của em không thể nhỏ hơn phần một đội làm được, nên đáp án đó sai. Lựa chọn C, 2/6, còn nhỏ hơn nữa. Lựa chọn A, 3/9, là cộng cả hai số trên và cả hai số dưới.
5/6. Cắt cái pizza thành sáu làm mỗi miếng của đội A thành hai miếng. Vậy 2/3 thành 4/6. Bốn phần sáu cộng một phần sáu là năm phần sáu. Còn thiếu một miếng trong sáu miếng. Điều đó hợp lý, vì riêng đội A đã sút được nhiều hơn một nửa rồi.
RODADA 1Encontrar o denominador comum
NA TELA — O seu quadro diz, em inglês: “A soccer player makes 1/2 of her penalty kicks in the first half and 1/4 in the second half.” Quer dizer: uma jogadora acertou 1/2 das cobranças de pênalti no primeiro tempo e 1/4 no segundo tempo. Acima dessa frase há quatro bolas, embaixo das palavras “Tap the pieces to count!”, que querem dizer: toque nas partes para contar. Embaixo há quatro respostas. A é 2/4. B é 3/4. C é 2/6. D é 1/6.
Leia em voz alta. No primeiro tempo ela acertou 1/2 das cobranças. No segundo tempo ela acertou 1/4. Você quer o jogo inteiro, então some as duas frações. Agora olhe os números de baixo, os denominadores. Um é 2. O outro é 4. Eles são diferentes, e é aí que está toda a dificuldade: não dá para somar pedaços de tamanhos diferentes. Então você deixa os pedaços do mesmo tamanho. Conte de 2 em 2: 2, 4. Conte de 4 em 4: 4, 8. As duas contagens caem no 4, então 4 é o tamanho que você vai usar. Transforme 1/2 em quartos. Corte cada metade em duas e você fica com 2 pedaços de 4, então 1/2 é a mesma quantidade que 2/4. Dá no mesmo que multiplicar o de cima e o de baixo por 2: 1 vezes 2 é 2, e 2 vezes 2 é 4. Agora os pedaços são iguais. Some os de cima: 2 mais 1 é 3. Mantenha o 4 embaixo, porque os pedaços não mudaram de tamanho. Ela acertou 3/4 das cobranças de pênalti.
Olhe na tela para os dois números de baixo, o 2 e o 4, e pergunte a si mesmo em qual número os dois cabem. Depois pare e espere antes de responder. Não deixe ninguém responder no seu lugar. Quando você chegar no 4, diga em voz alta no que 1/2 se transforma, antes de tocar em qualquer alternativa. Se você quiser tocar nas quatro bolas para ir contando os quartos, pode tocar; as bolas estão ali para isso.
O erro que você precisa vigiar é somar tudo de frente: 1 mais 1 em cima e 2 mais 4 embaixo, o que dá 2/6. É exatamente por isso que 2/6 está ali na alternativa C. Pergunte a si mesmo: por que você não pode simplesmente somar o 2 com o 4 lá embaixo? Se você conseguir dizer que os pedaços têm tamanhos diferentes até você transformar as metades em quartos, você entendeu. Se você responder 2/4, você fez a conversão certa e esqueceu de somar o segundo tempo.
3/4. A metade dela era dois quartos. Mais um quarto dá três quartos. O 4 de baixo continua igual, porque os pedaços nunca mudaram de tamanho. Você só contou mais pedaços.
RODADA 2Somar com denominadores diferentes
NA TELA — O seu quadro diz: “One team scores 1/3 of their penalty kicks on Monday and 1/6 on Tuesday. What fraction did they score in both days?” Quer dizer: um time acertou 1/3 das cobranças de pênalti na segunda-feira e 1/6 na terça-feira; nos dois dias juntos, que fração eles acertaram? Acima há seis bolas. As quatro respostas são A 1/9, B 1/2, C 2/9 e D 2/6.
Na segunda foi 1/3. Na terça foi 1/6. Você quer os dois dias juntos. Olhe os números de baixo. Um 3 e um 6. De novo são diferentes. Volte para a pizza. Corte ela em três pedaços grandes. É isso que o 3 de baixo quer dizer: a pizza inteira foi cortada em três. A segunda-feira é um desses pedaços grandes. Agora imagine a mesma pizza cortada em seis pedaços pequenos. É isso que o 6 quer dizer. A terça-feira é um desses pedaços pequenos. Um pedaço grande e um pedaço pequeno ainda não têm um nome só. Então corte cada pedaço grande ao meio. Agora a pizza está em seis pedaços. Todos são do mesmo tamanho. O pedaço grande da segunda virou dois desses pedaços. Então um terço é dois sextos. Agora tudo está em sextos. Dois sextos mais um sexto é três sextos. Agora olhe a sua pizza. Três pedaços de seis. Isso é meia pizza. Então 3/6 e 1/2 são a mesma quantidade. A sua lista de respostas traz 1/2. Eles acertaram 1/2 das cobranças.
Desenhe um círculo e corte ele em seis pedaços. Pinte dois pedaços para a segunda-feira. Pinte um pedaço para a terça-feira. Agora veja quanto do círculo você pintou.
O erro fácil é somar também os números de baixo: 1 mais 1 em cima, 3 mais 6 embaixo, o que dá 2/9. Confira. 2/9 é maior ou menor que o 1/3 que eles acertaram na segunda? Menor. A terça acrescentou cobranças, então o total não pode diminuir, e essa resposta está errada. A alternativa D, 2/6, é só a segunda-feira depois que você cortou. A terça-feira ainda está faltando nela. E se no seu papel apareceu 3/6, não fique procurando 3/6 na lista. O seu círculo já mostrou que três pedaços de seis é a metade.
1/2. O 1/3 da segunda são dois dos seis pedaços. Então a segunda é 2/6. Dois sextos mais um sexto é três sextos. Três pedaços de seis é meia pizza. Então 3/6 tem um nome mais curto. Esse nome é 1/2.
RODADA 3Três frações, um só objetivo
NA TELA — O seu quadro diz: “Alex makes 1/2 of penalty kicks in practice, 1/4 in warm-up, and 1/8 in the game. What fraction did Alex make total?” Quer dizer: Alex acerta 1/2 das cobranças de pênalti no treino, 1/4 no aquecimento e 1/8 no jogo; no total, que fração Alex acertou? Acima há oito bolas. As quatro respostas são A 3/8, B 7/8, C 4/8 e D 3/14.
Desta vez são três números, não dois. O treino foi 1/2. O aquecimento foi 1/4. O jogo foi 1/8. Olhe os números de baixo. Um 2, um 4 e um 8. Os três são diferentes. Pegue a pizza de novo. Corte ela em oito pedaços. É isso que o 8 de baixo quer dizer: a pizza inteira foi cortada em oito. O jogo é um desses oito pedaços. Agora ache os outros dois na mesma pizza. Meia pizza são quatro desses pedaços. Então 1/2 é 4/8. Um quarto são dois desses pedaços. Então 1/4 é 2/8. Agora tudo está em oitavos. Conte o que Alex acertou. Quatro pedaços, depois mais dois, depois mais um. Isso dá sete pedaços. Olhe a pizza. Sobrou um pedaço. Alex acertou 7/8 das cobranças.
Desenhe um círculo e corte ele em oito pedaços. Pinte quatro pedaços para o treino. Pinte mais dois para o aquecimento. Pinte mais um para o jogo. Agora olhe o pedaço que você não pintou.
O erro fácil é somar os três de frente: 1 mais 1 mais 1 em cima, 2 mais 4 mais 8 embaixo, o que dá 3/14. Confira com o seu desenho. Você pintou quase a pizza inteira. 3/14 é quase uma pizza inteira? Não, é bem pouquinho. Então essa resposta está errada. A alternativa C, 4/8, é só a metade do treino. A alternativa A, 3/8, é o que dá quando você soma os de cima antes de os pedaços ficarem do mesmo tamanho.
7/8. Meia pizza é 4/8. Um quarto dela é 2/8. O último pedaço já é 1/8. Quatro com dois com um dá sete pedaços. Um pedaço de oito continua sem pintar. É exatamente essa a cara de 7/8.
RODADA 4Pênaltis de dois times
NA TELA — O seu quadro diz: “Team A makes 2/3 of their penalty kicks and Team B makes 1/6 of theirs. What fraction of all kicks did both teams make together?” Quer dizer: o time A acerta 2/3 das cobranças de pênalti dele e o time B acerta 1/6 das dele; juntos, que fração de todas as cobranças os dois times acertaram? Acima há seis bolas. As quatro respostas são A 3/9, B 3/6, C 2/6 e D 5/6.
O time A acertou 2/3. O time B acertou 1/6. Você quer os dois times juntos. Olhe os números de baixo. Um 3 e um 6. Diferentes de novo. Corte a pizza em três pedaços grandes. O 3 de baixo quer dizer que a pizza inteira foi cortada em três. O 2 de cima quer dizer que o time A tem dois desses pedaços. Essa é a parte para ir devagar. O número de baixo é o tamanho. O número de cima é quantos. Agora corte cada pedaço grande ao meio. Agora a pizza está em seis pedaços. O time A tinha dois pedaços grandes. Cada um virou dois pedaços pequenos. Dois com dois é quatro. Então 2/3 é 4/6. Repare que o número de cima também mudou, de 2 para 4. Ele tem que mudar mesmo. Você cortou os pedaços do time A, não só os pedaços vazios. Agora tudo está em sextos. Quatro sextos mais um sexto é cinco sextos. Juntos, os dois times acertaram 5/6.
Desenhe um círculo e corte ele em três pedaços. Pinte dois deles para o time A. Agora corte cada pedaço ao meio, inclusive os pintados. Conte os seus pedaços pintados. Você tem que ficar com quatro.
O erro fácil é mudar o número de baixo e deixar o de cima parado. Isso transforma 2/3 em 2/6. Somando o time B, você cai em 3/6, que é a alternativa B. Confira. Só o time A já acertou 2/3, e isso é mais que a metade. 3/6 é mais que a metade? Não, é exatamente a metade. O seu total não pode ser menor do que um time sozinho acertou, então essa resposta está errada. A alternativa C, 2/6, é menor ainda. A alternativa A, 3/9, é somar os dois de cima e os dois de baixo.
5/6. Cortar a pizza em seis transformou cada pedaço do time A em dois pedaços. Então 2/3 virou 4/6. Quatro sextos mais um sexto é cinco sextos. Falta um pedaço de seis. Isso faz sentido, porque só o time A já tinha acertado mais que a metade.
USMLE Step 1 · Immunology
The board stays in the language the exam is given in. The script comes to you in yours.El tablero se queda en el idioma en que se presenta el examen. El guion le llega a usted en el suyo.The board stays in the language the exam is given in. The script comes to you in the one you read best.El tablero se queda en el idioma en que se presenta el examen. El guion te llega a ti en el que leas mejor.Bảng vẫn giữ nguyên ngôn ngữ của kỳ thi. Kịch bản đến với thầy/cô bằng ngôn ngữ của thầy/cô.Bảng vẫn giữ nguyên ngôn ngữ của kỳ thi. Kịch bản đến với em bằng ngôn ngữ em đọc tốt nhất.
This is what an instructor or tutor reads through with them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Coach Mia đọc to phần Nói — đúng những lời được in bên dưới.Mia đọc to phần Nói cho em nghe — đúng những lời được in bên dưới.
ROUND 1Localize the Defect
ON SCREEN — Board shows: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” This round has no manipulative. Four answer choices sit below the vignette — A is C1 esterase inhibitor, B is C3, C is C5 through C9, D is Mannose-binding lectin.
“Read the vignette out loud, then stop before you look at the choices. Three findings decide this one, and all three are in the stem. First the presentation: a 19-year-old man, fever, headache, a stiff neck. That is meningitis. Second the history: this is his second culture-confirmed meningococcal meningitis in three years. One episode is bad luck. Two of the same organism is a defect. Third the labs: CH50 is markedly reduced, and C3 and C4 are both normal. Now say what each test means before you use it. CH50 asks whether serum can run the classical cascade all the way through, C1 to C9, so one missing piece anywhere drops it. That makes CH50 sensitive and useless for location on its own. C3 and C4 are the locators. Classical activation consumes C4, so a low C4 points upstream at C1, C2, or C4 itself. C3 sits where all three pathways meet, so a low C3 points at C3 or at the loop that amplifies it. Here both locators are normal. Nothing upstream is being consumed, and yet the cascade still cannot finish. The only place left is downstream of the convertases — the terminal components.”
Have them draw three columns — CH50, C3, C4 — and write out the four patterns before they read a single answer choice: all three low, C4 low alone, C3 low alone, and CH50 low with C3 and C4 normal. Then ask them to point at the row this vignette is on. Make them commit to the row before they commit to a letter.
The mistake to watch for is treating a low CH50 as the diagnosis and picking C3 because C3 is the complement number everyone remembers. CH50 names no component by itself. Ask them: what would C3 and C4 look like if the defect were upstream? If they answer that at least one of the two would be consumed and therefore low, the pattern is theirs. Mannose-binding lectin is the same trap from the other side — a lectin-pathway defect does not flatten a classical-pathway screen.
C5 through C9. A markedly low CH50 with normal C3 and C4 places the loss downstream of the convertases, in the terminal pathway. Deficiency of C5 through C9 prevents assembly of the membrane attack complex, and Neisseria species depend on that complex for clearance — which is why the presentation is recurrent meningococcal disease rather than infection with every encapsulated organism.
RONDA 1Ubicar el defecto
EN PANTALLA — El tablero muestra, en inglés, tal como sale en el examen: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” Es decir: un hombre de 19 años ingresa con fiebre, dolor de cabeza y rigidez de cuello; es su segundo episodio de meningitis meningocócica confirmada por cultivo en tres años; el CH50 está muy bajo y el C3 y el C4 están normales. Esta ronda no lleva material para tocar. Debajo de la viñeta hay cuatro opciones: la A es C1 esterase inhibitor, la B es C3, la C es C5 through C9 y la D es Mannose-binding lectin.
“Lee la viñeta en voz alta y detente antes de mirar las opciones. Aquí todo se resuelve con tres datos, y los tres están en el enunciado. Primero el cuadro clínico: un hombre de 19 años con fiebre, dolor de cabeza y rigidez de cuello. Eso es meningitis. Segundo, el antecedente: es su segundo episodio de meningitis meningocócica confirmada por cultivo en tres años. Un episodio es mala suerte; dos del mismo germen es un defecto. Tercero, el laboratorio: el CH50 está muy bajo y el C3 y el C4 están normales. Ahora di qué mide cada prueba antes de usarla. El CH50 pregunta si el suero puede recorrer toda la vía clásica, del C1 al C9; si falta una sola pieza en cualquier punto, el CH50 baja. Por eso el CH50 es muy sensible y por sí solo no ubica nada. Los que ubican son el C3 y el C4. La activación clásica consume C4, así que un C4 bajo señala hacia arriba: C1, C2 o el propio C4. El C3 está donde se juntan las tres vías, así que un C3 bajo señala al C3 mismo o al circuito que lo amplifica. Aquí los dos localizadores están normales. No se está consumiendo nada antes y, aun así, la cascada no llega al final. El único lugar que queda es después de las convertasas: los componentes terminales.”
Pídale que dibuje tres columnas —CH50, C3 y C4— y que escriba los cuatro patrones posibles antes de leer una sola opción: los tres bajos, solo el C4 bajo, solo el C3 bajo, y el CH50 bajo con el C3 y el C4 normales. Después pídale que señale en cuál de esas cuatro filas cae esta viñeta. Que se comprometa con la fila antes de comprometerse con una letra.
El error que hay que vigilar es leer el CH50 bajo como si fuera el diagnóstico y elegir C3 porque el C3 es el número de complemento que todo el mundo recuerda. El CH50 por sí solo no nombra ningún componente. Pregúntele: ¿cómo estarían el C3 y el C4 si el defecto estuviera más arriba? Si le contesta que al menos uno de los dos estaría consumido y por lo tanto bajo, ya domina el patrón. La opción de la lectina de unión a manosa es la misma trampa por el otro lado: un defecto de la vía de las lectinas no tumba una prueba de la vía clásica.
C5 through C9, es decir del C5 al C9. Un CH50 muy bajo con C3 y C4 normales coloca la pérdida después de las convertasas, en la vía terminal. Sin los componentes C5 a C9 no se puede armar el complejo de ataque a la membrana, y las especies de Neisseria dependen de ese complejo para ser eliminadas. Por eso el cuadro es enfermedad meningocócica recurrente y no una infección por cualquier germen encapsulado.
VÒNG 1Định vị khiếm khuyết
TRÊN MÀN HÌNH — Bảng hiển thị bằng tiếng Anh, đúng như trong đề thi: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” Nghĩa là: một nam bệnh nhân 19 tuổi nhập viện vì sốt, đau đầu và cứng gáy; đây là đợt viêm màng não do não mô cầu thứ hai được xác định bằng cấy trong ba năm; CH50 giảm rõ rệt, còn C3 và C4 huyết thanh trong giới hạn bình thường. Vòng này không có học cụ để thao tác. Dưới ca lâm sàng có bốn lựa chọn: A là C1 esterase inhibitor, B là C3, C là C5 through C9 và D là Mannose-binding lectin.
“Em hãy đọc to ca lâm sàng, rồi dừng lại trước khi nhìn các lựa chọn. Bài này được quyết định bởi ba dữ kiện, và cả ba đều nằm trong phần đề. Thứ nhất là bệnh cảnh: nam, 19 tuổi, sốt, đau đầu, cứng gáy. Đó là viêm màng não. Thứ hai là tiền sử: đây là đợt viêm màng não do não mô cầu thứ hai được xác định bằng cấy trong ba năm. Một đợt là không may. Hai đợt cùng một tác nhân là một khiếm khuyết. Thứ ba là xét nghiệm: CH50 giảm rõ rệt, còn C3 và C4 đều bình thường. Bây giờ hãy nói rõ từng xét nghiệm đo cái gì trước khi dùng nó. CH50 hỏi xem huyết thanh có chạy được trọn vẹn con đường cổ điển hay không, từ C1 đến C9, nên chỉ cần thiếu một mắt xích ở bất cứ đâu là CH50 tụt xuống. Vì vậy CH50 rất nhạy nhưng tự nó thì vô dụng trong việc định vị. C3 và C4 mới là hai chỉ số định vị. Hoạt hóa theo con đường cổ điển tiêu thụ C4, nên C4 thấp chỉ về phía trên: C1, C2 hoặc chính C4. C3 nằm ở chỗ cả ba con đường gặp nhau, nên C3 thấp chỉ vào chính C3 hoặc vào vòng khuếch đại của nó. Ở đây cả hai chỉ số định vị đều bình thường. Không có gì ở phía trên bị tiêu thụ, vậy mà chuỗi phản ứng vẫn không đi đến cuối được. Chỗ duy nhất còn lại là các thành phần tận cùng nằm phía sau các convertase.”
Thầy/cô cho học viên kẻ ba cột — CH50, C3, C4 — và viết ra bốn kiểu hình trước khi đọc bất kỳ lựa chọn nào: cả ba đều thấp; chỉ C4 thấp; chỉ C3 thấp; và CH50 thấp trong khi C3 với C4 bình thường. Sau đó yêu cầu học viên chỉ vào hàng ứng với ca lâm sàng này. Bắt học viên chốt hàng trước khi chốt một chữ cái.
Lỗi cần để ý là coi CH50 thấp như một chẩn đoán rồi chọn C3, chỉ vì C3 là con số bổ thể ai cũng nhớ. Tự CH50 không gọi tên được thành phần nào. Thầy/cô hãy hỏi: nếu khiếm khuyết nằm ở phía trên thì C3 và C4 sẽ như thế nào? Nếu học viên trả lời rằng ít nhất một trong hai sẽ bị tiêu thụ và do đó thấp, thì kiểu hình đã thuộc về học viên. Lectin gắn mannose là đúng cái bẫy đó nhìn từ phía bên kia — khiếm khuyết con đường lectin không làm sập một xét nghiệm sàng lọc của con đường cổ điển.
C5 through C9, tức là từ C5 đến C9. CH50 giảm rõ rệt kèm C3 và C4 bình thường đặt tổn thất ở phía sau các convertase, trong con đường tận cùng. Thiếu hụt C5 đến C9 làm không lắp ráp được phức hợp tấn công màng, mà các loài Neisseria lại phụ thuộc vào phức hợp đó để bị tiêu diệt. Vì vậy bệnh cảnh là bệnh do não mô cầu tái phát chứ không phải nhiễm trùng bởi mọi vi khuẩn có vỏ.
ROUND 1Localize the Defect
ON SCREEN — Your board says: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” This round has no manipulative. Four answer choices sit below the vignette — A is C1 esterase inhibitor, B is C3, C is C5 through C9, D is Mannose-binding lectin.
Read the vignette out loud, then stop before you look at the choices. Three findings decide this one, and all three are in the stem. First the presentation: a 19-year-old man, fever, headache, a stiff neck. That is meningitis. Second the history: this is his second culture-confirmed meningococcal meningitis in three years. One episode is bad luck. Two of the same organism is a defect. Third the labs: CH50 is markedly reduced, and C3 and C4 are both normal. Now say what each test means before you use it. CH50 asks whether serum can run the classical cascade all the way through, C1 to C9, so one missing piece anywhere drops it. That makes CH50 sensitive and useless for location on its own. C3 and C4 are the locators. Classical activation consumes C4, so a low C4 points upstream at C1, C2, or C4 itself. C3 sits where all three pathways meet, so a low C3 points at C3 or at the loop that amplifies it. Here both locators are normal. Nothing upstream is being consumed, and yet the cascade still cannot finish. The only place left is downstream of the convertases — the terminal components.
Draw three columns — CH50, C3, C4 — and write out the four patterns before you read a single answer choice: all three low, C4 low alone, C3 low alone, and CH50 low with C3 and C4 normal. Then find the row this vignette is on and point at it. Commit to the row before you commit to a letter.
Don't stop at the low CH50 and pick C3 because it's the complement number everyone remembers. CH50 names no component by itself. Ask yourself: what would C3 and C4 look like if the defect were upstream? If you can answer that at least one of the two would be consumed and therefore low, the pattern is yours. Mannose-binding lectin is the same trap from the other side — a lectin-pathway defect does not flatten a classical-pathway screen.
C5 through C9. A markedly low CH50 with normal C3 and C4 places the loss downstream of the convertases, in the terminal pathway. Deficiency of C5 through C9 prevents assembly of the membrane attack complex, and Neisseria species depend on that complex for clearance — which is why the presentation is recurrent meningococcal disease rather than infection with every encapsulated organism.
RONDA 1Ubicar el defecto
EN PANTALLA — Tu tablero muestra, en inglés, tal como sale en el examen: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” Es decir: un hombre de 19 años ingresa con fiebre, dolor de cabeza y rigidez de cuello; es su segundo episodio de meningitis meningocócica confirmada por cultivo en tres años; el CH50 está muy bajo y el C3 y el C4 están normales. Esta ronda no lleva material para tocar. Debajo de la viñeta hay cuatro opciones: la A es C1 esterase inhibitor, la B es C3, la C es C5 through C9 y la D es Mannose-binding lectin.
Lee la viñeta en voz alta y detente antes de mirar las opciones. Aquí todo se resuelve con tres datos, y los tres están en el enunciado. Primero el cuadro clínico: un hombre de 19 años con fiebre, dolor de cabeza y rigidez de cuello. Eso es meningitis. Segundo, el antecedente: es su segundo episodio de meningitis meningocócica confirmada por cultivo en tres años. Un episodio es mala suerte; dos del mismo germen es un defecto. Tercero, el laboratorio: el CH50 está muy bajo y el C3 y el C4 están normales. Ahora di qué mide cada prueba antes de usarla. El CH50 pregunta si el suero puede recorrer toda la vía clásica, del C1 al C9; si falta una sola pieza en cualquier punto, el CH50 baja. Por eso el CH50 es muy sensible y por sí solo no ubica nada. Los que ubican son el C3 y el C4. La activación clásica consume C4, así que un C4 bajo señala hacia arriba: C1, C2 o el propio C4. El C3 está donde se juntan las tres vías, así que un C3 bajo señala al C3 mismo o al circuito que lo amplifica. Aquí los dos localizadores están normales. No se está consumiendo nada antes y, aun así, la cascada no llega al final. El único lugar que queda es después de las convertasas: los componentes terminales.
Dibuja tres columnas —CH50, C3 y C4— y escribe los cuatro patrones posibles antes de leer una sola opción: los tres bajos, solo el C4 bajo, solo el C3 bajo, y el CH50 bajo con el C3 y el C4 normales. Después señala en cuál de esas cuatro filas cae esta viñeta. Comprométete con la fila antes de comprometerte con una letra.
No te quedes en el CH50 bajo y elijas C3 porque el C3 es el número de complemento que todo el mundo recuerda. El CH50 por sí solo no nombra ningún componente. Pregúntate: ¿cómo estarían el C3 y el C4 si el defecto estuviera más arriba? Si puedes contestar que al menos uno de los dos estaría consumido y por lo tanto bajo, ya dominas el patrón. La opción de la lectina de unión a manosa es la misma trampa por el otro lado: un defecto de la vía de las lectinas no tumba una prueba de la vía clásica.
C5 through C9, es decir del C5 al C9. Un CH50 muy bajo con C3 y C4 normales coloca la pérdida después de las convertasas, en la vía terminal. Sin los componentes C5 a C9 no se puede armar el complejo de ataque a la membrana, y las especies de Neisseria dependen de ese complejo para ser eliminadas. Por eso el cuadro es enfermedad meningocócica recurrente y no una infección por cualquier germen encapsulado.
VÒNG 1Định vị khiếm khuyết
TRÊN MÀN HÌNH — Bảng của em hiển thị bằng tiếng Anh, đúng như trong đề thi: “A 19-year-old man is admitted with fever, headache, and nuchal rigidity. This is his second episode of culture-confirmed meningococcal meningitis in three years. CH50 is markedly reduced; serum C3 and C4 are within normal limits. Which component is most likely deficient?” Nghĩa là: một nam bệnh nhân 19 tuổi nhập viện vì sốt, đau đầu và cứng gáy; đây là đợt viêm màng não do não mô cầu thứ hai được xác định bằng cấy trong ba năm; CH50 giảm rõ rệt, còn C3 và C4 huyết thanh trong giới hạn bình thường. Vòng này không có học cụ để thao tác. Dưới ca lâm sàng có bốn lựa chọn: A là C1 esterase inhibitor, B là C3, C là C5 through C9 và D là Mannose-binding lectin.
Em hãy đọc to ca lâm sàng, rồi dừng lại trước khi nhìn các lựa chọn. Bài này được quyết định bởi ba dữ kiện, và cả ba đều nằm trong phần đề. Thứ nhất là bệnh cảnh: nam, 19 tuổi, sốt, đau đầu, cứng gáy. Đó là viêm màng não. Thứ hai là tiền sử: đây là đợt viêm màng não do não mô cầu thứ hai được xác định bằng cấy trong ba năm. Một đợt là không may. Hai đợt cùng một tác nhân là một khiếm khuyết. Thứ ba là xét nghiệm: CH50 giảm rõ rệt, còn C3 và C4 đều bình thường. Bây giờ hãy nói rõ từng xét nghiệm đo cái gì trước khi dùng nó. CH50 hỏi xem huyết thanh có chạy được trọn vẹn con đường cổ điển hay không, từ C1 đến C9, nên chỉ cần thiếu một mắt xích ở bất cứ đâu là CH50 tụt xuống. Vì vậy CH50 rất nhạy nhưng tự nó thì vô dụng trong việc định vị. C3 và C4 mới là hai chỉ số định vị. Hoạt hóa theo con đường cổ điển tiêu thụ C4, nên C4 thấp chỉ về phía trên: C1, C2 hoặc chính C4. C3 nằm ở chỗ cả ba con đường gặp nhau, nên C3 thấp chỉ vào chính C3 hoặc vào vòng khuếch đại của nó. Ở đây cả hai chỉ số định vị đều bình thường. Không có gì ở phía trên bị tiêu thụ, vậy mà chuỗi phản ứng vẫn không đi đến cuối được. Chỗ duy nhất còn lại là các thành phần tận cùng nằm phía sau các convertase.
Em hãy kẻ ba cột — CH50, C3, C4 — và viết ra bốn kiểu hình trước khi đọc bất kỳ lựa chọn nào: cả ba đều thấp; chỉ C4 thấp; chỉ C3 thấp; và CH50 thấp trong khi C3 với C4 bình thường. Sau đó em hãy tìm và chỉ vào hàng ứng với ca lâm sàng này. Hãy chốt hàng trước khi chốt một chữ cái.
Đừng dừng ở CH50 thấp rồi chọn C3, chỉ vì C3 là con số bổ thể ai cũng nhớ. Tự CH50 không gọi tên được thành phần nào. Em hãy tự hỏi: nếu khiếm khuyết nằm ở phía trên thì C3 và C4 sẽ như thế nào? Nếu em trả lời được rằng ít nhất một trong hai sẽ bị tiêu thụ và do đó thấp, thì kiểu hình đã thuộc về em. Lectin gắn mannose là đúng cái bẫy đó nhìn từ phía bên kia — khiếm khuyết con đường lectin không làm sập một xét nghiệm sàng lọc của con đường cổ điển.
C5 through C9, tức là từ C5 đến C9. CH50 giảm rõ rệt kèm C3 và C4 bình thường đặt tổn thất ở phía sau các convertase, trong con đường tận cùng. Thiếu hụt C5 đến C9 làm không lắp ráp được phức hợp tấn công màng, mà các loài Neisseria lại phụ thuộc vào phức hợp đó để bị tiêu diệt. Vì vậy bệnh cảnh là bệnh do não mô cầu tái phát chứ không phải nhiễm trùng bởi mọi vi khuẩn có vỏ.
SAT Math · Algebra
The board stays in the language the exam is given in. The script comes to you in yours.El tablero se queda en el idioma en que se presenta el examen. El guion le llega a usted en el suyo.The board stays in the language the exam is given in. The script comes to you in the one you read best.El tablero se queda en el idioma en que se presenta el examen. El guion te llega a ti en el que leas mejor.
This is what an instructor or tutor reads through with them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.
ROUND 1Identify the Rate of Change
ON SCREEN — Board shows: “A print shop charges a fixed setup fee plus a per-poster rate. The total cost in dollars for n posters is modeled by C(n) = 1.75n + 40. Which of the following is the best interpretation of the number 1.75 in this context?” The instruction above it reads “Select the best interpretation of the coefficient.” This round has no manipulative. Four answer choices sit below the item — A is The total cost, in dollars, of one poster including setup; B is The cost, in dollars, of each additional poster; C is The number of posters that can be printed for one dollar; D is The fixed setup fee, in dollars.
“Read the model out loud before you read a single choice. C of n equals 1.75n plus 40, and C is a total cost in dollars for n posters. Two numbers are doing two different jobs in there. Ask what happens to the total when n goes up by one. Run it: at 10 posters the cost is 17.50 plus 40, which is 57.50. At 11 posters it is 19.25 plus 40, which is 59.25. The difference is 1.75. Do it again from 40 posters to 41 and the difference is still 1.75. That is what a rate of change is — the amount the total moves for one more poster, and it is the same every time because the model is linear. Notice that the 40 never moved in either subtraction. It is paid once, so it cannot be a per-poster amount. Check the units too: the total is dollars and n is posters, so the coefficient is dollars per poster. Now say what 1.75 is in the shop's own words: one more poster costs a dollar seventy-five. Say it before you look down at the choices.”
Cover the four choices with a hand or a sheet of paper before they read them. Have them write two lines — C(10) and C(11) — worked out to an actual number, then subtract and say the difference out loud. Ask them to write the units next to it: dollars per poster, not dollars and not posters. Uncover the choices only after they have said the number and the units.
The mistake to watch for is reading 1.75 as “the price of a poster” and reaching for A, because one poster does cost money. A is the total for one poster, 1.75 plus 40, or 41.75 — a number this model does produce, but not the one the coefficient names. D swaps the two numbers outright; the fixed setup fee is the 40. C inverts the rate: one dollar buys about 0.57 of a poster, and 1 divided by 1.75 is where that choice comes from. Ask them: if the shop printed exactly one more poster, by how many dollars does the total change? If they answer a dollar seventy-five without touching the 40, the coefficient is theirs.
The cost, in dollars, of each additional poster. In C(n) = 1.75n + 40 the coefficient of n is the amount C changes when n increases by 1, so 1.75 is the cost of one more poster. The 40 is the fixed setup fee, paid once no matter how many posters are printed, and the total for a single poster is 41.75, not 1.75.
ROUND 2Identify the Initial Value
ON SCREEN — Board shows: “Using the same model C(n) = 1.75n + 40, which statement best describes the meaning of 40?” The instruction above it reads “Select the best interpretation of the constant term.” This round has no manipulative. Four answer choices sit below the item — A is The cost, in dollars, when no posters are printed; B is The maximum number of posters the shop will print; C is The cost, in dollars, per poster after the first; D is The number of dollars saved by ordering in bulk.
“Same model, other number. Last round we asked what changes when n goes up by one. This time ask the opposite question: what is the cost before anything happens at all? Put n equal to zero into the model and read it. 1.75 times 0 is 0, so C of 0 is 40. Nothing has been printed and the shop is already owed 40 dollars. That is the initial value — where the line starts, the y-intercept, the fee you pay for walking in the door. Now do a units check on all four choices before you weigh any of them, because two of them will fall on units alone. The 40 came out of a cost formula, so 40 is dollars. A choice that counts posters is answering a different question, and so is a choice about dollars saved, because nothing in this model gives a discount — the rate is 1.75 per poster from the first poster to the thousandth. Say it plainly: 40 dollars is what the job costs before a single poster exists.”
Have them write C(0) and evaluate it on paper before they read a choice. Then read the four choices aloud to them and have them say one word after each — “dollars” or “posters” — labelling what that choice is counting. Ask them which label the number 40 has to carry. Two choices are gone before they have reasoned about the shop at all.
The mistake to watch for is carrying round one's answer onto the wrong number and picking C, since “per poster after the first” is a real-sounding phrase for a per-poster rate — but that rate is 1.75, and it applies to every poster including the first. B reads 40 as a count and turns it into a print cap the model does not contain; nothing in C(n) = 1.75n + 40 limits n. D invents a bulk discount, which a linear model with one fixed rate cannot have. Ask them: what is n when nothing is printed, and what does the formula give you there? If they say n is zero and the formula gives 40 dollars, the constant term is theirs.
The cost, in dollars, when no posters are printed. Substituting n = 0 gives C(0) = 40, so 40 is the cost before any poster is printed — the fixed setup fee. It is a dollar amount, not a count of posters, which rules out the two options that describe quantities.
ROUND 3Solve the Model for n
ON SCREEN — Board shows: “A student organization has $250 to spend at this print shop, where C(n) = 1.75n + 40. What is the greatest number of posters the organization can order?” The instruction above it reads “Solve for the requested quantity.” This round has no manipulative. Four answer choices sit below the item — A is 120, B is 144, C is 165, D is 166.
“Two rounds of reading the model, and now we use it. The organization has 250 dollars for the whole order, and the setup fee comes out of that same 250 — once, before any poster is printed. Write the sentence as an equation: total cost equals 250, so 1.75n plus 40 equals 250. Take the fee off the top first. Subtract 40 from both sides and you get 1.75n equals 210. Now divide: 210 divided by 1.75 is 120. Before you pick it, check whether 120 is actually affordable. 1.75 times 120 is 210, plus the 40 fee is exactly 250 — it fits to the cent, with nothing left over. The question asks for the greatest number, so ask whether one more fits: 121 posters would cost 251.75, which is over budget by a dollar seventy-five. So the answer is 120, and it is exact rather than rounded.”
Have them write the equation 1.75n + 40 = 250 before they look at any of the four numbers, and say out loud what the 250 is — a budget for the whole order, fee included. Then have them check their own answer by putting it back into the model and reading the total off the page. Ask them to price 121 posters too, so the word “greatest” has been tested rather than assumed.
The mistake to watch for is dividing the budget by the rate without taking the setup fee out first: 250 divided by 1.75 is about 142.9, and 144 is the choice sitting in that neighbourhood waiting for it. Reversing the sign on the fee — adding it instead of subtracting it — gives 290 divided by 1.75, or about 165.7, and that one slip produces both remaining choices: 165 if it is floored, 166 if it is rounded up. Ask them: whose money does the 40 come out of? If they can say the fee is paid from the same 250 before a single poster is printed, the setup step is theirs. If they land on 120 but hesitate over whether it should be 119, have them price 120 exactly — it comes to 250 on the nose, so nothing needs to be handed back.
120. Set 1.75n + 40 = 250, so 1.75n = 210 and n = 120. Because 120 is exactly achievable — 210 in posters plus the 40 fee is 250 — the greatest whole number of posters is 120, and 121 at 251.75 is over budget. Dividing 250 by 1.75 without first removing the setup fee gives about 142.9, which is the source of the nearby incorrect values.
ROUND 4Compare Two Linear Models
ON SCREEN — Board shows: “A second shop charges no setup fee and $2.25 per poster, so its cost is D(n) = 2.25n. For what number of posters do the two shops charge the same amount?” The instruction above it reads “Select the value at which the two models agree.” This round has no manipulative. Four answer choices sit below the item — A is 18, B is 40, C is 80, D is 89.
“Two shops now, and two models. The first charges 1.75 a poster plus a 40 dollar fee. The second charges 2.25 a poster and no fee — a worse rate with a better start. “Charge the same amount” means the two costs are equal, so set the expressions equal to each other: 1.75n plus 40 equals 2.25n. Collect the n terms. Subtract 1.75n from both sides: the left is just 40, and the right is 0.50n. So 40 equals 0.50n, and n is 80. Now read what that arithmetic means, because the SAT will ask it the other way round one day. The first shop's 40 dollar fee is being paid off at 50 cents a poster — 50 cents is what you save on each poster by taking the lower rate. It takes 80 of those savings to cover 40 dollars. Check it: at 80 posters the first shop is 140 plus 40, which is 180, and the second is 2.25 times 80, which is also 180.”
Have them write the two models one under the other and circle what is different — a fee on one, a higher rate on the other. Before they choose, have them price both shops at an easy number, say 20 posters, and say out loud which shop is cheaper there. Then ask which shop has to win a very large order, and why. A learner who can name the winner on both sides of the crossing rarely picks the wrong crossing.
The mistake to watch for is dividing the fee by the wrong number: 40 divided by 2.25 is about 17.8, and that is where 18 comes from — it pays the fee off at the second shop's whole rate instead of at the gap between the two rates. 40 is the fee itself, sitting in the list to be copied straight out of the stem by anyone who stops reading at “break even.” A decimal slip in the subtraction — taking 2.25 minus 1.75 as 0.45 rather than 0.50 — gives 40 divided by 0.45, or about 88.9, which is the neighbour of 89. Ask them: how much does each poster save you at the first shop, and how many of those savings pay off a 40 dollar fee? If they answer fifty cents and eighty posters, the comparison is theirs.
80. Set 1.75n + 40 = 2.25n. Subtracting 1.75n gives 40 = 0.50n, so n = 80, and both shops charge 180 dollars there. Below 80 posters the second shop is cheaper because there is no setup fee; above 80 the first shop is cheaper because its per-poster rate is lower.
RONDA 1Identificar la tasa de variación
EN PANTALLA — El tablero muestra, en inglés, tal como sale en el examen: “A print shop charges a fixed setup fee plus a per-poster rate. The total cost in dollars for n posters is modeled by C(n) = 1.75n + 40. Which of the following is the best interpretation of the number 1.75 in this context?” Es decir: una imprenta cobra una cuota fija de preparación más una tarifa por cartel, el costo total en dólares de n carteles se modela con C(n) = 1.75n + 40, y se pregunta qué representa el 1.75. La instrucción de arriba dice “Select the best interpretation of the coefficient.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es The total cost, in dollars, of one poster including setup; la B es The cost, in dollars, of each additional poster; la C es The number of posters that can be printed for one dollar; y la D es The fixed setup fee, in dollars.
“Lee el modelo en voz alta antes de mirar una sola opción. C de n es igual a 1.75n más 40, donde C es el costo total en dólares de n carteles. Ahí hay dos números haciendo dos trabajos distintos. Pregúntate qué le pasa al total cuando n aumenta en uno. Compruébalo: con 10 carteles el costo es 17.50 más 40, o sea 57.50. Con 11 carteles es 19.25 más 40, o sea 59.25. La diferencia es 1.75. Hazlo otra vez, de 40 carteles a 41, y la diferencia sigue siendo 1.75. Eso es una tasa de variación: lo que se mueve el total por un cartel más, y es igual siempre porque el modelo es lineal. Fíjate en que el 40 no se movió en ninguna de las dos restas. Se paga una sola vez, así que no puede ser un monto por cartel. Revisa también las unidades: el total está en dólares y n son carteles, así que el coeficiente está en dólares por cartel. Ahora di qué es 1.75 en las palabras de la imprenta: cada cartel adicional cuesta un dólar con setenta y cinco.”
Tape las cuatro opciones con la mano o con una hoja antes de que las lea. Pídale que escriba dos renglones —C(10) y C(11)— resueltos hasta llegar a un número, que reste y que diga la diferencia en voz alta. Pídale que anote las unidades al lado: dólares por cartel, no dólares y no carteles. Destape las opciones solo cuando ya haya dicho el número y las unidades.
El error que hay que vigilar es leer el 1.75 como “el precio de un cartel” y lanzarse a la A, porque al final un cartel sí cuesta dinero. La A es el total de un cartel, 1.75 más 40, es decir 41.75: un número que el modelo sí da, pero no el que nombra el coeficiente. La D cambia los dos números de lugar; la cuota fija de preparación es el 40. La C invierte la tasa: con un dólar alcanza para unos 0.57 de cartel, y de ahí sale esa opción, de dividir 1 entre 1.75. Pregúntele: si la imprenta imprimiera exactamente un cartel más, ¿en cuántos dólares cambia el total? Si le contesta un dólar con setenta y cinco sin tocar el 40, ya domina el coeficiente.
The cost, in dollars, of each additional poster, es decir el costo en dólares de cada cartel adicional. En C(n) = 1.75n + 40, el coeficiente de n es cuánto cambia C cuando n aumenta en 1, así que 1.75 es lo que cuesta un cartel más. El 40 es la cuota fija de preparación, que se paga una sola vez por muchos carteles que se impriman, y el total de un solo cartel es 41.75, no 1.75.
RONDA 2Identificar el valor inicial
EN PANTALLA — El tablero muestra, en inglés: “Using the same model C(n) = 1.75n + 40, which statement best describes the meaning of 40?” Es decir: con el mismo modelo, ¿qué significa el 40? La instrucción de arriba dice “Select the best interpretation of the constant term.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es The cost, in dollars, when no posters are printed; la B es The maximum number of posters the shop will print; la C es The cost, in dollars, per poster after the first; y la D es The number of dollars saved by ordering in bulk.
“Mismo modelo, otro número. En la ronda anterior preguntamos qué cambia cuando n aumenta en uno. Ahora pregunta lo contrario: ¿cuánto cuesta antes de que pase nada? Sustituye n igual a cero en el modelo y lee el resultado. 1.75 por 0 es 0, así que C de 0 es 40. Todavía no se ha impreso nada y la imprenta ya cobra 40 dólares. Ese es el valor inicial: donde arranca la recta, la ordenada al origen, lo que cuesta nada más entrar por la puerta. Ahora revisa las unidades de las cuatro opciones antes de sopesar cualquiera, porque dos se caen solo por ahí. El 40 salió de una fórmula de costo, así que el 40 son dólares. Una opción que cuenta carteles está contestando otra pregunta, y lo mismo pasa con una opción de dólares ahorrados, porque este modelo no da ningún descuento: la tarifa es 1.75 por cartel desde el primero hasta el número mil. Dilo con todas las palabras: 40 dólares es lo que cuesta el trabajo antes de que exista un solo cartel.”
Pídale que escriba C(0) y que lo evalúe en el papel antes de leer una opción. Después lea usted las cuatro opciones en voz alta y pídale que diga una sola palabra después de cada una —“dólares” o “carteles”— para etiquetar qué está contando esa opción. Pregúntele cuál de las dos etiquetas le toca al 40. Dos opciones se van antes de haber razonado nada sobre la imprenta.
El error que hay que vigilar es arrastrar la respuesta de la ronda 1 al número equivocado y elegir la C, porque “por cartel después del primero” suena a tarifa por cartel; pero esa tarifa es 1.75 y se aplica a todos los carteles, incluido el primero. La B lee el 40 como una cantidad de carteles y lo convierte en un tope de impresión que el modelo no tiene: nada en C(n) = 1.75n + 40 limita a n. La D inventa un descuento por volumen, que un modelo lineal con una sola tarifa fija no puede dar. Pregúntele: ¿cuánto vale n cuando no se imprime nada, y qué da la fórmula ahí? Si le contesta que n es cero y que la fórmula da 40 dólares, ya domina el término constante.
The cost, in dollars, when no posters are printed, es decir el costo en dólares cuando no se imprime ningún cartel. Al sustituir n = 0 queda C(0) = 40, así que el 40 es el costo antes de imprimir el primer cartel: la cuota fija de preparación. Es una cantidad de dólares y no un número de carteles, y eso descarta las dos opciones que hablan de cantidades.
RONDA 3Despejar n en el modelo
EN PANTALLA — El tablero muestra, en inglés: “A student organization has $250 to spend at this print shop, where C(n) = 1.75n + 40. What is the greatest number of posters the organization can order?” Es decir: una organización estudiantil tiene 250 dólares para gastar en esa imprenta y se pregunta cuál es la mayor cantidad de carteles que puede pedir. La instrucción de arriba dice “Solve for the requested quantity.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es 120, la B es 144, la C es 165 y la D es 166.
“Dos rondas leyendo el modelo, y ahora lo usamos. La organización tiene 250 dólares para todo el pedido, y la cuota de preparación sale de esos mismos 250, una sola vez, antes de imprimir un cartel. Escribe la frase como ecuación: el costo total es igual a 250, así que 1.75n más 40 es igual a 250. Primero quita la cuota. Resta 40 en los dos lados y te queda 1.75n igual a 210. Ahora divide: 210 entre 1.75 es 120. Antes de marcarlo, comprueba que 120 de verdad alcanza. 1.75 por 120 es 210, más los 40 de la cuota son exactamente 250: alcanza al centavo, sin que sobre nada. La pregunta pide la mayor cantidad, así que averigua si cabe uno más: 121 carteles costarían 251.75, o sea un dólar con setenta y cinco de más. Entonces la respuesta es 120, y es exacta, no redondeada.”
Pídale que escriba la ecuación 1.75n + 40 = 250 antes de mirar cualquiera de los cuatro números, y que diga en voz alta qué son esos 250: el presupuesto de todo el pedido, cuota incluida. Después pídale que verifique su propia respuesta sustituyéndola en el modelo y leyendo el total en la hoja. Pídale también que calcule cuánto costarían 121 carteles, para que la palabra “mayor” quede comprobada y no supuesta.
El error que hay que vigilar es dividir el presupuesto entre la tarifa sin quitar antes la cuota de preparación: 250 entre 1.75 da como 142.9, y la opción 144 está puesta justo en ese vecindario esperando ese descuido. Equivocarse en el signo de la cuota —sumarla en vez de restarla— da 290 entre 1.75, o sea como 165.7, y ese solo resbalón produce las otras dos opciones: 165 si se trunca y 166 si se redondea hacia arriba. Pregúntele: ¿de qué bolsillo salen esos 40 dólares? Si le contesta que la cuota se paga de los mismos 250 antes de imprimir un cartel, ya domina ese primer paso. Y si llega a 120 pero duda si debería ser 119, pídale que cotice 120 exacto: da 250 justos, así que no hay nada que devolver.
120. Planteamos 1.75n + 40 = 250, de donde 1.75n = 210 y n = 120. Como 120 se alcanza exacto —210 de carteles más los 40 de la cuota son 250—, la mayor cantidad entera de carteles es 120, y 121 ya cuesta 251.75, por encima del presupuesto. Dividir 250 entre 1.75 sin quitar primero la cuota da como 142.9, y de ahí salen los valores incorrectos que están cerca.
RONDA 4Comparar dos modelos lineales
EN PANTALLA — El tablero muestra, en inglés: “A second shop charges no setup fee and $2.25 per poster, so its cost is D(n) = 2.25n. For what number of posters do the two shops charge the same amount?” Es decir: una segunda imprenta no cobra cuota de preparación y cobra 2.25 por cartel, con costo D(n) = 2.25n, y se pregunta con cuántos carteles las dos imprentas cobran lo mismo. La instrucción de arriba dice “Select the value at which the two models agree.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es 18, la B es 40, la C es 80 y la D es 89.
“Ahora hay dos imprentas y dos modelos. La primera cobra 1.75 por cartel más 40 dólares de cuota. La segunda cobra 2.25 por cartel y ninguna cuota: peor tarifa, mejor arranque. Que “cobren lo mismo” quiere decir que los dos costos son iguales, así que iguala las dos expresiones: 1.75n más 40 es igual a 2.25n. Junta los términos con n. Resta 1.75n en los dos lados: a la izquierda queda solo 40 y a la derecha 0.50n. Entonces 40 es igual a 0.50n, y n es 80. Ahora lee qué significa esa cuenta, porque algún día te la van a preguntar al revés. Los 40 dólares de cuota de la primera imprenta se van pagando a 50 centavos por cartel, y esos 50 centavos son justo lo que te ahorras en cada cartel por la tarifa más baja. Hacen falta 80 de esos ahorros para cubrir 40 dólares. Compruébalo: con 80 carteles la primera cobra 140 más 40, o sea 180, y la segunda cobra 2.25 por 80, que también son 180.”
Pídale que escriba los dos modelos uno debajo del otro y que encierre en un círculo lo que cambia: una cuota en uno, una tarifa más alta en el otro. Antes de que elija, pídale que cotice las dos imprentas con un número fácil, por ejemplo 20 carteles, y que diga en voz alta cuál sale más barata ahí. Después pregúntele cuál imprenta tiene que ganar en un pedido enorme, y por qué. Quien puede nombrar al ganador en los dos lados del cruce casi nunca se equivoca de cruce.
El error que hay que vigilar es dividir la cuota entre el número equivocado: 40 entre 2.25 da como 17.8, y de ahí sale el 18, porque paga la cuota con la tarifa completa de la segunda imprenta en lugar de con la diferencia entre las dos tarifas. El 40 es la cuota misma, puesta en la lista para que la copie del enunciado quien deje de leer en “recuperar la cuota”. Un resbalón con los decimales en la resta —tomar 2.25 menos 1.75 como 0.45 en vez de 0.50— da 40 entre 0.45, o sea como 88.9, que es el vecino del 89. Pregúntele: ¿cuánto se ahorra en cada cartel con la primera imprenta, y cuántos de esos ahorros pagan una cuota de 40 dólares? Si le contesta cincuenta centavos y ochenta carteles, ya domina la comparación.
80. Planteamos 1.75n + 40 = 2.25n. Al restar 1.75n queda 40 = 0.50n, así que n = 80, y ahí las dos imprentas cobran 180 dólares. Con menos de 80 carteles sale más barata la segunda, porque no tiene cuota de preparación; con más de 80 sale más barata la primera, porque su tarifa por cartel es más baja.
ROUND 1Identify the Rate of Change
ON SCREEN — Your board says: “A print shop charges a fixed setup fee plus a per-poster rate. The total cost in dollars for n posters is modeled by C(n) = 1.75n + 40. Which of the following is the best interpretation of the number 1.75 in this context?” The instruction above it reads “Select the best interpretation of the coefficient.” This round has no manipulative. Four answer choices sit below the item — A is The total cost, in dollars, of one poster including setup; B is The cost, in dollars, of each additional poster; C is The number of posters that can be printed for one dollar; D is The fixed setup fee, in dollars.
Read the model out loud before you read a single choice. C of n equals 1.75n plus 40, and C is a total cost in dollars for n posters. Two numbers are doing two different jobs in there. Ask what happens to the total when n goes up by one. Run it: at 10 posters the cost is 17.50 plus 40, which is 57.50. At 11 posters it is 19.25 plus 40, which is 59.25. The difference is 1.75. Do it again from 40 posters to 41 and the difference is still 1.75. That is what a rate of change is — the amount the total moves for one more poster, and it is the same every time because the model is linear. Notice that the 40 never moved in either subtraction. It is paid once, so it cannot be a per-poster amount. Check the units too: the total is dollars and n is posters, so the coefficient is dollars per poster. Now say what 1.75 is in the shop's own words: one more poster costs a dollar seventy-five. Say it before you look down at the choices.
Cover the four choices with your hand or a sheet of paper before you read them. Write two lines — C(10) and C(11) — worked out to an actual number, then subtract and say the difference out loud. Write the units next to it: dollars per poster, not dollars and not posters. Uncover the choices only after you have said the number and the units.
Don't read 1.75 as “the price of a poster” and reach for A because one poster does cost money. A is the total for one poster, 1.75 plus 40, or 41.75 — a number this model does produce, but not the one the coefficient names. D swaps the two numbers outright; the fixed setup fee is the 40. C inverts the rate: one dollar buys about 0.57 of a poster, and 1 divided by 1.75 is where that choice comes from. Ask yourself: if the shop printed exactly one more poster, by how many dollars does the total change? If you answer a dollar seventy-five without touching the 40, the coefficient is yours.
The cost, in dollars, of each additional poster. In C(n) = 1.75n + 40 the coefficient of n is the amount C changes when n increases by 1, so 1.75 is the cost of one more poster. The 40 is the fixed setup fee, paid once no matter how many posters are printed, and the total for a single poster is 41.75, not 1.75.
ROUND 2Identify the Initial Value
ON SCREEN — Your board says: “Using the same model C(n) = 1.75n + 40, which statement best describes the meaning of 40?” The instruction above it reads “Select the best interpretation of the constant term.” This round has no manipulative. Four answer choices sit below the item — A is The cost, in dollars, when no posters are printed; B is The maximum number of posters the shop will print; C is The cost, in dollars, per poster after the first; D is The number of dollars saved by ordering in bulk.
Same model, other number. Last round you asked what changes when n goes up by one. This time ask the opposite question: what is the cost before anything happens at all? Put n equal to zero into the model and read it. 1.75 times 0 is 0, so C of 0 is 40. Nothing has been printed and the shop is already owed 40 dollars. That is the initial value — where the line starts, the y-intercept, the fee you pay for walking in the door. Now do a units check on all four choices before you weigh any of them, because two of them will fall on units alone. The 40 came out of a cost formula, so 40 is dollars. A choice that counts posters is answering a different question, and so is a choice about dollars saved, because nothing in this model gives a discount — the rate is 1.75 per poster from the first poster to the thousandth. Say it plainly: 40 dollars is what the job costs before a single poster exists.
Write C(0) and evaluate it on paper before you read a choice. Then read the four choices aloud and say one word after each — “dollars” or “posters” — labelling what that choice is counting. Then ask which label the number 40 has to carry. Two choices are gone before you have reasoned about the shop at all.
Don't carry round one's answer onto the wrong number and pick C, since “per poster after the first” is a real-sounding phrase for a per-poster rate — but that rate is 1.75, and it applies to every poster including the first. B reads 40 as a count and turns it into a print cap the model does not contain; nothing in C(n) = 1.75n + 40 limits n. D invents a bulk discount, which a linear model with one fixed rate cannot have. Ask yourself: what is n when nothing is printed, and what does the formula give you there? If you say n is zero and the formula gives 40 dollars, the constant term is yours.
The cost, in dollars, when no posters are printed. Substituting n = 0 gives C(0) = 40, so 40 is the cost before any poster is printed — the fixed setup fee. It is a dollar amount, not a count of posters, which rules out the two options that describe quantities.
ROUND 3Solve the Model for n
ON SCREEN — Your board says: “A student organization has $250 to spend at this print shop, where C(n) = 1.75n + 40. What is the greatest number of posters the organization can order?” The instruction above it reads “Solve for the requested quantity.” This round has no manipulative. Four answer choices sit below the item — A is 120, B is 144, C is 165, D is 166.
Two rounds of reading the model, and now you use it. The organization has 250 dollars for the whole order, and the setup fee comes out of that same 250 — once, before any poster is printed. Write the sentence as an equation: total cost equals 250, so 1.75n plus 40 equals 250. Take the fee off the top first. Subtract 40 from both sides and you get 1.75n equals 210. Now divide: 210 divided by 1.75 is 120. Before you pick it, check whether 120 is actually affordable. 1.75 times 120 is 210, plus the 40 fee is exactly 250 — it fits to the cent, with nothing left over. The question asks for the greatest number, so ask whether one more fits: 121 posters would cost 251.75, which is over budget by a dollar seventy-five. So the answer is 120, and it is exact rather than rounded.
Write the equation 1.75n + 40 = 250 before you look at any of the four numbers, and say out loud what the 250 is — a budget for the whole order, fee included. Then check your own answer by putting it back into the model and reading the total off the page. Price 121 posters too, so the word “greatest” has been tested rather than assumed.
Don't divide the budget by the rate without taking the setup fee out first: 250 divided by 1.75 is about 142.9, and 144 is the choice sitting in that neighbourhood waiting for it. Reversing the sign on the fee — adding it instead of subtracting it — gives 290 divided by 1.75, or about 165.7, and that one slip produces both remaining choices: 165 if it is floored, 166 if it is rounded up. Ask yourself: whose money does the 40 come out of? If you can say the fee is paid from the same 250 before a single poster is printed, the setup step is yours. If you land on 120 but hesitate over whether it should be 119, price 120 exactly — it comes to 250 on the nose, so nothing needs to be handed back.
120. Set 1.75n + 40 = 250, so 1.75n = 210 and n = 120. Because 120 is exactly achievable — 210 in posters plus the 40 fee is 250 — the greatest whole number of posters is 120, and 121 at 251.75 is over budget. Dividing 250 by 1.75 without first removing the setup fee gives about 142.9, which is the source of the nearby incorrect values.
ROUND 4Compare Two Linear Models
ON SCREEN — Your board says: “A second shop charges no setup fee and $2.25 per poster, so its cost is D(n) = 2.25n. For what number of posters do the two shops charge the same amount?” The instruction above it reads “Select the value at which the two models agree.” This round has no manipulative. Four answer choices sit below the item — A is 18, B is 40, C is 80, D is 89.
Two shops now, and two models. The first charges 1.75 a poster plus a 40 dollar fee. The second charges 2.25 a poster and no fee — a worse rate with a better start. “Charge the same amount” means the two costs are equal, so set the expressions equal to each other: 1.75n plus 40 equals 2.25n. Collect the n terms. Subtract 1.75n from both sides: the left is just 40, and the right is 0.50n. So 40 equals 0.50n, and n is 80. Now read what that arithmetic means, because the SAT will ask it the other way round one day. The first shop's 40 dollar fee is being paid off at 50 cents a poster — 50 cents is what you save on each poster by taking the lower rate. It takes 80 of those savings to cover 40 dollars. Check it: at 80 posters the first shop is 140 plus 40, which is 180, and the second is 2.25 times 80, which is also 180.
Write the two models one under the other and circle what is different — a fee on one, a higher rate on the other. Before you choose, price both shops at an easy number, say 20 posters, and say out loud which shop is cheaper there. Then ask which shop has to win a very large order, and why. If you can name the winner on both sides of the crossing you will rarely pick the wrong crossing.
Don't divide the fee by the wrong number: 40 divided by 2.25 is about 17.8, and that is where 18 comes from — it pays the fee off at the second shop's whole rate instead of at the gap between the two rates. 40 is the fee itself, sitting in the list to be copied straight out of the stem by anyone who stops reading at “break even.” A decimal slip in the subtraction — taking 2.25 minus 1.75 as 0.45 rather than 0.50 — gives 40 divided by 0.45, or about 88.9, which is the neighbour of 89. Ask yourself: how much does each poster save you at the first shop, and how many of those savings pay off a 40 dollar fee? If you answer fifty cents and eighty posters, the comparison is yours.
80. Set 1.75n + 40 = 2.25n. Subtracting 1.75n gives 40 = 0.50n, so n = 80, and both shops charge 180 dollars there. Below 80 posters the second shop is cheaper because there is no setup fee; above 80 the first shop is cheaper because its per-poster rate is lower.
RONDA 1Identificar la tasa de variación
EN PANTALLA — Tu tablero muestra, en inglés, tal como sale en el examen: “A print shop charges a fixed setup fee plus a per-poster rate. The total cost in dollars for n posters is modeled by C(n) = 1.75n + 40. Which of the following is the best interpretation of the number 1.75 in this context?” Es decir: una imprenta cobra una cuota fija de preparación más una tarifa por cartel, el costo total en dólares de n carteles se modela con C(n) = 1.75n + 40, y se pregunta qué representa el 1.75. La instrucción de arriba dice “Select the best interpretation of the coefficient.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es The total cost, in dollars, of one poster including setup; la B es The cost, in dollars, of each additional poster; la C es The number of posters that can be printed for one dollar; y la D es The fixed setup fee, in dollars.
Lee el modelo en voz alta antes de mirar una sola opción. C de n es igual a 1.75n más 40, donde C es el costo total en dólares de n carteles. Ahí hay dos números haciendo dos trabajos distintos. Pregúntate qué le pasa al total cuando n aumenta en uno. Compruébalo: con 10 carteles el costo es 17.50 más 40, o sea 57.50. Con 11 carteles es 19.25 más 40, o sea 59.25. La diferencia es 1.75. Hazlo otra vez, de 40 carteles a 41, y la diferencia sigue siendo 1.75. Eso es una tasa de variación: lo que se mueve el total por un cartel más, y es igual siempre porque el modelo es lineal. Fíjate en que el 40 no se movió en ninguna de las dos restas. Se paga una sola vez, así que no puede ser un monto por cartel. Revisa también las unidades: el total está en dólares y n son carteles, así que el coeficiente está en dólares por cartel. Ahora di qué es 1.75 en las palabras de la imprenta: cada cartel adicional cuesta un dólar con setenta y cinco.
Tapa las cuatro opciones con la mano o con una hoja antes de leerlas. Escribe dos renglones —C(10) y C(11)— resueltos hasta llegar a un número, resta y di la diferencia en voz alta. Anota las unidades al lado: dólares por cartel, no dólares y no carteles. Destapa las opciones solo cuando ya hayas dicho el número y las unidades.
No leas el 1.75 como “el precio de un cartel” ni te lances a la A porque al final un cartel sí cuesta dinero. La A es el total de un cartel, 1.75 más 40, es decir 41.75: un número que el modelo sí da, pero no el que nombra el coeficiente. La D cambia los dos números de lugar; la cuota fija de preparación es el 40. La C invierte la tasa: con un dólar alcanza para unos 0.57 de cartel, y de ahí sale esa opción, de dividir 1 entre 1.75. Pregúntate: si la imprenta imprimiera exactamente un cartel más, ¿en cuántos dólares cambia el total? Si contestas un dólar con setenta y cinco sin tocar el 40, ya dominas el coeficiente.
The cost, in dollars, of each additional poster, es decir el costo en dólares de cada cartel adicional. En C(n) = 1.75n + 40, el coeficiente de n es cuánto cambia C cuando n aumenta en 1, así que 1.75 es lo que cuesta un cartel más. El 40 es la cuota fija de preparación, que se paga una sola vez por muchos carteles que se impriman, y el total de un solo cartel es 41.75, no 1.75.
RONDA 2Identificar el valor inicial
EN PANTALLA — Tu tablero muestra, en inglés: “Using the same model C(n) = 1.75n + 40, which statement best describes the meaning of 40?” Es decir: con el mismo modelo, ¿qué significa el 40? La instrucción de arriba dice “Select the best interpretation of the constant term.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es The cost, in dollars, when no posters are printed; la B es The maximum number of posters the shop will print; la C es The cost, in dollars, per poster after the first; y la D es The number of dollars saved by ordering in bulk.
Mismo modelo, otro número. En la ronda anterior preguntaste qué cambia cuando n aumenta en uno. Ahora pregunta lo contrario: ¿cuánto cuesta antes de que pase nada? Sustituye n igual a cero en el modelo y lee el resultado. 1.75 por 0 es 0, así que C de 0 es 40. Todavía no se ha impreso nada y la imprenta ya cobra 40 dólares. Ese es el valor inicial: donde arranca la recta, la ordenada al origen, lo que cuesta nada más entrar por la puerta. Ahora revisa las unidades de las cuatro opciones antes de sopesar cualquiera, porque dos se caen solo por ahí. El 40 salió de una fórmula de costo, así que el 40 son dólares. Una opción que cuenta carteles está contestando otra pregunta, y lo mismo pasa con una opción de dólares ahorrados, porque este modelo no da ningún descuento: la tarifa es 1.75 por cartel desde el primero hasta el número mil. Dilo con todas las palabras: 40 dólares es lo que cuesta el trabajo antes de que exista un solo cartel.
Escribe C(0) y evalúalo en el papel antes de leer una opción. Después lee las cuatro opciones en voz alta y di una sola palabra después de cada una —“dólares” o “carteles”— para etiquetar qué está contando esa opción. Luego pregúntate cuál de las dos etiquetas le toca al 40. Dos opciones se van antes de haber razonado nada sobre la imprenta.
No arrastres la respuesta de la ronda 1 al número equivocado ni elijas la C porque “por cartel después del primero” suena a tarifa por cartel; esa tarifa es 1.75 y se aplica a todos los carteles, incluido el primero. La B lee el 40 como una cantidad de carteles y lo convierte en un tope de impresión que el modelo no tiene: nada en C(n) = 1.75n + 40 limita a n. La D inventa un descuento por volumen, que un modelo lineal con una sola tarifa fija no puede dar. Pregúntate: ¿cuánto vale n cuando no se imprime nada, y qué da la fórmula ahí? Si contestas que n es cero y que la fórmula da 40 dólares, ya dominas el término constante.
The cost, in dollars, when no posters are printed, es decir el costo en dólares cuando no se imprime ningún cartel. Al sustituir n = 0 queda C(0) = 40, así que el 40 es el costo antes de imprimir el primer cartel: la cuota fija de preparación. Es una cantidad de dólares y no un número de carteles, y eso descarta las dos opciones que hablan de cantidades.
RONDA 3Despejar n en el modelo
EN PANTALLA — Tu tablero muestra, en inglés: “A student organization has $250 to spend at this print shop, where C(n) = 1.75n + 40. What is the greatest number of posters the organization can order?” Es decir: una organización estudiantil tiene 250 dólares para gastar en esa imprenta y se pregunta cuál es la mayor cantidad de carteles que puede pedir. La instrucción de arriba dice “Solve for the requested quantity.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es 120, la B es 144, la C es 165 y la D es 166.
Dos rondas leyendo el modelo, y ahora lo usas. La organización tiene 250 dólares para todo el pedido, y la cuota de preparación sale de esos mismos 250, una sola vez, antes de imprimir un cartel. Escribe la frase como ecuación: el costo total es igual a 250, así que 1.75n más 40 es igual a 250. Primero quita la cuota. Resta 40 en los dos lados y te queda 1.75n igual a 210. Ahora divide: 210 entre 1.75 es 120. Antes de marcarlo, comprueba que 120 de verdad alcanza. 1.75 por 120 es 210, más los 40 de la cuota son exactamente 250: alcanza al centavo, sin que sobre nada. La pregunta pide la mayor cantidad, así que averigua si cabe uno más: 121 carteles costarían 251.75, o sea un dólar con setenta y cinco de más. Entonces la respuesta es 120, y es exacta, no redondeada.
Escribe la ecuación 1.75n + 40 = 250 antes de mirar cualquiera de los cuatro números, y di en voz alta qué son esos 250: el presupuesto de todo el pedido, cuota incluida. Después verifica tu propia respuesta sustituyéndola en el modelo y leyendo el total en la hoja. Calcula también cuánto costarían 121 carteles, para que la palabra “mayor” quede comprobada y no supuesta.
No dividas el presupuesto entre la tarifa sin quitar antes la cuota de preparación: 250 entre 1.75 da como 142.9, y la opción 144 está puesta justo en ese vecindario esperando ese descuido. Equivocarte en el signo de la cuota —sumarla en vez de restarla— da 290 entre 1.75, o sea como 165.7, y ese solo resbalón produce las otras dos opciones: 165 si se trunca y 166 si se redondea hacia arriba. Pregúntate: ¿de qué bolsillo salen esos 40 dólares? Si contestas que la cuota se paga de los mismos 250 antes de imprimir un cartel, ya dominas ese primer paso. Y si llegas a 120 pero dudas si debería ser 119, cotiza 120 exacto: da 250 justos, así que no hay nada que devolver.
120. Planteamos 1.75n + 40 = 250, de donde 1.75n = 210 y n = 120. Como 120 se alcanza exacto —210 de carteles más los 40 de la cuota son 250—, la mayor cantidad entera de carteles es 120, y 121 ya cuesta 251.75, por encima del presupuesto. Dividir 250 entre 1.75 sin quitar primero la cuota da como 142.9, y de ahí salen los valores incorrectos que están cerca.
RONDA 4Comparar dos modelos lineales
EN PANTALLA — Tu tablero muestra, en inglés: “A second shop charges no setup fee and $2.25 per poster, so its cost is D(n) = 2.25n. For what number of posters do the two shops charge the same amount?” Es decir: una segunda imprenta no cobra cuota de preparación y cobra 2.25 por cartel, con costo D(n) = 2.25n, y se pregunta con cuántos carteles las dos imprentas cobran lo mismo. La instrucción de arriba dice “Select the value at which the two models agree.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es 18, la B es 40, la C es 80 y la D es 89.
Ahora hay dos imprentas y dos modelos. La primera cobra 1.75 por cartel más 40 dólares de cuota. La segunda cobra 2.25 por cartel y ninguna cuota: peor tarifa, mejor arranque. Que “cobren lo mismo” quiere decir que los dos costos son iguales, así que iguala las dos expresiones: 1.75n más 40 es igual a 2.25n. Junta los términos con n. Resta 1.75n en los dos lados: a la izquierda queda solo 40 y a la derecha 0.50n. Entonces 40 es igual a 0.50n, y n es 80. Ahora lee qué significa esa cuenta, porque algún día te la van a preguntar al revés. Los 40 dólares de cuota de la primera imprenta se van pagando a 50 centavos por cartel, y esos 50 centavos son justo lo que te ahorras en cada cartel por la tarifa más baja. Hacen falta 80 de esos ahorros para cubrir 40 dólares. Compruébalo: con 80 carteles la primera cobra 140 más 40, o sea 180, y la segunda cobra 2.25 por 80, que también son 180.
Escribe los dos modelos uno debajo del otro y encierra en un círculo lo que cambia: una cuota en uno, una tarifa más alta en el otro. Antes de elegir, cotiza las dos imprentas con un número fácil, por ejemplo 20 carteles, y di en voz alta cuál sale más barata ahí. Después pregúntate cuál imprenta tiene que ganar en un pedido enorme, y por qué. Si puedes nombrar al ganador en los dos lados del cruce casi nunca te equivocarás de cruce.
No dividas la cuota entre el número equivocado: 40 entre 2.25 da como 17.8, y de ahí sale el 18, porque paga la cuota con la tarifa completa de la segunda imprenta en lugar de con la diferencia entre las dos tarifas. El 40 es la cuota misma, puesta en la lista para que la copie del enunciado quien deje de leer en “recuperar la cuota”. Un resbalón con los decimales en la resta —tomar 2.25 menos 1.75 como 0.45 en vez de 0.50— da 40 entre 0.45, o sea como 88.9, que es el vecino del 89. Pregúntate: ¿cuánto te ahorras en cada cartel con la primera imprenta, y cuántos de esos ahorros pagan una cuota de 40 dólares? Si contestas cincuenta centavos y ochenta carteles, ya dominas la comparación.
80. Planteamos 1.75n + 40 = 2.25n. Al restar 1.75n queda 40 = 0.50n, así que n = 80, y ahí las dos imprentas cobran 180 dólares. Con menos de 80 carteles sale más barata la segunda, porque no tiene cuota de preparación; con más de 80 sale más barata la primera, porque su tarifa por cartel es más baja.
MCAT · Biochemistry
The board stays in the language the exam is given in. The script comes to you in yours.El tablero se queda en el idioma en que se presenta el examen. El guion le llega a usted en el suyo.The board stays in the language the exam is given in. The script comes to you in the one you read best.El tablero se queda en el idioma en que se presenta el examen. El guion te llega a ti en el que leas mejor.
This is what an instructor or tutor reads through with them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.
ROUND 1Classify From Km and Vmax
ON SCREEN — Board shows: “An enzyme is assayed alone and again with inhibitor X at fixed concentration. Vmax is unchanged at 42 µmol/min in both runs, while the apparent Km rises from 0.8 mM to 3.2 mM. Which mode of inhibition does X exhibit?” The instruction above it reads “Use the reported constants to classify the inhibitor.” This round has no manipulative. Four answer choices sit below the item — A is Competitive, B is Noncompetitive, C is Uncompetitive, D is Irreversible covalent.
“Read the two numbers before you read the choices, and say what each one measures. Vmax is the ceiling — the velocity the enzyme reaches when substrate is saturating. Km is the substrate concentration that gets you to half of that ceiling, so it is a statement about apparent affinity, not about speed. Now the data. Vmax is 42 micromoles per minute in both runs, unchanged. Apparent Km went from 0.8 to 3.2 millimolar, four times higher. Turn that into a sentence about the enzyme: it still reaches full speed, but it needs four times as much substrate to get halfway there. Then ask what kind of inhibitor lets you reach the same ceiling. One you can outcompete. If the inhibitor sits in the active site of free enzyme, piling on substrate wins the site back, and at saturation the inhibitor is effectively gone — same Vmax. What it costs you is substrate: you need more of it to reach half-maximal velocity, so apparent Km rises. Unchanged Vmax with a raised apparent Km is the competitive signature.”
Have them draw a two-by-two table — Km up or unchanged against Vmax down or unchanged — and fill in the four modes from memory before they read a single answer choice. Then have them put a finger on the cell this data sits in and say the mode out loud. If they cannot fill the table, that is the gap to work on, and this item is the place to find it rather than the place to fix it.
The mistake to watch for is reading a fourfold rise in Km as “the inhibitor hurt the enzyme” and choosing B, noncompetitive, because it sounds like the general-purpose answer. Noncompetitive is the opposite pattern — Vmax falls and Km does not move — and it is round three of this board, not this one. C, uncompetitive, lowers both constants, so the unchanged 42 rules it out on its own. D takes enzyme out of the pool permanently, and no amount of substrate recovers it, so Vmax would have to fall as well. Ask them: what would raising substrate concentration do to each of these four inhibitors? If they can say that only the competitive one loses its grip as substrate rises, the pattern is theirs.
Competitive. A competitive inhibitor binds the free enzyme at the active site, so raising substrate concentration outcompetes it and the same maximal velocity is still reached — Vmax is unchanged at 42 µmol/min. More substrate is required to reach half-maximal velocity, so the apparent Km rises from 0.8 to 3.2 mM. Unchanged Vmax with increased apparent Km is the competitive signature.
ROUND 2Predict the Lineweaver-Burk Plot
ON SCREEN — Board shows: “The same two data sets are replotted as 1/v against 1/[S]. Compared with the uninhibited enzyme, how do the lines for inhibitor X differ?” The instruction above it reads “Select the plot behavior consistent with the data above.” This round has no manipulative. Four answer choices sit below the item — A is Same y-intercept, steeper slope; B is Same slope, higher y-intercept; C is Higher y-intercept and shallower slope; D is Identical slope and identical y-intercept.
“Do not picture the plot yet. Write the equation of the line first. Take the Michaelis-Menten equation, flip both sides, and you get one over v equals Km over Vmax times one over substrate, plus one over Vmax. That is y equals mx plus b, so name the parts out loud: the y-intercept is one over Vmax, and the slope is Km over Vmax. Both of those are quantities you already read in round one. Vmax did not change, so one over Vmax did not change, so the two lines cross the y-axis at the same point. Km went up fourfold and Vmax stayed put, so Km over Vmax went up fourfold: the inhibited line is steeper. Now say the picture before you look at it — two lines meeting on the y-axis, the inhibited one climbing faster. That is what people mean when they say competitive inhibition converges on the y-axis, and you just derived it instead of memorizing it.”
Have them write the double-reciprocal form from memory and label the intercept and the slope with what each one equals — one over Vmax and Km over Vmax — before they read a choice. Then have them sketch both lines on scratch paper, uninhibited first, and say which axis the two meet on. Ask them to state which of the two constants they would need in order to move the intercept.
The mistake to watch for is reaching for a remembered picture instead of the algebra and picking B because the inhibited line “should be higher.” B — same slope, higher y-intercept — is the noncompetitive picture, where Vmax falls and Km holds, which is round three's data and not this round's. C raises the intercept and lowers the slope, meaning one over Vmax rose while Km over Vmax fell; that needs both constants to drop, which is round four's uncompetitive case. D says nothing changed, which contradicts the fourfold Km already given in the stem. Ask them: which quantity is the y-intercept, and did that quantity change between the two runs? If they answer one over Vmax, and no, the plot follows without a picture.
Same y-intercept, steeper slope. On a Lineweaver-Burk plot the y-intercept is 1/Vmax and the slope is Km/Vmax. A competitive inhibitor leaves Vmax unchanged, so the y-intercept is unchanged, and raises apparent Km, so the slope increases. The lines therefore converge on the y-axis.
ROUND 3A Second Inhibitor
ON SCREEN — Board shows: “Inhibitor Y is tested on the same enzyme. Vmax falls from 42 to 21 µmol/min while the apparent Km remains 0.8 mM. Which statement best describes how Y binds?” The instruction above it reads “Classify the second inhibitor from its constants.” This round has no manipulative. Four answer choices sit below the item — A is It competes with substrate for the active site; B is It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site; C is It binds only the enzyme-substrate complex; D is It forms an irreversible covalent bond with the catalytic residue.
“New inhibitor, same enzyme, and this time the choices are mechanisms rather than labels — every one of them says where the inhibitor binds. Start from the data anyway. Vmax fell by half, 42 down to 21. Apparent Km did not move; it is still 0.8 millimolar. Say what each of those means on its own. A halved Vmax says half the catalytic capacity is gone and substrate cannot buy it back. An unchanged Km says the enzyme that is still working needs the same substrate concentration to reach half speed, so what is left behaves completely normally. Put them together: something is removing a fixed fraction of active enzyme at every substrate concentration, and leaving the rest untouched. That is what happens when the inhibitor does not care whether substrate is bound — it binds free enzyme and the enzyme-substrate complex with the same affinity, at a site that is not the active site. Substrate cannot displace it, because the two are not competing for the same place.”
Before they read the choices, have them write two sentences with the numbers in them: what happened to Vmax, and what happened to Km. Then ask them which of the two would have to change if substrate could displace this inhibitor. Make them predict the mechanism in their own words first — the choices in this round are long, and a learner who reads them cold will match on wording rather than on the data.
The mistake to watch for is picking A because the word inhibitor cues “active site.” Round one already showed what active-site competition looks like — Vmax unchanged, Km up — and both halves of that are wrong here. C, binding only the enzyme-substrate complex, is uncompetitive, which lowers Km as well; round four is that exact case, so the unchanged 0.8 rules it out. D deserves a full sentence, because an irreversible inhibitor also lowers Vmax while leaving the surviving enzyme's Km alone, so these numbers do not exclude it on the arithmetic. What excludes it is that nothing in the stem tests reversibility: a fixed inhibitor concentration with a measurable apparent Km names a mode of reversible inhibition, while covalent modification is a chemical claim this data has not earned. Ask them: what experiment would separate a noncompetitive inhibitor from an irreversible one? If they say dilute or dialyse the sample and see whether activity comes back, they are reading mechanism rather than matching words.
It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site. Reduced Vmax with unchanged Km is the classic noncompetitive pattern. Binding at an allosteric site with equal affinity for E and ES removes a constant fraction of active enzyme regardless of substrate concentration, so maximal velocity falls from 42 to 21 µmol/min while the substrate concentration giving half-maximal velocity is unchanged at 0.8 mM.
ROUND 4The Uncompetitive Case
ON SCREEN — Board shows: “Inhibitor Z binds only to the enzyme-substrate complex. Which combination of apparent constants is expected?” The instruction above it reads “Select the pattern that identifies uncompetitive inhibition.” This round has no manipulative. Four answer choices sit below the item — A is Km increases, Vmax unchanged; B is Km unchanged, Vmax unchanged; C is Km decreases, Vmax decreases; D is Km increases, Vmax increases.
“This is round one run backwards: you are given the mechanism and asked for the numbers. Z binds only the enzyme-substrate complex — only ES, never free enzyme. Take the two consequences one at a time. First, any ES that Z grabs is pulled out of the catalytic cycle and cannot go on to make product, so the ceiling comes down and Vmax falls. Most people get that far. Second, think about the equilibrium. Free enzyme plus substrate sits in balance with ES. If ES is being consumed by something else, that balance shifts toward making more ES — Le Chatelier, inside an enzyme. So the enzyme looks as though it grabs substrate more readily than before, and the concentration needed for half-maximal velocity goes down: apparent Km falls. Both constants drop, and that is the whole point of this item. A falling Vmax alone does not name the mode. Noncompetitive drops Vmax and leaves Km. Uncompetitive drops both.”
Have them write the two-step chain out for themselves before they look at the choices: ES is removed, so Vmax does what — and the E plus S equilibrium shifts which way, so apparent Km does what. Then have them say what happens to the ratio Km over Vmax, and whether the Lineweaver-Burk lines for this inhibitor would be parallel to the uninhibited line. That last question ties this round back to round two.
The mistake to watch for is stopping after the first consequence — Vmax falls, so reach for a falling Vmax with Km untouched, which is round three's noncompetitive pattern rather than this one. A is round one's competitive answer, Km up with Vmax unchanged, sitting here to catch pattern-matching on the word inhibitor. B says nothing changes, which cannot be true of something that removes ES from the cycle. D has Vmax rising, which no inhibitor does. The counterintuitive half is the falling Km, so probe exactly there. Ask them: if something is constantly removing ES, which way does the E plus S to ES equilibrium shift, and does the enzyme then look hungrier or less hungry for substrate? If they answer toward ES and hungrier, the drop in apparent Km follows.
Km decreases, Vmax decreases. Binding only to ES pulls that complex out of the catalytic cycle, lowering Vmax. Removing ES also shifts the E + S ⇌ ES equilibrium toward complex formation, so the enzyme behaves as though its substrate affinity increased and the apparent Km falls. Both constants decrease, which distinguishes uncompetitive from noncompetitive inhibition.
RONDA 1Clasificar a partir de Km y Vmax
EN PANTALLA — El tablero muestra, en inglés, tal como sale en el examen: “An enzyme is assayed alone and again with inhibitor X at fixed concentration. Vmax is unchanged at 42 µmol/min in both runs, while the apparent Km rises from 0.8 mM to 3.2 mM. Which mode of inhibition does X exhibit?” Es decir: se ensaya una enzima sola y otra vez con el inhibidor X a concentración fija; la Vmax se mantiene en 42 µmol/min en las dos corridas y la Km aparente sube de 0.8 mM a 3.2 mM; se pregunta qué tipo de inhibición presenta X. La instrucción de arriba dice “Use the reported constants to classify the inhibitor.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Competitive, la B es Noncompetitive, la C es Uncompetitive y la D es Irreversible covalent.
“Lee los dos números antes de mirar las opciones y di qué mide cada uno. La Vmax es el techo: la velocidad que alcanza la enzima cuando el sustrato es saturante. La Km es la concentración de sustrato con la que llegas a la mitad de ese techo, así que habla de afinidad aparente, no de rapidez. Ahora los datos. La Vmax es 42 micromoles por minuto en las dos corridas: no cambió. La Km aparente pasó de 0.8 a 3.2 milimolar, cuatro veces más. Conviértelo en una frase sobre la enzima: sigue llegando a toda velocidad, pero necesita cuatro veces más sustrato para llegar a la mitad del camino. Y pregúntate qué clase de inhibidor te deja alcanzar el mismo techo. Uno al que le puedes ganar por cantidad. Si el inhibidor está en el sitio activo de la enzima libre, echar más sustrato le recupera el sitio, y en saturación el inhibidor prácticamente desaparece: la misma Vmax. Lo que te cuesta es sustrato, porque necesitas más para llegar a la velocidad semimáxima, y por eso sube la Km aparente. Vmax sin cambios con Km aparente elevada es la firma de la inhibición competitiva.”
Pídale que dibuje una tabla de dos por dos —Km elevada o sin cambio contra Vmax disminuida o sin cambio— y que ubique de memoria los cuatro tipos de inhibición antes de leer una sola opción. Después pídale que ponga el dedo en la casilla donde caen estos datos y que diga el tipo en voz alta. Si no puede llenar la tabla, ahí está el hueco que hay que trabajar, y esta pregunta sirve para encontrarlo, no para taparlo.
El error que hay que vigilar es leer que la Km se cuadruplicó como “el inhibidor dañó la enzima” y elegir la B, Noncompetitive, porque suena a la respuesta que sirve para todo. La inhibición no competitiva es el patrón contrario —la Vmax baja y la Km no se mueve— y es la ronda 3 de este tablero, no esta. La C, Uncompetitive, es decir acompetitiva, baja las dos constantes, así que la Vmax intacta de 42 la descarta sola. La D saca enzima del juego de forma permanente y ningún exceso de sustrato la recupera, así que la Vmax también tendría que bajar. Pregúntele: ¿qué le haría a cada uno de estos cuatro inhibidores subir la concentración de sustrato? Si le contesta que solo el competitivo pierde su agarre cuando sube el sustrato, ya domina el patrón.
Competitive, es decir inhibición competitiva. Un inhibidor competitivo se une a la enzima libre en el sitio activo, así que al subir la concentración de sustrato se le gana por cantidad y se sigue alcanzando la misma velocidad máxima: la Vmax se queda en 42 µmol/min. Hace falta más sustrato para llegar a la velocidad semimáxima, así que la Km aparente sube de 0.8 a 3.2 mM. Vmax sin cambios con Km aparente aumentada es la firma de la inhibición competitiva.
RONDA 2Predecir la gráfica de Lineweaver-Burk
EN PANTALLA — El tablero muestra, en inglés: “The same two data sets are replotted as 1/v against 1/[S]. Compared with the uninhibited enzyme, how do the lines for inhibitor X differ?” Es decir: los mismos dos conjuntos de datos se grafican ahora como 1/v contra 1/[S], y se pregunta en qué se diferencian las rectas del inhibidor X respecto a la enzima sin inhibidor. La instrucción de arriba dice “Select the plot behavior consistent with the data above.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Same y-intercept, steeper slope; la B es Same slope, higher y-intercept; la C es Higher y-intercept and shallower slope; y la D es Identical slope and identical y-intercept.
“Todavía no te imagines la gráfica. Primero escribe la ecuación de la recta. Toma la ecuación de Michaelis-Menten, invierte los dos lados y te queda uno sobre v igual a Km sobre Vmax por uno sobre sustrato, más uno sobre Vmax. Eso es y igual a mx más b, así que nombra las partes en voz alta: la ordenada al origen es uno sobre Vmax y la pendiente es Km sobre Vmax. Las dos son cantidades que ya leíste en la ronda 1. La Vmax no cambió, entonces uno sobre Vmax no cambió, entonces las dos rectas cortan el eje y en el mismo punto. La Km se cuadruplicó y la Vmax se quedó igual, entonces Km sobre Vmax se cuadruplicó: la recta con inhibidor es más empinada. Ahora describe el dibujo antes de verlo: dos rectas que se juntan en el eje y, y la del inhibidor sube más rápido. Eso es lo que quiere decir que la inhibición competitiva converge en el eje y, y acabas de deducirlo en vez de memorizarlo.”
Pídale que escriba de memoria la forma de dobles recíprocos y que etiquete la ordenada al origen y la pendiente con lo que vale cada una —uno sobre Vmax y Km sobre Vmax— antes de leer una opción. Después pídale que dibuje las dos rectas en una hoja, primero la de la enzima sin inhibidor, y que diga en qué eje se juntan. Pregúntele cuál de las dos constantes tendría que cambiar para mover la ordenada al origen.
El error que hay que vigilar es echar mano de un dibujo memorizado en lugar del álgebra y elegir la B porque la recta con inhibidor “debería ir más arriba”. La B —misma pendiente, ordenada al origen más alta— es el dibujo de la inhibición no competitiva, donde baja la Vmax y la Km se queda; esos son los datos de la ronda 3, no los de esta. La C sube la ordenada y baja la pendiente, o sea que uno sobre Vmax subió mientras Km sobre Vmax bajó; para eso tienen que bajar las dos constantes, que es el caso acompetitivo de la ronda 4. La D dice que nada cambió, y eso contradice la Km cuadruplicada que ya venía en el enunciado. Pregúntele: ¿qué cantidad es la ordenada al origen, y esa cantidad cambió entre las dos corridas? Si le contesta uno sobre Vmax, y que no cambió, la gráfica sale sin necesidad de imaginarla.
Same y-intercept, steeper slope, es decir la misma ordenada al origen con una pendiente más empinada. En una gráfica de Lineweaver-Burk la ordenada al origen es 1/Vmax y la pendiente es Km/Vmax. Un inhibidor competitivo deja la Vmax igual, así que la ordenada al origen no cambia, y sube la Km aparente, así que la pendiente aumenta. Por eso las rectas convergen en el eje y.
RONDA 3Un segundo inhibidor
EN PANTALLA — El tablero muestra, en inglés: “Inhibitor Y is tested on the same enzyme. Vmax falls from 42 to 21 µmol/min while the apparent Km remains 0.8 mM. Which statement best describes how Y binds?” Es decir: se prueba el inhibidor Y en la misma enzima, la Vmax baja de 42 a 21 µmol/min y la Km aparente se mantiene en 0.8 mM; se pregunta qué enunciado describe mejor cómo se une Y. La instrucción de arriba dice “Classify the second inhibitor from its constants.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es It competes with substrate for the active site; la B es It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site; la C es It binds only the enzyme-substrate complex; y la D es It forms an irreversible covalent bond with the catalytic residue.
“Otro inhibidor, la misma enzima, y esta vez las opciones son mecanismos y no etiquetas: cada una dice dónde se une el inhibidor. Aun así, empieza por los datos. La Vmax cayó a la mitad, de 42 a 21. La Km aparente no se movió: sigue en 0.8 milimolar. Di qué significa cada cosa por separado. Que la Vmax se reduzca a la mitad significa que se perdió la mitad de la capacidad catalítica y que el sustrato no la puede recuperar. Que la Km no cambie significa que la enzima que sigue trabajando necesita la misma concentración de sustrato para llegar a media velocidad, o sea que lo que queda funciona con toda normalidad. Júntalo: algo está quitando una fracción fija de enzima activa a cualquier concentración de sustrato y está dejando intacto el resto. Eso pasa cuando al inhibidor le da igual si el sustrato está unido: se une a la enzima libre y al complejo enzima-sustrato con la misma afinidad, en un sitio que no es el sitio activo. El sustrato no lo puede desplazar, porque no se están peleando el mismo lugar.”
Antes de que lea las opciones, pídale que escriba dos frases con los números adentro: qué pasó con la Vmax y qué pasó con la Km. Después pregúntele cuál de las dos tendría que cambiar si el sustrato pudiera desplazar a este inhibidor. Hágalo predecir el mecanismo con sus propias palabras primero: las opciones de esta ronda son largas, y quien las lee en frío acaba emparejando por redacción y no por los datos.
El error que hay que vigilar es elegir la A porque la palabra inhibidor lleva de la mano al “sitio activo”. La ronda 1 ya mostró cómo se ve la competencia por el sitio activo —Vmax igual, Km arriba— y aquí las dos mitades están al revés. La C, unirse solo al complejo enzima-sustrato, es la inhibición acompetitiva, que además baja la Km; la ronda 4 es exactamente ese caso, así que la Km intacta de 0.8 la descarta. La D merece una frase completa, porque un inhibidor irreversible también baja la Vmax y deja igual la Km de la enzima que sobrevive, así que con la aritmética no queda descartada. Lo que la descarta es que nada en el enunciado pone a prueba la reversibilidad: una concentración fija de inhibidor con una Km aparente medible nombra un tipo de inhibición reversible, mientras que la modificación covalente es una afirmación química que estos datos no alcanzan a sostener. Pregúntele: ¿qué experimento separaría a un inhibidor no competitivo de uno irreversible? Si le contesta que diluir o dializar la muestra y ver si la actividad regresa, está leyendo mecanismo y no emparejando palabras.
It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site, es decir que se une con la misma afinidad a la enzima libre y al complejo enzima-sustrato, en un sitio distinto del activo. Vmax reducida con Km sin cambios es el patrón clásico de la inhibición no competitiva. Unirse a un sitio alostérico con la misma afinidad por E y por ES quita una fracción constante de enzima activa a cualquier concentración de sustrato, así que la velocidad máxima cae de 42 a 21 µmol/min mientras la concentración de sustrato que da la velocidad semimáxima se queda en 0.8 mM.
RONDA 4El caso acompetitivo
EN PANTALLA — El tablero muestra, en inglés: “Inhibitor Z binds only to the enzyme-substrate complex. Which combination of apparent constants is expected?” Es decir: el inhibidor Z se une únicamente al complejo enzima-sustrato, y se pregunta qué combinación de constantes aparentes se espera. La instrucción de arriba dice “Select the pattern that identifies uncompetitive inhibition.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Km increases, Vmax unchanged; la B es Km unchanged, Vmax unchanged; la C es Km decreases, Vmax decreases; y la D es Km increases, Vmax increases.
“Esta es la ronda 1 al revés: te dan el mecanismo y te piden los números. Z se une únicamente al complejo enzima-sustrato, solo a ES, nunca a la enzima libre. Toma las dos consecuencias una por una. Primero, cada ES que Z agarra sale del ciclo catalítico y ya no va a formar producto, así que el techo baja y la Vmax cae. Hasta ahí llega casi todo el mundo. Segundo, piensa en el equilibrio. La enzima libre más el sustrato están en equilibrio con ES. Si algo está consumiendo ES, ese equilibrio se desplaza hacia formar más ES: Le Châtelier, dentro de una enzima. Entonces la enzima parece agarrar el sustrato con más facilidad que antes, y la concentración que hace falta para la velocidad semimáxima baja: la Km aparente baja. Las dos constantes bajan, y de eso se trata esta pregunta. Que la Vmax caiga no basta para nombrar el tipo. La no competitiva baja la Vmax y deja la Km. La acompetitiva baja las dos.”
Pídale que escriba la cadena de dos pasos por su cuenta antes de mirar las opciones: se quita ES, entonces la Vmax hace qué; y el equilibrio de E más S se desplaza hacia dónde, entonces la Km aparente hace qué. Después pídale que diga qué pasa con el cociente Km sobre Vmax y si las rectas de Lineweaver-Burk de este inhibidor quedarían paralelas a la de la enzima sin inhibidor. Esa última pregunta amarra esta ronda con la ronda 2.
El error que hay que vigilar es detenerse en la primera consecuencia: baja la Vmax, entonces buscar una Vmax que baja con la Km intacta, que es el patrón no competitivo de la ronda 3 y no el de esta. La A es la respuesta competitiva de la ronda 1, Km arriba con Vmax igual, puesta aquí para atrapar a quien empareja con la palabra inhibidor. La B dice que nada cambia, y eso no puede ser cierto de algo que saca ES del ciclo. La D pone la Vmax subiendo, y eso no lo hace ningún inhibidor. La mitad contraintuitiva es la Km que baja, así que pregunte justo ahí. Pregúntele: si algo está quitando ES todo el tiempo, ¿hacia dónde se desplaza el equilibrio de E más S hacia ES, y entonces la enzima se ve con más hambre o con menos hambre de sustrato? Si le contesta hacia ES y con más hambre, la bajada de la Km aparente sale sola.
Km decreases, Vmax decreases, es decir que bajan las dos: la Km y la Vmax. Unirse solo a ES saca ese complejo del ciclo catalítico y por eso baja la velocidad máxima. Quitar ES también desplaza el equilibrio E + S ⇌ ES hacia la formación del complejo, así que la enzima se comporta como si hubiera aumentado su afinidad por el sustrato y la Km aparente baja. Que bajen las dos constantes es lo que distingue a la inhibición acompetitiva de la no competitiva.
ROUND 1Classify From Km and Vmax
ON SCREEN — Your board says: “An enzyme is assayed alone and again with inhibitor X at fixed concentration. Vmax is unchanged at 42 µmol/min in both runs, while the apparent Km rises from 0.8 mM to 3.2 mM. Which mode of inhibition does X exhibit?” The instruction above it reads “Use the reported constants to classify the inhibitor.” This round has no manipulative. Four answer choices sit below the item — A is Competitive, B is Noncompetitive, C is Uncompetitive, D is Irreversible covalent.
Read the two numbers before you read the choices, and say what each one measures. Vmax is the ceiling — the velocity the enzyme reaches when substrate is saturating. Km is the substrate concentration that gets you to half of that ceiling, so it is a statement about apparent affinity, not about speed. Now the data. Vmax is 42 micromoles per minute in both runs, unchanged. Apparent Km went from 0.8 to 3.2 millimolar, four times higher. Turn that into a sentence about the enzyme: it still reaches full speed, but it needs four times as much substrate to get halfway there. Then ask what kind of inhibitor lets you reach the same ceiling. One you can outcompete. If the inhibitor sits in the active site of free enzyme, piling on substrate wins the site back, and at saturation the inhibitor is effectively gone — same Vmax. What it costs you is substrate: you need more of it to reach half-maximal velocity, so apparent Km rises. Unchanged Vmax with a raised apparent Km is the competitive signature.
Draw a two-by-two table — Km up or unchanged against Vmax down or unchanged — and fill in the four modes from memory before you read a single answer choice. Then put a finger on the cell this data sits in and say the mode out loud. If you cannot fill the table, that is the gap to work on, and this item is the place to find it rather than the place to fix it.
Don't read a fourfold rise in Km as “the inhibitor hurt the enzyme” and choose B, noncompetitive, because it sounds like the general-purpose answer. Noncompetitive is the opposite pattern — Vmax falls and Km does not move — and it is round three of this board, not this one. C, uncompetitive, lowers both constants, so the unchanged 42 rules it out on its own. D takes enzyme out of the pool permanently, and no amount of substrate recovers it, so Vmax would have to fall as well. Ask yourself: what would raising substrate concentration do to each of these four inhibitors? If you can say that only the competitive one loses its grip as substrate rises, the pattern is yours.
Competitive. A competitive inhibitor binds the free enzyme at the active site, so raising substrate concentration outcompetes it and the same maximal velocity is still reached — Vmax is unchanged at 42 µmol/min. More substrate is required to reach half-maximal velocity, so the apparent Km rises from 0.8 to 3.2 mM. Unchanged Vmax with increased apparent Km is the competitive signature.
ROUND 2Predict the Lineweaver-Burk Plot
ON SCREEN — Your board says: “The same two data sets are replotted as 1/v against 1/[S]. Compared with the uninhibited enzyme, how do the lines for inhibitor X differ?” The instruction above it reads “Select the plot behavior consistent with the data above.” This round has no manipulative. Four answer choices sit below the item — A is Same y-intercept, steeper slope; B is Same slope, higher y-intercept; C is Higher y-intercept and shallower slope; D is Identical slope and identical y-intercept.
Do not picture the plot yet. Write the equation of the line first. Take the Michaelis-Menten equation, flip both sides, and you get one over v equals Km over Vmax times one over substrate, plus one over Vmax. That is y equals mx plus b, so name the parts out loud: the y-intercept is one over Vmax, and the slope is Km over Vmax. Both of those are quantities you already read in round one. Vmax did not change, so one over Vmax did not change, so the two lines cross the y-axis at the same point. Km went up fourfold and Vmax stayed put, so Km over Vmax went up fourfold: the inhibited line is steeper. Now say the picture before you look at it — two lines meeting on the y-axis, the inhibited one climbing faster. That is what people mean when they say competitive inhibition converges on the y-axis, and you just derived it instead of memorizing it.
Write the double-reciprocal form from memory and label the intercept and the slope with what each one equals — one over Vmax and Km over Vmax — before you read a choice. Then sketch both lines on scratch paper, uninhibited first, and say which axis the two meet on. State which of the two constants you would need in order to move the intercept.
Don't reach for a remembered picture instead of the algebra and pick B because the inhibited line “should be higher.” B — same slope, higher y-intercept — is the noncompetitive picture, where Vmax falls and Km holds, which is round three's data and not this round's. C raises the intercept and lowers the slope, meaning one over Vmax rose while Km over Vmax fell; that needs both constants to drop, which is round four's uncompetitive case. D says nothing changed, which contradicts the fourfold Km already given in the stem. Ask yourself: which quantity is the y-intercept, and did that quantity change between the two runs? If you answer one over Vmax, and no, the plot follows without a picture.
Same y-intercept, steeper slope. On a Lineweaver-Burk plot the y-intercept is 1/Vmax and the slope is Km/Vmax. A competitive inhibitor leaves Vmax unchanged, so the y-intercept is unchanged, and raises apparent Km, so the slope increases. The lines therefore converge on the y-axis.
ROUND 3A Second Inhibitor
ON SCREEN — Your board says: “Inhibitor Y is tested on the same enzyme. Vmax falls from 42 to 21 µmol/min while the apparent Km remains 0.8 mM. Which statement best describes how Y binds?” The instruction above it reads “Classify the second inhibitor from its constants.” This round has no manipulative. Four answer choices sit below the item — A is It competes with substrate for the active site; B is It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site; C is It binds only the enzyme-substrate complex; D is It forms an irreversible covalent bond with the catalytic residue.
New inhibitor, same enzyme, and this time the choices are mechanisms rather than labels — every one of them says where the inhibitor binds. Start from the data anyway. Vmax fell by half, 42 down to 21. Apparent Km did not move; it is still 0.8 millimolar. Say what each of those means on its own. A halved Vmax says half the catalytic capacity is gone and substrate cannot buy it back. An unchanged Km says the enzyme that is still working needs the same substrate concentration to reach half speed, so what is left behaves completely normally. Put them together: something is removing a fixed fraction of active enzyme at every substrate concentration, and leaving the rest untouched. That is what happens when the inhibitor does not care whether substrate is bound — it binds free enzyme and the enzyme-substrate complex with the same affinity, at a site that is not the active site. Substrate cannot displace it, because the two are not competing for the same place.
Before you read the choices, write two sentences with the numbers in them: what happened to Vmax, and what happened to Km. Then ask which of the two would have to change if substrate could displace this inhibitor. Predict the mechanism in your own words first — the choices in this round are long, and reading them cold makes you match on wording rather than on the data.
Don't pick A because the word inhibitor cues “active site.” Round one already showed what active-site competition looks like — Vmax unchanged, Km up — and both halves of that are wrong here. C, binding only the enzyme-substrate complex, is uncompetitive, which lowers Km as well; round four is that exact case, so the unchanged 0.8 rules it out. D deserves a full sentence, because an irreversible inhibitor also lowers Vmax while leaving the surviving enzyme's Km alone, so these numbers do not exclude it on the arithmetic. What excludes it is that nothing in the stem tests reversibility: a fixed inhibitor concentration with a measurable apparent Km names a mode of reversible inhibition, while covalent modification is a chemical claim this data has not earned. Ask yourself: what experiment would separate a noncompetitive inhibitor from an irreversible one? If you say dilute or dialyse the sample and see whether activity comes back, you are reading mechanism rather than matching words.
It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site. Reduced Vmax with unchanged Km is the classic noncompetitive pattern. Binding at an allosteric site with equal affinity for E and ES removes a constant fraction of active enzyme regardless of substrate concentration, so maximal velocity falls from 42 to 21 µmol/min while the substrate concentration giving half-maximal velocity is unchanged at 0.8 mM.
ROUND 4The Uncompetitive Case
ON SCREEN — Your board says: “Inhibitor Z binds only to the enzyme-substrate complex. Which combination of apparent constants is expected?” The instruction above it reads “Select the pattern that identifies uncompetitive inhibition.” This round has no manipulative. Four answer choices sit below the item — A is Km increases, Vmax unchanged; B is Km unchanged, Vmax unchanged; C is Km decreases, Vmax decreases; D is Km increases, Vmax increases.
This is round one run backwards: you are given the mechanism and asked for the numbers. Z binds only the enzyme-substrate complex — only ES, never free enzyme. Take the two consequences one at a time. First, any ES that Z grabs is pulled out of the catalytic cycle and cannot go on to make product, so the ceiling comes down and Vmax falls. Most people get that far. Second, think about the equilibrium. Free enzyme plus substrate sits in balance with ES. If ES is being consumed by something else, that balance shifts toward making more ES — Le Chatelier, inside an enzyme. So the enzyme looks as though it grabs substrate more readily than before, and the concentration needed for half-maximal velocity goes down: apparent Km falls. Both constants drop, and that is the whole point of this item. A falling Vmax alone does not name the mode. Noncompetitive drops Vmax and leaves Km. Uncompetitive drops both.
Write the two-step chain out for yourself before you look at the choices: ES is removed, so Vmax does what — and the E plus S equilibrium shifts which way, so apparent Km does what. Then say what happens to the ratio Km over Vmax, and whether the Lineweaver-Burk lines for this inhibitor would be parallel to the uninhibited line. That last question ties this round back to round two.
Don't stop after the first consequence — Vmax falls, so you reach for a falling Vmax with Km untouched, which is round three's noncompetitive pattern rather than this one. A is round one's competitive answer, Km up with Vmax unchanged, sitting here to catch pattern-matching on the word inhibitor. B says nothing changes, which cannot be true of something that removes ES from the cycle. D has Vmax rising, which no inhibitor does. The counterintuitive half is the falling Km, so probe exactly there. Ask yourself: if something is constantly removing ES, which way does the E plus S to ES equilibrium shift, and does the enzyme then look hungrier or less hungry for substrate? If you answer toward ES and hungrier, the drop in apparent Km follows.
Km decreases, Vmax decreases. Binding only to ES pulls that complex out of the catalytic cycle, lowering Vmax. Removing ES also shifts the E + S ⇌ ES equilibrium toward complex formation, so the enzyme behaves as though its substrate affinity increased and the apparent Km falls. Both constants decrease, which distinguishes uncompetitive from noncompetitive inhibition.
RONDA 1Clasificar a partir de Km y Vmax
EN PANTALLA — Tu tablero muestra, en inglés, tal como sale en el examen: “An enzyme is assayed alone and again with inhibitor X at fixed concentration. Vmax is unchanged at 42 µmol/min in both runs, while the apparent Km rises from 0.8 mM to 3.2 mM. Which mode of inhibition does X exhibit?” Es decir: se ensaya una enzima sola y otra vez con el inhibidor X a concentración fija; la Vmax se mantiene en 42 µmol/min en las dos corridas y la Km aparente sube de 0.8 mM a 3.2 mM; se pregunta qué tipo de inhibición presenta X. La instrucción de arriba dice “Use the reported constants to classify the inhibitor.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Competitive, la B es Noncompetitive, la C es Uncompetitive y la D es Irreversible covalent.
Lee los dos números antes de mirar las opciones y di qué mide cada uno. La Vmax es el techo: la velocidad que alcanza la enzima cuando el sustrato es saturante. La Km es la concentración de sustrato con la que llegas a la mitad de ese techo, así que habla de afinidad aparente, no de rapidez. Ahora los datos. La Vmax es 42 micromoles por minuto en las dos corridas: no cambió. La Km aparente pasó de 0.8 a 3.2 milimolar, cuatro veces más. Conviértelo en una frase sobre la enzima: sigue llegando a toda velocidad, pero necesita cuatro veces más sustrato para llegar a la mitad del camino. Y pregúntate qué clase de inhibidor te deja alcanzar el mismo techo. Uno al que le puedes ganar por cantidad. Si el inhibidor está en el sitio activo de la enzima libre, echar más sustrato le recupera el sitio, y en saturación el inhibidor prácticamente desaparece: la misma Vmax. Lo que te cuesta es sustrato, porque necesitas más para llegar a la velocidad semimáxima, y por eso sube la Km aparente. Vmax sin cambios con Km aparente elevada es la firma de la inhibición competitiva.
Dibuja una tabla de dos por dos —Km elevada o sin cambio contra Vmax disminuida o sin cambio— y ubica de memoria los cuatro tipos de inhibición antes de leer una sola opción. Después pon el dedo en la casilla donde caen estos datos y di el tipo en voz alta. Si no puedes llenar la tabla, ahí está el hueco que hay que trabajar, y esta pregunta sirve para encontrarlo, no para taparlo.
No leas que la Km se cuadruplicó como “el inhibidor dañó la enzima” ni elijas la B, Noncompetitive, porque suena a la respuesta que sirve para todo. La inhibición no competitiva es el patrón contrario —la Vmax baja y la Km no se mueve— y es la ronda 3 de este tablero, no esta. La C, Uncompetitive, es decir acompetitiva, baja las dos constantes, así que la Vmax intacta de 42 la descarta sola. La D saca enzima del juego de forma permanente y ningún exceso de sustrato la recupera, así que la Vmax también tendría que bajar. Pregúntate: ¿qué le haría a cada uno de estos cuatro inhibidores subir la concentración de sustrato? Si contestas que solo el competitivo pierde su agarre cuando sube el sustrato, ya dominas el patrón.
Competitive, es decir inhibición competitiva. Un inhibidor competitivo se une a la enzima libre en el sitio activo, así que al subir la concentración de sustrato se le gana por cantidad y se sigue alcanzando la misma velocidad máxima: la Vmax se queda en 42 µmol/min. Hace falta más sustrato para llegar a la velocidad semimáxima, así que la Km aparente sube de 0.8 a 3.2 mM. Vmax sin cambios con Km aparente aumentada es la firma de la inhibición competitiva.
RONDA 2Predecir la gráfica de Lineweaver-Burk
EN PANTALLA — Tu tablero muestra, en inglés: “The same two data sets are replotted as 1/v against 1/[S]. Compared with the uninhibited enzyme, how do the lines for inhibitor X differ?” Es decir: los mismos dos conjuntos de datos se grafican ahora como 1/v contra 1/[S], y se pregunta en qué se diferencian las rectas del inhibidor X respecto a la enzima sin inhibidor. La instrucción de arriba dice “Select the plot behavior consistent with the data above.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Same y-intercept, steeper slope; la B es Same slope, higher y-intercept; la C es Higher y-intercept and shallower slope; y la D es Identical slope and identical y-intercept.
Todavía no te imagines la gráfica. Primero escribe la ecuación de la recta. Toma la ecuación de Michaelis-Menten, invierte los dos lados y te queda uno sobre v igual a Km sobre Vmax por uno sobre sustrato, más uno sobre Vmax. Eso es y igual a mx más b, así que nombra las partes en voz alta: la ordenada al origen es uno sobre Vmax y la pendiente es Km sobre Vmax. Las dos son cantidades que ya leíste en la ronda 1. La Vmax no cambió, entonces uno sobre Vmax no cambió, entonces las dos rectas cortan el eje y en el mismo punto. La Km se cuadruplicó y la Vmax se quedó igual, entonces Km sobre Vmax se cuadruplicó: la recta con inhibidor es más empinada. Ahora describe el dibujo antes de verlo: dos rectas que se juntan en el eje y, y la del inhibidor sube más rápido. Eso es lo que quiere decir que la inhibición competitiva converge en el eje y, y acabas de deducirlo en vez de memorizarlo.
Escribe de memoria la forma de dobles recíprocos y etiqueta la ordenada al origen y la pendiente con lo que vale cada una —uno sobre Vmax y Km sobre Vmax— antes de leer una opción. Después dibuja las dos rectas en una hoja, primero la de la enzima sin inhibidor, y di en qué eje se juntan. Pregúntate cuál de las dos constantes tendría que cambiar para mover la ordenada al origen.
No eches mano de un dibujo memorizado en lugar del álgebra ni elijas la B porque la recta con inhibidor “debería ir más arriba”. La B —misma pendiente, ordenada al origen más alta— es el dibujo de la inhibición no competitiva, donde baja la Vmax y la Km se queda; esos son los datos de la ronda 3, no los de esta. La C sube la ordenada y baja la pendiente, o sea que uno sobre Vmax subió mientras Km sobre Vmax bajó; para eso tienen que bajar las dos constantes, que es el caso acompetitivo de la ronda 4. La D dice que nada cambió, y eso contradice la Km cuadruplicada que ya venía en el enunciado. Pregúntate: ¿qué cantidad es la ordenada al origen, y esa cantidad cambió entre las dos corridas? Si contestas uno sobre Vmax, y que no cambió, la gráfica sale sin necesidad de imaginarla.
Same y-intercept, steeper slope, es decir la misma ordenada al origen con una pendiente más empinada. En una gráfica de Lineweaver-Burk la ordenada al origen es 1/Vmax y la pendiente es Km/Vmax. Un inhibidor competitivo deja la Vmax igual, así que la ordenada al origen no cambia, y sube la Km aparente, así que la pendiente aumenta. Por eso las rectas convergen en el eje y.
RONDA 3Un segundo inhibidor
EN PANTALLA — Tu tablero muestra, en inglés: “Inhibitor Y is tested on the same enzyme. Vmax falls from 42 to 21 µmol/min while the apparent Km remains 0.8 mM. Which statement best describes how Y binds?” Es decir: se prueba el inhibidor Y en la misma enzima, la Vmax baja de 42 a 21 µmol/min y la Km aparente se mantiene en 0.8 mM; se pregunta qué enunciado describe mejor cómo se une Y. La instrucción de arriba dice “Classify the second inhibitor from its constants.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es It competes with substrate for the active site; la B es It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site; la C es It binds only the enzyme-substrate complex; y la D es It forms an irreversible covalent bond with the catalytic residue.
Otro inhibidor, la misma enzima, y esta vez las opciones son mecanismos y no etiquetas: cada una dice dónde se une el inhibidor. Aun así, empieza por los datos. La Vmax cayó a la mitad, de 42 a 21. La Km aparente no se movió: sigue en 0.8 milimolar. Di qué significa cada cosa por separado. Que la Vmax se reduzca a la mitad significa que se perdió la mitad de la capacidad catalítica y que el sustrato no la puede recuperar. Que la Km no cambie significa que la enzima que sigue trabajando necesita la misma concentración de sustrato para llegar a media velocidad, o sea que lo que queda funciona con toda normalidad. Júntalo: algo está quitando una fracción fija de enzima activa a cualquier concentración de sustrato y está dejando intacto el resto. Eso pasa cuando al inhibidor le da igual si el sustrato está unido: se une a la enzima libre y al complejo enzima-sustrato con la misma afinidad, en un sitio que no es el sitio activo. El sustrato no lo puede desplazar, porque no se están peleando el mismo lugar.
Antes de leer las opciones, escribe dos frases con los números adentro: qué pasó con la Vmax y qué pasó con la Km. Después pregúntate cuál de las dos tendría que cambiar si el sustrato pudiera desplazar a este inhibidor. Predice el mecanismo con tus propias palabras primero: las opciones de esta ronda son largas, y leerlas en frío te hace emparejar por redacción y no por los datos.
No elijas la A porque la palabra inhibidor lleva de la mano al “sitio activo”. La ronda 1 ya mostró cómo se ve la competencia por el sitio activo —Vmax igual, Km arriba— y aquí las dos mitades están al revés. La C, unirse solo al complejo enzima-sustrato, es la inhibición acompetitiva, que además baja la Km; la ronda 4 es exactamente ese caso, así que la Km intacta de 0.8 la descarta. La D merece una frase completa, porque un inhibidor irreversible también baja la Vmax y deja igual la Km de la enzima que sobrevive, así que con la aritmética no queda descartada. Lo que la descarta es que nada en el enunciado pone a prueba la reversibilidad: una concentración fija de inhibidor con una Km aparente medible nombra un tipo de inhibición reversible, mientras que la modificación covalente es una afirmación química que estos datos no alcanzan a sostener. Pregúntate: ¿qué experimento separaría a un inhibidor no competitivo de uno irreversible? Si contestas que diluir o dializar la muestra y ver si la actividad regresa, estás leyendo mecanismo y no emparejando palabras.
It binds free enzyme and enzyme-substrate complex with equal affinity, away from the active site, es decir que se une con la misma afinidad a la enzima libre y al complejo enzima-sustrato, en un sitio distinto del activo. Vmax reducida con Km sin cambios es el patrón clásico de la inhibición no competitiva. Unirse a un sitio alostérico con la misma afinidad por E y por ES quita una fracción constante de enzima activa a cualquier concentración de sustrato, así que la velocidad máxima cae de 42 a 21 µmol/min mientras la concentración de sustrato que da la velocidad semimáxima se queda en 0.8 mM.
RONDA 4El caso acompetitivo
EN PANTALLA — Tu tablero muestra, en inglés: “Inhibitor Z binds only to the enzyme-substrate complex. Which combination of apparent constants is expected?” Es decir: el inhibidor Z se une únicamente al complejo enzima-sustrato, y se pregunta qué combinación de constantes aparentes se espera. La instrucción de arriba dice “Select the pattern that identifies uncompetitive inhibition.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Km increases, Vmax unchanged; la B es Km unchanged, Vmax unchanged; la C es Km decreases, Vmax decreases; y la D es Km increases, Vmax increases.
Esta es la ronda 1 al revés: te dan el mecanismo y te piden los números. Z se une únicamente al complejo enzima-sustrato, solo a ES, nunca a la enzima libre. Toma las dos consecuencias una por una. Primero, cada ES que Z agarra sale del ciclo catalítico y ya no va a formar producto, así que el techo baja y la Vmax cae. Hasta ahí llega casi todo el mundo. Segundo, piensa en el equilibrio. La enzima libre más el sustrato están en equilibrio con ES. Si algo está consumiendo ES, ese equilibrio se desplaza hacia formar más ES: Le Châtelier, dentro de una enzima. Entonces la enzima parece agarrar el sustrato con más facilidad que antes, y la concentración que hace falta para la velocidad semimáxima baja: la Km aparente baja. Las dos constantes bajan, y de eso se trata esta pregunta. Que la Vmax caiga no basta para nombrar el tipo. La no competitiva baja la Vmax y deja la Km. La acompetitiva baja las dos.
Escribe la cadena de dos pasos por tu cuenta antes de mirar las opciones: se quita ES, entonces la Vmax hace qué; y el equilibrio de E más S se desplaza hacia dónde, entonces la Km aparente hace qué. Después di qué pasa con el cociente Km sobre Vmax y si las rectas de Lineweaver-Burk de este inhibidor quedarían paralelas a la de la enzima sin inhibidor. Esa última pregunta amarra esta ronda con la ronda 2.
No te detengas en la primera consecuencia: baja la Vmax, entonces buscas una Vmax que baja con la Km intacta, que es el patrón no competitivo de la ronda 3 y no el de esta. La A es la respuesta competitiva de la ronda 1, Km arriba con Vmax igual, puesta aquí para atrapar a quien empareja con la palabra inhibidor. La B dice que nada cambia, y eso no puede ser cierto de algo que saca ES del ciclo. La D pone la Vmax subiendo, y eso no lo hace ningún inhibidor. La mitad contraintuitiva es la Km que baja, así que indaga justo ahí. Pregúntate: si algo está quitando ES todo el tiempo, ¿hacia dónde se desplaza el equilibrio de E más S hacia ES, y entonces la enzima se ve con más hambre o con menos hambre de sustrato? Si contestas hacia ES y con más hambre, la bajada de la Km aparente sale sola.
Km decreases, Vmax decreases, es decir que bajan las dos: la Km y la Vmax. Unirse solo a ES saca ese complejo del ciclo catalítico y por eso baja la velocidad máxima. Quitar ES también desplaza el equilibrio E + S ⇌ ES hacia la formación del complejo, así que la enzima se comporta como si hubiera aumentado su afinidad por el sustrato y la Km aparente baja. Que bajen las dos constantes es lo que distingue a la inhibición acompetitiva de la no competitiva.
CPA · FAR
The board stays in the language the exam is given in. The script comes to you in yours.El tablero se queda en el idioma en que se presenta el examen. El guion le llega a usted en el suyo.The board stays in the language the exam is given in. The script comes to you in the one you read best.El tablero se queda en el idioma en que se presenta el examen. El guion te llega a ti en el que leas mejor.
This is what an instructor or tutor reads through with them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.
ROUND 1Count the Performance Obligations
ON SCREEN — Board shows: “On January 1, a software vendor signs a $120,000 contract that provides a perpetual software license delivered immediately, one year of technical support beginning January 1, and a separately priced installation service that a competitor also performs. Each item has an observable standalone selling price. How many distinct performance obligations does the contract contain?” The instruction above it reads “Select the number of distinct performance obligations.” This round has no manipulative. Four answer choices sit below the item — A is One, B is Two, C is Three, D is Four.
“Read the contract out loud before you read a single choice, and count promises rather than dollars. Three things are being handed to this customer: a perpetual license delivered on January 1, twelve months of technical support, and an installation service. Now put each one through the two-part test, in order, because a promise only becomes its own performance obligation when it passes both. First test: can the customer benefit from it on its own, or with resources it already has? Second test: is the promise separately identifiable within this contract, or is the vendor really using it as an input to one combined output? The license passes both — it is delivered and it works. Support passes both — it is a distinct service across a period, and the stem tells you every item here has an observable standalone selling price. Installation is the one worth slowing down on, and the stem hands you the deciding fact: a competitor also performs it. If somebody outside can do that work, the vendor is not integrating it into a single combined deliverable, so it stands on its own. Three promises, three obligations.”
Have them list the three promises down the left of a page and write the two tests across the top, then say yes or no in each of the six cells out loud before they read a choice. Ask them to name the one phrase in the stem that settles installation. Make them commit to a count on paper first — the choices here are bare numbers, and a bare number is easy to agree with after the fact.
The mistake to watch for is folding installation into the license because software has to be installed before anyone can use it, which lands on B, two. That instinct is the second test run from memory instead of from the stem, and the phrase “a competitor also performs” is sitting there precisely to defeat it. A, one, is the same error taken all the way, treating the whole contract as a single combined output. D, four, comes from counting the January 1 delivery as a promise separate from the license itself, or from splitting the support into a service and a period. Ask them: what would have to be true about the installation for it not to be distinct? If they answer that only this vendor could perform it, and that it turns the software into something the customer could not otherwise get, the second test is theirs.
Three. A promise is a distinct performance obligation when the customer can benefit from it on its own and it is separately identifiable within the contract. The license, the support and the installation each pass both tests, and installation passes the second one visibly, because a competitor can perform it — which shows the vendor is not integrating the three into one combined output.
ROUND 2Allocate the Transaction Price
ON SCREEN — Board shows: “The standalone selling prices are $100,000 for the license, $30,000 for the support, and $20,000 for the installation, a total of $150,000. The contract price is $120,000. What amount of the transaction price is allocated to the license?” The instruction above it reads “Compute the amount allocated to the license.” This round has no manipulative. Four answer choices sit below the item — A is $80,000, B is $96,000, C is $100,000, D is $120,000.
“Three obligations from round one, and now a number has to be attached to each. Write the two totals side by side before anything else: the standalone selling prices add to 150,000, and the customer is paying 120,000. The gap is 30,000, and this whole round turns on where that 30,000 goes. It goes everywhere, in proportion. A discount given on a bundle is a discount on the bundle, not a markdown on whichever item you choose to discount. So build the ratio out of standalone prices only: the license is 100,000 of the 150,000, which is two-thirds. Two-thirds of the 120,000 the customer is actually paying is 80,000. Say the sentence that produces — the license carries two-thirds of the price because it carries two-thirds of the standalone value. Then finish the other two the same way to prove the method closes: 30 over 150 of 120,000 is 24,000 for support, 20 over 150 is 16,000 for installation, and 80 plus 24 plus 16 is 120,000 exactly.”
Have them write the allocation as a fraction times a price — standalone over total standalone, times the contract price — and label all three numbers in words before computing anything. Then have them allocate all three obligations rather than only the license, and add the three results. Keep the choices covered until the three add to 120,000. A method that does not close is the fastest thing in the world to catch on paper.
The mistake to watch for is not allocating at all and taking C, 100,000, because that is the license's own number on the price list. Have them read the consequence out loud: it leaves 20,000 to cover a support obligation worth 30,000 and an installation worth 20,000, which hands the customer's entire discount to two items and none to the license. B is subtler and worth naming precisely — the bundle discount is 20 percent of 150,000, and applying that 20 percent to the contract price gives 120,000 times 0.8, or 96,000: the right rate on the wrong base. D allocates the whole contract to one of three obligations. Ask them: what has to sit in the denominator of the ratio, and why is it not 120,000? If they answer that the denominator is the sum of the standalone selling prices, because that is what the proportions are measured against, the allocation is theirs.
$80,000. The transaction price is allocated in proportion to standalone selling prices, so the $30,000 aggregate discount is spread across all three obligations rather than assigned to any one of them: 100,000 ÷ 150,000 × 120,000 = $80,000. Allocating the full $100,000 to the license would leave only $20,000 for the other two obligations and ignore the proportional requirement.
ROUND 3Time the Recognition
ON SCREEN — Board shows: “The license transfers on January 1 and installation is completed on January 15. Support is provided ratably over the twelve months of the year. Using the allocation above, how much revenue is recognized for the quarter ended March 31?” The instruction above it reads “Determine revenue recognized for the quarter.” This round has no manipulative. Four answer choices sit below the item — A is $80,000, B is $96,000, C is $102,000, D is $120,000.
“Two rounds of taking the contract apart, and now every piece gets a date. Go through the three obligations one at a time and ask each the same question: is this satisfied at a point in time, or over time? The license transfers on January 1 — a point in time, inside the quarter, so all 80,000 is earned. Installation is completed on January 15 — also a point in time, also inside the quarter, so all of its allocation is earned, and that allocation is 20 over 150 of 120,000, which is 16,000. Support is the one that behaves differently: it is provided ratably across twelve months, so control transfers continuously and the revenue follows the passage of time. Its allocation is 30 over 150 of 120,000, or 24,000, and three of those twelve months are gone by March 31 — a quarter of the year, so a quarter of 24,000, which is 6,000. Now add only what has actually been earned: 80,000 plus 16,000 plus 6,000.”
Have them draw a three-row schedule — obligation, allocated amount, point in time or over time — and fill in all nine cells before they read a choice. Then have them write the fraction of the year elapsed at March 31 next to the one row that needs it. Ask them to say, in their own words, what makes the support row behave differently from the other two before they add anything up.
The mistake to watch for is letting the January 1 cash decide the answer and taking D, 120,000, because the customer has already paid in full. Cash is not the trigger; transfer of control is, and one of the three obligations is only a quarter transferred by March 31. B, 96,000, is the near miss to probe hardest: it is 80,000 plus 16,000, the two point-in-time obligations recognized correctly and the support forgotten altogether. A, 80,000, stops after the license and drops installation too, usually because January 15 gets read as a date outside the quarter. Ask them: how much of the support has the vendor delivered by March 31, and what is that worth? If they answer three months of twelve, so a quarter of 24,000, the timing is theirs.
$102,000. The license and the installation are satisfied at points in time inside the quarter, so their full allocated amounts are recognized — $80,000 plus $16,000, since installation is 20,000 ÷ 150,000 × 120,000 = $16,000. Support is satisfied over time, so one quarter of its allocated $24,000 is recognized, or $6,000. Total: 80,000 + 16,000 + 6,000 = $102,000.
ROUND 4Classify the Unrecognized Amount
ON SCREEN — Board shows: “The customer paid the full $120,000 on January 1. How should the portion of the support allocation not yet recognized be presented in the March 31 balance sheet?” The instruction above it reads “Select the correct presentation at March 31.” This round has no manipulative. Four answer choices sit below the item — A is As a contract asset, B is As a contract liability, C is As a reduction of accounts receivable, D is As an addition to retained earnings.
“Round three said what was earned. This round asks what to do with the rest, and the whole thing is settled by which came first — the cash or the performance. Say the two facts back to me. The customer paid all 120,000 on January 1, and on March 31 the vendor still owes nine months of support. So the money arrived before the work was done. Money held against work not yet done is an obligation, and an obligation belongs on the right-hand side of the balance sheet. That is exactly what a contract liability is: consideration received, or unconditionally receivable, before the related performance obligation has been satisfied. The name most people learn first is deferred revenue, and it is the same balance. Now put a number on it before you look down at the choices. The support allocation was 24,000, and 6,000 of it was recognized in the quarter, so 18,000 is still owed as service. Say it as a sentence — eighteen thousand dollars of the customer's money is sitting there against support the vendor has not performed yet.”
Have them write the January 1 journal entry before they read a choice — cash 120,000 on one side, and what on the other — and then the entry that runs at each month end. Ask them which balance that second entry is draining. Have them carry round three's numbers through to the closing balance and say the figure out loud, so the presentation question is being answered about an amount they have already found for themselves.
The mistake to watch for is reaching for A because “contract” and “asset” both sound like the vocabulary being tested. A contract asset is the mirror image of this fact pattern — performance first, unconditional right to payment not yet earned — so ask which side got ahead here, the vendor or the customer, and A goes away. C assumes there is a receivable to net against, but the customer paid cash on January 1, so accounts receivable is zero and there is nothing to reduce. D takes revenue the vendor has not earned and puts it straight into equity, which is the error the five-step model exists to prevent. Ask them: what does the vendor still owe on March 31, and does owing something make it an asset or a liability? If they answer nine months of support, therefore a liability, the presentation is theirs.
As a contract liability. Consideration has been received but the related performance obligation has not yet been satisfied, which is the definition of a contract liability — the deferred revenue balance. A contract asset arises in the opposite case, when performance precedes an unconditional right to payment. The unrecognized support balance is the $24,000 allocation less the $6,000 recognized, or $18,000.
RONDA 1Contar las obligaciones de desempeño
EN PANTALLA — El tablero muestra, en inglés, tal como sale en el examen: “On January 1, a software vendor signs a $120,000 contract that provides a perpetual software license delivered immediately, one year of technical support beginning January 1, and a separately priced installation service that a competitor also performs. Each item has an observable standalone selling price. How many distinct performance obligations does the contract contain?” Es decir: el 1 de enero un proveedor de software firma un contrato de 120,000 dólares que incluye una licencia perpetua entregada de inmediato, un año de soporte técnico que arranca el 1 de enero y un servicio de instalación con precio aparte que también presta un competidor; cada elemento tiene un precio de venta independiente observable, y se pregunta cuántas obligaciones de desempeño distintas contiene el contrato. La instrucción de arriba dice “Select the number of distinct performance obligations.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es One, la B es Two, la C es Three y la D es Four.
“Lee el contrato en voz alta antes de mirar una sola opción, y cuenta promesas, no dólares. Al cliente le están entregando tres cosas: una licencia perpetua que se entrega el 1 de enero, doce meses de soporte técnico y un servicio de instalación. Ahora pásale a cada una la prueba de dos partes, en ese orden, porque una promesa se vuelve obligación de desempeño solo si pasa las dos. Primera parte: ¿el cliente puede aprovecharla por sí sola, o junto con recursos que ya tiene? Segunda parte: ¿la promesa se puede identificar por separado dentro de este contrato, o el proveedor la está usando como insumo de un solo producto combinado? La licencia pasa las dos: se entrega y funciona. El soporte pasa las dos: es un servicio distinto a lo largo de un periodo, y el enunciado dice que cada elemento tiene un precio de venta independiente observable. La instalación es la que hay que pensar con calma, y el enunciado te regala el dato que decide: un competidor también la presta. Si alguien de fuera puede hacer ese trabajo, el proveedor no la está integrando en un solo entregable combinado, así que se sostiene sola. Tres promesas, tres obligaciones.”
Pídale que anote las tres promesas en una columna a la izquierda de la hoja y las dos pruebas arriba, y que diga en voz alta sí o no en cada una de las seis casillas antes de leer una opción. Pregúntele cuál es la frase exacta del enunciado que resuelve el caso de la instalación. Que se comprometa primero con un número en el papel: aquí las opciones son números pelados, y a un número pelado es facilísimo darle la razón después.
El error que hay que vigilar es meter la instalación dentro de la licencia porque un software hay que instalarlo antes de poder usarlo, y eso cae en la B, Two. Ese impulso es la segunda prueba hecha de memoria y no con el enunciado: la frase “a competitor also performs” está puesta ahí justamente para desarmarlo. La A, One, es el mismo error llevado hasta el final, tratar todo el contrato como un único producto combinado. La D, Four, sale de contar la entrega del 1 de enero como una promesa aparte de la licencia misma, o de partir el soporte en un servicio y un periodo. Pregúntele: ¿qué tendría que pasar con la instalación para que no fuera distinta? Si le contesta que solo este proveedor pudiera prestarla, y que convirtiera el software en algo que el cliente no podría obtener de otra manera, ya domina la segunda prueba.
Three, es decir tres. Una promesa es una obligación de desempeño distinta cuando el cliente puede aprovecharla por sí sola y además se puede identificar por separado dentro del contrato. La licencia, el soporte y la instalación pasan las dos pruebas, y la instalación pasa la segunda de forma visible, porque un competidor puede prestarla: eso demuestra que el proveedor no está integrando las tres en un solo producto combinado.
RONDA 2Asignar el precio de la transacción
EN PANTALLA — El tablero muestra, en inglés: “The standalone selling prices are $100,000 for the license, $30,000 for the support, and $20,000 for the installation, a total of $150,000. The contract price is $120,000. What amount of the transaction price is allocated to the license?” Es decir: los precios de venta independientes son 100,000 dólares la licencia, 30,000 el soporte y 20,000 la instalación, en total 150,000, mientras que el precio del contrato es 120,000, y se pregunta cuánto del precio de la transacción se asigna a la licencia. La instrucción de arriba dice “Compute the amount allocated to the license.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es $80,000, la B es $96,000, la C es $100,000 y la D es $120,000.
“Ya tenemos tres obligaciones de la ronda anterior, y ahora hay que ponerle un número a cada una. Antes de cualquier otra cosa, escribe los dos totales uno al lado del otro: los precios de venta independientes suman 150,000 y el cliente está pagando 120,000. La diferencia es 30,000, y toda la ronda se decide en a dónde van esos 30,000. Van a todas partes, en proporción. Un descuento que se da sobre un paquete es un descuento del paquete, no una rebaja al elemento que uno elija rebajar. Así que arma la razón solo con los precios independientes: la licencia es 100,000 de los 150,000, o sea dos tercios. Dos tercios de los 120,000 que el cliente realmente paga son 80,000. Di la frase que sale de ahí: la licencia se lleva dos tercios del precio porque aporta dos tercios del valor independiente. Después termina las otras dos igual, para comprobar que el método cierra: 30 sobre 150 de 120,000 son 24,000 del soporte, 20 sobre 150 son 16,000 de la instalación, y 80 más 24 más 16 son exactamente 120,000.”
Pídale que escriba la asignación como una fracción por un precio —precio independiente sobre suma de precios independientes, por el precio del contrato— y que etiquete con palabras los tres números antes de calcular nada. Después pídale que asigne las tres obligaciones y no solo la licencia, y que sume los tres resultados. Mantenga tapadas las opciones hasta que los tres sumen 120,000. Un método que no cierra es lo más rápido de cachar en el papel.
El error que hay que vigilar es no asignar nada y tomar la C, 100,000, porque ese es el número de la licencia en la lista de precios. Pídale que lea en voz alta la consecuencia: deja 20,000 para cubrir un soporte que vale 30,000 y una instalación que vale 20,000, es decir, le regala todo el descuento del cliente a dos elementos y nada a la licencia. La B es más fina y vale la pena nombrarla con precisión: el descuento del paquete es 20 por ciento de 150,000, y aplicar ese 20 por ciento al precio del contrato da 120,000 por 0.8, o sea 96,000; la tasa correcta sobre la base equivocada. La D le asigna todo el contrato a una de tres obligaciones. Pregúntele: ¿qué tiene que ir en el denominador de la razón, y por qué no es 120,000? Si le contesta que el denominador es la suma de los precios de venta independientes, porque contra eso se miden las proporciones, ya domina la asignación.
$80,000. El precio de la transacción se asigna en proporción a los precios de venta independientes, así que el descuento agregado de 30,000 dólares se reparte entre las tres obligaciones en lugar de cargarse a una sola: 100,000 ÷ 150,000 × 120,000 = 80,000 dólares. Asignarle los 100,000 completos a la licencia dejaría apenas 20,000 para las otras dos obligaciones y pasaría por encima del requisito de proporcionalidad.
RONDA 3Ubicar el reconocimiento en el tiempo
EN PANTALLA — El tablero muestra, en inglés: “The license transfers on January 1 and installation is completed on January 15. Support is provided ratably over the twelve months of the year. Using the allocation above, how much revenue is recognized for the quarter ended March 31?” Es decir: la licencia se transfiere el 1 de enero, la instalación se termina el 15 de enero y el soporte se presta de forma lineal a lo largo de los doce meses del año, y con la asignación anterior se pregunta cuánto ingreso se reconoce en el trimestre que cierra el 31 de marzo. La instrucción de arriba dice “Determine revenue recognized for the quarter.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es $80,000, la B es $96,000, la C es $102,000 y la D es $120,000.
“Dos rondas desarmando el contrato, y ahora cada pieza recibe una fecha. Recorre las tres obligaciones una por una y hazle a cada una la misma pregunta: ¿esto se satisface en un momento determinado o a lo largo del tiempo? La licencia se transfiere el 1 de enero: un momento determinado, dentro del trimestre, así que se ganan los 80,000 completos. La instalación se termina el 15 de enero: también un momento determinado, también dentro del trimestre, así que se gana toda su asignación, y esa asignación es 20 sobre 150 de 120,000, o sea 16,000. El soporte es el que se porta distinto: se presta de forma lineal a lo largo de doce meses, así que el control se transfiere de manera continua y el ingreso sigue el paso del tiempo. Su asignación es 30 sobre 150 de 120,000, o sea 24,000, y al 31 de marzo ya pasaron tres de esos doce meses: un cuarto del año, así que un cuarto de 24,000, que son 6,000. Ahora suma solo lo que de verdad se ganó: 80,000 más 16,000 más 6,000.”
Pídale que dibuje un cuadro de tres renglones —obligación, monto asignado, y en un momento determinado o a lo largo del tiempo— y que llene las nueve casillas antes de leer una opción. Después pídale que escriba, junto al único renglón que lo necesita, qué fracción del año transcurrió al 31 de marzo. Pregúntele con sus propias palabras qué hace que el renglón del soporte se porte distinto de los otros dos, antes de que sume nada.
El error que hay que vigilar es dejar que el cobro del 1 de enero decida la respuesta y tomar la D, 120,000, porque el cliente ya pagó todo. El cobro no es el disparador; el disparador es la transferencia del control, y al 31 de marzo una de las tres obligaciones apenas se transfirió en una cuarta parte. La B, 96,000, es el casi acierto que hay que sondear con más fuerza: son 80,000 más 16,000, es decir, las dos obligaciones de momento determinado bien reconocidas y el soporte olvidado por completo. La A, 80,000, se detiene en la licencia y además deja fuera la instalación, casi siempre porque el 15 de enero se lee como una fecha fuera del trimestre. Pregúntele: ¿cuánto soporte entregó el proveedor al 31 de marzo, y cuánto vale eso? Si le contesta tres meses de doce, o sea un cuarto de 24,000, ya domina el momento del reconocimiento.
$102,000. La licencia y la instalación se satisfacen en momentos determinados dentro del trimestre, así que se reconocen sus montos asignados completos —80,000 más 16,000, porque la instalación es 20,000 ÷ 150,000 × 120,000 = 16,000 dólares—. El soporte se satisface a lo largo del tiempo, así que se reconoce un cuarto de sus 24,000 asignados, o sea 6,000. Total: 80,000 + 16,000 + 6,000 = 102,000 dólares.
RONDA 4Clasificar el monto no reconocido
EN PANTALLA — El tablero muestra, en inglés: “The customer paid the full $120,000 on January 1. How should the portion of the support allocation not yet recognized be presented in the March 31 balance sheet?” Es decir: el cliente pagó los 120,000 dólares completos el 1 de enero, y se pregunta cómo debe presentarse en el balance del 31 de marzo la parte de la asignación del soporte que todavía no se reconoce. La instrucción de arriba dice “Select the correct presentation at March 31.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es As a contract asset, la B es As a contract liability, la C es As a reduction of accounts receivable y la D es As an addition to retained earnings.
“La ronda anterior dijo qué se ganó. Esta pregunta qué se hace con el resto, y todo se decide por qué llegó primero: el dinero o el desempeño. Repíteme los dos hechos. El cliente pagó los 120,000 completos el 1 de enero, y al 31 de marzo el proveedor todavía le debe nueve meses de soporte. Entonces el dinero llegó antes de que el trabajo estuviera hecho. Dinero retenido contra un trabajo que todavía no se hace es una obligación, y una obligación va del lado derecho del balance. Eso es exactamente un pasivo del contrato: contraprestación recibida, o exigible sin condiciones, antes de haber satisfecho la obligación de desempeño relacionada. El nombre que casi todos aprendemos primero es ingresos diferidos, y es el mismo saldo. Ahora ponle número antes de mirar las opciones. La asignación del soporte era 24,000, se reconocieron 6,000 en el trimestre, así que quedan 18,000 que todavía se deben en servicio. Dilo como frase: hay dieciocho mil dólares del cliente parados ahí, contra un soporte que el proveedor todavía no ha prestado.”
Pídale que escriba el asiento del 1 de enero antes de leer una opción —efectivo 120,000 de un lado, y qué del otro— y luego el asiento que corre al cierre de cada mes. Pregúntele qué saldo va vaciando ese segundo asiento. Pídale que arrastre los números de la ronda anterior hasta el saldo final y que diga la cifra en voz alta, para que la pregunta de presentación se conteste sobre un monto que ya encontró él mismo.
El error que hay que vigilar es lanzarse a la A porque “contract” y “asset” suenan las dos al vocabulario que se está evaluando. Un activo del contrato es la imagen invertida de este caso —primero el desempeño, y todavía sin derecho incondicional al cobro—, así que pregúntele quién se adelantó aquí, el proveedor o el cliente, y la A se cae sola. La C supone que hay una cuenta por cobrar contra la cual netear, pero el cliente pagó en efectivo el 1 de enero, así que las cuentas por cobrar están en cero y no hay nada que reducir. La D toma un ingreso que el proveedor no ha ganado y lo manda directo al patrimonio, que es justo el error que el modelo de cinco pasos existe para evitar. Pregúntele: ¿qué sigue debiendo el proveedor al 31 de marzo, y deber algo lo convierte en activo o en pasivo? Si le contesta nueve meses de soporte, y por lo tanto pasivo, ya domina la presentación.
As a contract liability, es decir como un pasivo del contrato. Se recibió la contraprestación pero todavía no se ha satisfecho la obligación de desempeño relacionada, y esa es la definición de pasivo del contrato: el saldo de ingresos diferidos. Un activo del contrato surge en el caso opuesto, cuando el desempeño va delante del derecho incondicional al cobro. El saldo de soporte no reconocido son los 24,000 asignados menos los 6,000 reconocidos, o sea 18,000 dólares.
ROUND 1Count the Performance Obligations
ON SCREEN — Your board says: “On January 1, a software vendor signs a $120,000 contract that provides a perpetual software license delivered immediately, one year of technical support beginning January 1, and a separately priced installation service that a competitor also performs. Each item has an observable standalone selling price. How many distinct performance obligations does the contract contain?” The instruction above it reads “Select the number of distinct performance obligations.” This round has no manipulative. Four answer choices sit below the item — A is One, B is Two, C is Three, D is Four.
Read the contract out loud before you read a single choice, and count promises rather than dollars. Three things are being handed to this customer: a perpetual license delivered on January 1, twelve months of technical support, and an installation service. Now put each one through the two-part test, in order, because a promise only becomes its own performance obligation when it passes both. First test: can the customer benefit from it on its own, or with resources it already has? Second test: is the promise separately identifiable within this contract, or is the vendor really using it as an input to one combined output? The license passes both — it is delivered and it works. Support passes both — it is a distinct service across a period, and the stem tells you every item here has an observable standalone selling price. Installation is the one worth slowing down on, and the stem hands you the deciding fact: a competitor also performs it. If somebody outside can do that work, the vendor is not integrating it into a single combined deliverable, so it stands on its own. Three promises, three obligations.
List the three promises down the left of a page and write the two tests across the top, then say yes or no in each of the six cells out loud before you read a choice. Name the one phrase in the stem that settles installation. Commit to a count on paper first — the choices here are bare numbers, and a bare number is very easy to agree with after the fact.
Don't fold installation into the license because software has to be installed before anyone can use it, which lands on B, two. That instinct is the second test run from memory instead of from the stem, and the phrase “a competitor also performs” is sitting there precisely to defeat it. A, one, is the same error taken all the way, treating the whole contract as a single combined output. D, four, comes from counting the January 1 delivery as a promise separate from the license itself, or from splitting the support into a service and a period. Ask yourself: what would have to be true about the installation for it not to be distinct? If you answer that only this vendor could perform it, and that it turns the software into something the customer could not otherwise get, the second test is yours.
Three. A promise is a distinct performance obligation when the customer can benefit from it on its own and it is separately identifiable within the contract. The license, the support and the installation each pass both tests, and installation passes the second one visibly, because a competitor can perform it — which shows the vendor is not integrating the three into one combined output.
ROUND 2Allocate the Transaction Price
ON SCREEN — Your board says: “The standalone selling prices are $100,000 for the license, $30,000 for the support, and $20,000 for the installation, a total of $150,000. The contract price is $120,000. What amount of the transaction price is allocated to the license?” The instruction above it reads “Compute the amount allocated to the license.” This round has no manipulative. Four answer choices sit below the item — A is $80,000, B is $96,000, C is $100,000, D is $120,000.
Three obligations from round one, and now a number has to be attached to each. Write the two totals side by side before anything else: the standalone selling prices add to 150,000, and the customer is paying 120,000. The gap is 30,000, and this whole round turns on where that 30,000 goes. It goes everywhere, in proportion. A discount given on a bundle is a discount on the bundle, not a markdown on whichever item you choose to discount. So build the ratio out of standalone prices only: the license is 100,000 of the 150,000, which is two-thirds. Two-thirds of the 120,000 the customer is actually paying is 80,000. Say the sentence that produces — the license carries two-thirds of the price because it carries two-thirds of the standalone value. Then finish the other two the same way to prove the method closes: 30 over 150 of 120,000 is 24,000 for support, 20 over 150 is 16,000 for installation, and 80 plus 24 plus 16 is 120,000 exactly.
Write the allocation as a fraction times a price — standalone over total standalone, times the contract price — and label all three numbers in words before computing anything. Then allocate all three obligations rather than only the license, and add the three results. Keep the choices covered until the three add to 120,000. A method that does not close is the fastest thing in the world to catch on paper.
Don't skip the allocation and take C, 100,000, because that is the license's own number on the price list. Read the consequence out loud: it leaves 20,000 to cover a support obligation worth 30,000 and an installation worth 20,000, which hands the customer's entire discount to two items and none to the license. B is subtler and worth naming precisely — the bundle discount is 20 percent of 150,000, and applying that 20 percent to the contract price gives 120,000 times 0.8, or 96,000: the right rate on the wrong base. D allocates the whole contract to one of three obligations. Ask yourself: what has to sit in the denominator of the ratio, and why is it not 120,000? If you answer that the denominator is the sum of the standalone selling prices, because that is what the proportions are measured against, the allocation is yours.
$80,000. The transaction price is allocated in proportion to standalone selling prices, so the $30,000 aggregate discount is spread across all three obligations rather than assigned to any one of them: 100,000 ÷ 150,000 × 120,000 = $80,000. Allocating the full $100,000 to the license would leave only $20,000 for the other two obligations and ignore the proportional requirement.
ROUND 3Time the Recognition
ON SCREEN — Your board says: “The license transfers on January 1 and installation is completed on January 15. Support is provided ratably over the twelve months of the year. Using the allocation above, how much revenue is recognized for the quarter ended March 31?” The instruction above it reads “Determine revenue recognized for the quarter.” This round has no manipulative. Four answer choices sit below the item — A is $80,000, B is $96,000, C is $102,000, D is $120,000.
Two rounds of taking the contract apart, and now every piece gets a date. Go through the three obligations one at a time and ask each the same question: is this satisfied at a point in time, or over time? The license transfers on January 1 — a point in time, inside the quarter, so all 80,000 is earned. Installation is completed on January 15 — also a point in time, also inside the quarter, so all of its allocation is earned, and that allocation is 20 over 150 of 120,000, which is 16,000. Support is the one that behaves differently: it is provided ratably across twelve months, so control transfers continuously and the revenue follows the passage of time. Its allocation is 30 over 150 of 120,000, or 24,000, and three of those twelve months are gone by March 31 — a quarter of the year, so a quarter of 24,000, which is 6,000. Now add only what has actually been earned: 80,000 plus 16,000 plus 6,000.
Draw a three-row schedule — obligation, allocated amount, point in time or over time — and fill in all nine cells before you read a choice. Then write the fraction of the year elapsed at March 31 next to the one row that needs it. Say, in your own words, what makes the support row behave differently from the other two before you add anything up.
Don't let the January 1 cash decide the answer and take D, 120,000, because the customer has already paid in full. Cash is not the trigger; transfer of control is, and one of the three obligations is only a quarter transferred by March 31. B, 96,000, is the near miss to probe hardest: it is 80,000 plus 16,000, the two point-in-time obligations recognized correctly and the support forgotten altogether. A, 80,000, stops after the license and drops installation too, usually because January 15 gets read as a date outside the quarter. Ask yourself: how much of the support has the vendor delivered by March 31, and what is that worth? If you answer three months of twelve, so a quarter of 24,000, the timing is yours.
$102,000. The license and the installation are satisfied at points in time inside the quarter, so their full allocated amounts are recognized — $80,000 plus $16,000, since installation is 20,000 ÷ 150,000 × 120,000 = $16,000. Support is satisfied over time, so one quarter of its allocated $24,000 is recognized, or $6,000. Total: 80,000 + 16,000 + 6,000 = $102,000.
ROUND 4Classify the Unrecognized Amount
ON SCREEN — Your board says: “The customer paid the full $120,000 on January 1. How should the portion of the support allocation not yet recognized be presented in the March 31 balance sheet?” The instruction above it reads “Select the correct presentation at March 31.” This round has no manipulative. Four answer choices sit below the item — A is As a contract asset, B is As a contract liability, C is As a reduction of accounts receivable, D is As an addition to retained earnings.
Round three said what was earned. This round asks what to do with the rest, and the whole thing is settled by which came first — the cash or the performance. Say the two facts out loud to yourself. The customer paid all 120,000 on January 1, and on March 31 the vendor still owes nine months of support. So the money arrived before the work was done. Money held against work not yet done is an obligation, and an obligation belongs on the right-hand side of the balance sheet. That is exactly what a contract liability is: consideration received, or unconditionally receivable, before the related performance obligation has been satisfied. The name most people learn first is deferred revenue, and it is the same balance. Now put a number on it before you look down at the choices. The support allocation was 24,000, and 6,000 of it was recognized in the quarter, so 18,000 is still owed as service. Say it as a sentence — eighteen thousand dollars of the customer's money is sitting there against support the vendor has not performed yet.
Write the January 1 journal entry before you read a choice — cash 120,000 on one side, and what on the other — and then the entry that runs at each month end. Say which balance that second entry is draining. Carry round three's numbers through to the closing balance and say the figure out loud, so the presentation question is being answered about an amount you have already found for yourself.
Don't reach for A because “contract” and “asset” both sound like the vocabulary being tested. A contract asset is the mirror image of this fact pattern — performance first, unconditional right to payment not yet earned — so ask which side got ahead here, the vendor or the customer, and A goes away. C assumes there is a receivable to net against, but the customer paid cash on January 1, so accounts receivable is zero and there is nothing to reduce. D takes revenue the vendor has not earned and puts it straight into equity, which is the error the five-step model exists to prevent. Ask yourself: what does the vendor still owe on March 31, and does owing something make it an asset or a liability? If you answer nine months of support, therefore a liability, the presentation is yours.
As a contract liability. Consideration has been received but the related performance obligation has not yet been satisfied, which is the definition of a contract liability — the deferred revenue balance. A contract asset arises in the opposite case, when performance precedes an unconditional right to payment. The unrecognized support balance is the $24,000 allocation less the $6,000 recognized, or $18,000.
RONDA 1Contar las obligaciones de desempeño
EN PANTALLA — Tu tablero muestra, en inglés, tal como sale en el examen: “On January 1, a software vendor signs a $120,000 contract that provides a perpetual software license delivered immediately, one year of technical support beginning January 1, and a separately priced installation service that a competitor also performs. Each item has an observable standalone selling price. How many distinct performance obligations does the contract contain?” Es decir: el 1 de enero un proveedor de software firma un contrato de 120,000 dólares que incluye una licencia perpetua entregada de inmediato, un año de soporte técnico que arranca el 1 de enero y un servicio de instalación con precio aparte que también presta un competidor; cada elemento tiene un precio de venta independiente observable, y se pregunta cuántas obligaciones de desempeño distintas contiene el contrato. La instrucción de arriba dice “Select the number of distinct performance obligations.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es One, la B es Two, la C es Three y la D es Four.
Lee el contrato en voz alta antes de mirar una sola opción, y cuenta promesas, no dólares. Al cliente le están entregando tres cosas: una licencia perpetua que se entrega el 1 de enero, doce meses de soporte técnico y un servicio de instalación. Ahora pásale a cada una la prueba de dos partes, en ese orden, porque una promesa se vuelve obligación de desempeño solo si pasa las dos. Primera parte: ¿el cliente puede aprovecharla por sí sola, o junto con recursos que ya tiene? Segunda parte: ¿la promesa se puede identificar por separado dentro de este contrato, o el proveedor la está usando como insumo de un solo producto combinado? La licencia pasa las dos: se entrega y funciona. El soporte pasa las dos: es un servicio distinto a lo largo de un periodo, y el enunciado dice que cada elemento tiene un precio de venta independiente observable. La instalación es la que hay que pensar con calma, y el enunciado te regala el dato que decide: un competidor también la presta. Si alguien de fuera puede hacer ese trabajo, el proveedor no la está integrando en un solo entregable combinado, así que se sostiene sola. Tres promesas, tres obligaciones.
Anota las tres promesas en una columna a la izquierda de la hoja y las dos pruebas arriba, y di en voz alta sí o no en cada una de las seis casillas antes de leer una opción. Busca cuál es la frase exacta del enunciado que resuelve el caso de la instalación. Comprométete primero con un número en el papel: aquí las opciones son números pelados, y a un número pelado es facilísimo darle la razón después.
No metas la instalación dentro de la licencia porque un software hay que instalarlo antes de poder usarlo: eso cae en la B, Two. Ese impulso es la segunda prueba hecha de memoria y no con el enunciado, y la frase “a competitor also performs” está puesta ahí justamente para desarmarlo. La A, One, es el mismo error llevado hasta el final, tratar todo el contrato como un único producto combinado. La D, Four, sale de contar la entrega del 1 de enero como una promesa aparte de la licencia misma, o de partir el soporte en un servicio y un periodo. Pregúntate: ¿qué tendría que pasar con la instalación para que no fuera distinta? Si contestas que solo este proveedor pudiera prestarla, y que convirtiera el software en algo que el cliente no podría obtener de otra manera, ya dominas la segunda prueba.
Three, es decir tres. Una promesa es una obligación de desempeño distinta cuando el cliente puede aprovecharla por sí sola y además se puede identificar por separado dentro del contrato. La licencia, el soporte y la instalación pasan las dos pruebas, y la instalación pasa la segunda de forma visible, porque un competidor puede prestarla: eso demuestra que el proveedor no está integrando las tres en un solo producto combinado.
RONDA 2Asignar el precio de la transacción
EN PANTALLA — Tu tablero muestra, en inglés: “The standalone selling prices are $100,000 for the license, $30,000 for the support, and $20,000 for the installation, a total of $150,000. The contract price is $120,000. What amount of the transaction price is allocated to the license?” Es decir: los precios de venta independientes son 100,000 dólares la licencia, 30,000 el soporte y 20,000 la instalación, en total 150,000, mientras que el precio del contrato es 120,000, y se pregunta cuánto del precio de la transacción se asigna a la licencia. La instrucción de arriba dice “Compute the amount allocated to the license.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es $80,000, la B es $96,000, la C es $100,000 y la D es $120,000.
Ya tenemos tres obligaciones de la ronda anterior, y ahora hay que ponerle un número a cada una. Antes de cualquier otra cosa, escribe los dos totales uno al lado del otro: los precios de venta independientes suman 150,000 y el cliente está pagando 120,000. La diferencia es 30,000, y toda la ronda se decide en a dónde van esos 30,000. Van a todas partes, en proporción. Un descuento que se da sobre un paquete es un descuento del paquete, no una rebaja al elemento que uno elija rebajar. Así que arma la razón solo con los precios independientes: la licencia es 100,000 de los 150,000, o sea dos tercios. Dos tercios de los 120,000 que el cliente realmente paga son 80,000. Di la frase que sale de ahí: la licencia se lleva dos tercios del precio porque aporta dos tercios del valor independiente. Después termina las otras dos igual, para comprobar que el método cierra: 30 sobre 150 de 120,000 son 24,000 del soporte, 20 sobre 150 son 16,000 de la instalación, y 80 más 24 más 16 son exactamente 120,000.
Escribe la asignación como una fracción por un precio —precio independiente sobre suma de precios independientes, por el precio del contrato— y etiqueta con palabras los tres números antes de calcular nada. Después asigna las tres obligaciones y no solo la licencia, y suma los tres resultados. Mantén tapadas las opciones hasta que los tres sumen 120,000. Un método que no cierra es lo más rápido de cachar en el papel.
No dejes de asignar para tomar la C, 100,000, porque ese es el número de la licencia en la lista de precios. Lee en voz alta la consecuencia: deja 20,000 para cubrir un soporte que vale 30,000 y una instalación que vale 20,000, es decir, le regala todo el descuento del cliente a dos elementos y nada a la licencia. La B es más fina y vale la pena nombrarla con precisión: el descuento del paquete es 20 por ciento de 150,000, y aplicar ese 20 por ciento al precio del contrato da 120,000 por 0.8, o sea 96,000; la tasa correcta sobre la base equivocada. La D le asigna todo el contrato a una de tres obligaciones. Pregúntate: ¿qué tiene que ir en el denominador de la razón, y por qué no es 120,000? Si contestas que el denominador es la suma de los precios de venta independientes, porque contra eso se miden las proporciones, ya dominas la asignación.
$80,000. El precio de la transacción se asigna en proporción a los precios de venta independientes, así que el descuento agregado de 30,000 dólares se reparte entre las tres obligaciones en lugar de cargarse a una sola: 100,000 ÷ 150,000 × 120,000 = 80,000 dólares. Asignarle los 100,000 completos a la licencia dejaría apenas 20,000 para las otras dos obligaciones y pasaría por encima del requisito de proporcionalidad.
RONDA 3Ubicar el reconocimiento en el tiempo
EN PANTALLA — Tu tablero muestra, en inglés: “The license transfers on January 1 and installation is completed on January 15. Support is provided ratably over the twelve months of the year. Using the allocation above, how much revenue is recognized for the quarter ended March 31?” Es decir: la licencia se transfiere el 1 de enero, la instalación se termina el 15 de enero y el soporte se presta de forma lineal a lo largo de los doce meses del año, y con la asignación anterior se pregunta cuánto ingreso se reconoce en el trimestre que cierra el 31 de marzo. La instrucción de arriba dice “Determine revenue recognized for the quarter.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es $80,000, la B es $96,000, la C es $102,000 y la D es $120,000.
Dos rondas desarmando el contrato, y ahora cada pieza recibe una fecha. Recorre las tres obligaciones una por una y hazle a cada una la misma pregunta: ¿esto se satisface en un momento determinado o a lo largo del tiempo? La licencia se transfiere el 1 de enero: un momento determinado, dentro del trimestre, así que se ganan los 80,000 completos. La instalación se termina el 15 de enero: también un momento determinado, también dentro del trimestre, así que se gana toda su asignación, y esa asignación es 20 sobre 150 de 120,000, o sea 16,000. El soporte es el que se porta distinto: se presta de forma lineal a lo largo de doce meses, así que el control se transfiere de manera continua y el ingreso sigue el paso del tiempo. Su asignación es 30 sobre 150 de 120,000, o sea 24,000, y al 31 de marzo ya pasaron tres de esos doce meses: un cuarto del año, así que un cuarto de 24,000, que son 6,000. Ahora suma solo lo que de verdad se ganó: 80,000 más 16,000 más 6,000.
Dibuja un cuadro de tres renglones —obligación, monto asignado, y en un momento determinado o a lo largo del tiempo— y llena las nueve casillas antes de leer una opción. Después escribe, junto al único renglón que lo necesita, qué fracción del año transcurrió al 31 de marzo. Di con tus propias palabras qué hace que el renglón del soporte se porte distinto de los otros dos, antes de sumar nada.
No dejes que el cobro del 1 de enero decida la respuesta ni tomes la D, 120,000, porque el cliente ya pagó todo. El cobro no es el disparador; el disparador es la transferencia del control, y al 31 de marzo una de las tres obligaciones apenas se transfirió en una cuarta parte. La B, 96,000, es el casi acierto que hay que sondear con más fuerza: son 80,000 más 16,000, es decir, las dos obligaciones de momento determinado bien reconocidas y el soporte olvidado por completo. La A, 80,000, se detiene en la licencia y además deja fuera la instalación, casi siempre porque el 15 de enero se lee como una fecha fuera del trimestre. Pregúntate: ¿cuánto soporte entregó el proveedor al 31 de marzo, y cuánto vale eso? Si contestas tres meses de doce, o sea un cuarto de 24,000, ya dominas el momento del reconocimiento.
$102,000. La licencia y la instalación se satisfacen en momentos determinados dentro del trimestre, así que se reconocen sus montos asignados completos —80,000 más 16,000, porque la instalación es 20,000 ÷ 150,000 × 120,000 = 16,000 dólares—. El soporte se satisface a lo largo del tiempo, así que se reconoce un cuarto de sus 24,000 asignados, o sea 6,000. Total: 80,000 + 16,000 + 6,000 = 102,000 dólares.
RONDA 4Clasificar el monto no reconocido
EN PANTALLA — Tu tablero muestra, en inglés: “The customer paid the full $120,000 on January 1. How should the portion of the support allocation not yet recognized be presented in the March 31 balance sheet?” Es decir: el cliente pagó los 120,000 dólares completos el 1 de enero, y se pregunta cómo debe presentarse en el balance del 31 de marzo la parte de la asignación del soporte que todavía no se reconoce. La instrucción de arriba dice “Select the correct presentation at March 31.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es As a contract asset, la B es As a contract liability, la C es As a reduction of accounts receivable y la D es As an addition to retained earnings.
La ronda anterior dijo qué se ganó. Esta pregunta qué se hace con el resto, y todo se decide por qué llegó primero: el dinero o el desempeño. Repite los dos hechos en voz alta. El cliente pagó los 120,000 completos el 1 de enero, y al 31 de marzo el proveedor todavía le debe nueve meses de soporte. Entonces el dinero llegó antes de que el trabajo estuviera hecho. Dinero retenido contra un trabajo que todavía no se hace es una obligación, y una obligación va del lado derecho del balance. Eso es exactamente un pasivo del contrato: contraprestación recibida, o exigible sin condiciones, antes de haber satisfecho la obligación de desempeño relacionada. El nombre que casi todos aprendemos primero es ingresos diferidos, y es el mismo saldo. Ahora ponle número antes de mirar las opciones. La asignación del soporte era 24,000, se reconocieron 6,000 en el trimestre, así que quedan 18,000 que todavía se deben en servicio. Dilo como frase: hay dieciocho mil dólares del cliente parados ahí, contra un soporte que el proveedor todavía no ha prestado.
Escribe el asiento del 1 de enero antes de leer una opción —efectivo 120,000 de un lado, y qué del otro— y luego el asiento que corre al cierre de cada mes. Di qué saldo va vaciando ese segundo asiento. Arrastra los números de la ronda anterior hasta el saldo final y di la cifra en voz alta, para que la pregunta de presentación se conteste sobre un monto que ya encontraste tú mismo.
No te lances a la A porque “contract” y “asset” suenan las dos al vocabulario que se está evaluando. Un activo del contrato es la imagen invertida de este caso —primero el desempeño, y todavía sin derecho incondicional al cobro—, así que pregúntate quién se adelantó aquí, el proveedor o el cliente, y la A se cae sola. La C supone que hay una cuenta por cobrar contra la cual netear, pero el cliente pagó en efectivo el 1 de enero, así que las cuentas por cobrar están en cero y no hay nada que reducir. La D toma un ingreso que el proveedor no ha ganado y lo manda directo al patrimonio, que es justo el error que el modelo de cinco pasos existe para evitar. Pregúntate: ¿qué sigue debiendo el proveedor al 31 de marzo, y deber algo lo convierte en activo o en pasivo? Si contestas nueve meses de soporte, y por lo tanto pasivo, ya dominas la presentación.
As a contract liability, es decir como un pasivo del contrato. Se recibió la contraprestación pero todavía no se ha satisfecho la obligación de desempeño relacionada, y esa es la definición de pasivo del contrato: el saldo de ingresos diferidos. Un activo del contrato surge en el caso opuesto, cuando el desempeño va delante del derecho incondicional al cobro. El saldo de soporte no reconocido son los 24,000 asignados menos los 6,000 reconocidos, o sea 18,000 dólares.
TOEFL iBT · Reading
The board stays in the language the exam is given in. The script comes to you in yours.El tablero se queda en el idioma en que se presenta el examen. El guion le llega a usted en el suyo.The board stays in the language the exam is given in. The script comes to you in the one you read best.El tablero se queda en el idioma en que se presenta el examen. El guion te llega a ti en el que leas mejor.
This is what an instructor or tutor reads through with them.
Coach Mia reads the Say block aloud — the same words printed below.Coach Mia lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.Mia reads the Say block aloud — the same words printed below.Mia te lee en voz alta el bloque Di: las mismas palabras que están escritas abajo.
ROUND 1Inference
ON SCREEN — Board shows the passage: “Cities are measurably warmer than the countryside around them. Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The resulting difference is largest on clear, still nights and can approach seven degrees Celsius.” The question beneath it reads “It can be inferred from the passage that the temperature difference between a city and its surroundings is smallest when”, and the instruction above it reads “Select the answer the passage supports.” This round has no manipulative. Four answer choices sit below the item — A is wind and cloud cover are substantial, B is the city contains a large amount of masonry, C is the surrounding countryside has little vegetation, D is solar radiation is at its daily maximum.
“Read the passage all the way through before you read a single choice, and then find the one sentence the question is really about. The question asks when the difference is smallest, and the passage never says smallest anywhere — it says largest. That gap is the whole item. Read that sentence again: the resulting difference is largest on clear, still nights. Two conditions, and they are the conditions for the maximum. An inference question wants what follows from the text rather than what the text states, so flip both conditions and say the result out loud. The opposite of clear is cloudy. The opposite of still is windy. So the difference should be smallest on a cloudy, windy night, and that is an inference you can defend by pointing at one sentence. Notice what you did not do. You did not reason about asphalt, or about trees, or about the countryside. The first two sentences explain why cities are warm at all. Only the third sentence says anything about how big the gap gets and under what conditions, so only the third sentence can answer a question about the size of the gap.”
Before they look at the choices, have them underline the one sentence in the passage that mentions the size of the difference, then write its two conditions in the margin as a pair. Next to that pair, have them write the opposite of each word — one word each, not a phrase. Ask them to say the finished inference as a full sentence out loud. That sentence, not the list of choices, is what they should be matching against.
The mistake to watch for is answering from the mechanism instead of from the third sentence, which pulls a learner straight to B. Masonry is a cause of the warming, so more of it makes the difference larger, not smaller — B is the right topic pointing the wrong way. D is the more attractive trap, because “largest at night” feels like it licenses “smallest at noon”; the passage never compares day with night by size, and midday is not the opposite of clear and still. C is the unsupported inference: countryside with little vegetation would plausibly narrow the gap in the real world, but the passage attaches vegetation only to cities and never links the countryside to the size of the difference. Ask them: which two words in the passage name the conditions for the largest difference, and what is the one-word opposite of each? If they answer clear and still, then cloudy and windy, the inference is theirs.
Wind and cloud cover are substantial. The passage states the difference is largest on clear, still nights, so the logical complement is that it is smallest when conditions are the opposite of clear and still — cloudy and windy. The other three options either restate a cause of the warming or introduce a condition the passage never links to the size of the difference.
ROUND 2Rhetorical Purpose
ON SCREEN — Board shows: “In the passage above, why does the author mention ‘the cooling that evaporation from leaves would otherwise provide’?” The instruction above it reads “Select the author's reason for including the detail.” The passage from round one stays on screen. This round has no manipulative. Four answer choices sit below the item — A is To argue that planting trees is the only effective response to urban heat, B is To identify a cooling process that is reduced when vegetation is scarce, C is To question whether asphalt absorbs as much radiation as is claimed, D is To explain why the difference is measured at night rather than at midday.
“This question type is not asking what the phrase means. It is asking what job the phrase is doing, so answer it grammatically before you answer it thematically. Find the phrase in the passage and read the whole clause it belongs to: while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The phrase is attached to the verb removes. Something is being taken away, and this phrase names what. So its job is to name the cooling that cities have lost, which makes it the second cause of the warming, sitting alongside the asphalt and masonry in the first cause. Say the function in four words before you look down: it names what's missing. Now hold that against the passage's tone. Nothing here recommends anything, nothing here doubts anything, and nothing here is arguing with an opponent. It explains a mechanism, sentence by sentence. A purpose that requires the author to be recommending, doubting or arguing cannot be right in a passage that is only explaining.”
Have them find the quoted phrase in the passage and read out the full clause around it, then say which verb the phrase depends on. Have them write the purpose in their own words in under ten words before they read the four choices. Then read the choices aloud and have them label each with one verb — argues, identifies, questions, explains — and check that verb against what the passage is actually doing.
The mistake to watch for is picking A because trees really are a remedy for urban heat, so a choice about planting them sounds like the point of the sentence. The passage makes no recommendation at all, and A also carries the word only, an absolute the text nowhere supports. C fails on stance rather than on topic: the passage asserts that asphalt and masonry absorb radiation, and asserting is the opposite of questioning. D is the closest miss, because the passage does mention night — but the timing lives in the third sentence, not in this clause, so D attaches the phrase to the wrong sentence. Ask them: which verb in the passage does this phrase depend on, and is the author recommending, doubting or explaining? If they answer removes, and explaining, the purpose is theirs.
To identify a cooling process that is reduced when vegetation is scarce. A rhetorical purpose question asks what work a detail does in the argument. The clause is attached to the scarcity of vegetation and names the mechanism that scarcity removes, so it supplies a second cause of the warming. The passage makes no recommendation, casts no doubt on absorption, and does not use the clause to justify when the difference is measured.
ROUND 3Vocabulary in Context
ON SCREEN — Board shows: “In the sentence ‘The resulting difference is largest on clear, still nights,’ the word ‘still’ is closest in meaning to”. The instruction above it reads “Select the closest meaning as used in the passage.” The passage from round one stays on screen. This round has no manipulative. Four answer choices sit below the item — A is silent, B is windless, C is continuing, D is cloudless.
“Do not answer this one from what the word usually means. A vocabulary-in-context item is decided by the sentence around the word, and the piece you need is only three words long: clear, still nights. Ask what kind of word clear is there. It is a weather word — no cloud. So still is sitting in a list beside a weather word, describing the same night, which tells you still is also being used about the atmosphere and not about sound. Now bring the mechanism back. The passage says the surfaces release heat slowly after sunset, and that heat has to stay near the ground for the difference to build up. What carries heat away from the ground? Moving air. So the night that produces the biggest difference is a night with no wind. Say the substitution out loud with the word in place: the difference is largest on clear, windless nights. It fits the weather list and it fits the physics. One more check before you commit — whatever you pick must not repeat clear, because a writer listing two conditions is naming two different things.”
Have them cover the choices with a hand and write their own one-word substitution first, from the sentence alone. Then have them read the sentence aloud four times, once with each choice slotted in, and reject any reading that either changes the topic from weather or repeats a word already in the list. Ask them which condition the mechanism in the second sentence actually needs the air to satisfy.
The mistake to watch for is A, silent, because still night is the collocation almost everyone has heard, and the familiar meaning arrives before the sentence has been read. Ask what silence would have to do with the temperature of a city and the trap becomes visible on its own. C, continuing, is a genuine sense of still — the adverb in still raining — and that is exactly why it is on the list; it cannot work as an adjective describing a night. D is the neat one to catch: cloudless is a real meaning of still's neighbour, not of still, and it duplicates clear, which is already in the sentence. Ask them: what is still paired with here, and what does the mechanism need the air to be doing? If they answer clear, a weather word, and not moving, the item is theirs.
Windless. Vocabulary items are answered from the surrounding sentence, not from the most familiar meaning of the word. Still is paired with clear to describe atmospheric conditions, and the passage's mechanism depends on air that does not mix, so windless fits. Continuing is a common alternative sense of the word but does not describe a night, and cloudless duplicates clear.
ROUND 4Sentence Simplification
ON SCREEN — Board shows: “Which of the following best expresses the essential information in this sentence: ‘Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide.’?” The instruction above it reads “Select the choice that best expresses the essential information.” This round has no manipulative. Four answer choices sit below the item — A is Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling, B is Asphalt absorbs more solar radiation than masonry does, so cities with more paving are warmer at night, C is Vegetation cools cities by evaporation, which is why urban surfaces release heat slowly after sunset, D is Solar radiation is absorbed by cities during the day but has little effect on nighttime temperatures.
“Simplification is an accounting problem, not a style contest. Before you read a choice, count the independent ideas in the original sentence. It has two, joined by the word while. Idea one: dark surfaces take in radiation all day and let it out slowly after sunset. Idea two: because there is little vegetation, the cooling that evaporation from leaves would give is gone. Write those as two short lines. Now the rule. A correct simplification keeps every main idea and adds nothing. Two ideas in means two ideas out, so any choice carrying one idea has dropped something, and any choice carrying a fact the original never stated has added something. Two failure modes, and you can test for both with your two lines. Read each choice against line one, then against line two, then ask whether it says anything that is in neither line. The right answer will be the dull one. It will read like your two lines pushed together, because that is exactly what it is supposed to be.”
Have them write the two ideas as two short numbered lines before they read a single choice — cover the choices with a hand while they do it. Then have them score each choice out loud with three marks: does it keep line one, does it keep line two, does it add anything that is in neither. Ask them to say which failure each rejected choice commits, dropped or invented, in one word.
The mistake to watch for is picking a choice because every word in it appears in the original, which is what makes C so effective: it contains vegetation, evaporation, cooling and release heat slowly, and it still reverses the causation, turning evaporative cooling into the reason surfaces release heat slowly rather than a separate lost mechanism. B invents a comparison the original never draws — the original groups asphalt and masonry together and never ranks them — and adds a claim about how much paving a city has. D contradicts the sentence outright, because the whole point of release it slowly after sunset is an effect on the night. Ask them: how many ideas were in the original, and does this choice have both of them and nothing extra? If they can name the dropped idea or the invented one for each of the three wrong choices, the question type is theirs.
Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling. A correct simplification keeps both main ideas and adds nothing. The original has two clauses — surfaces store and re-release heat, and missing vegetation removes evaporative cooling — and only this choice preserves both. The second invents a comparison between asphalt and masonry, the third reverses the causal relationship, and the fourth contradicts the sentence.
RONDA 1Inferencia
EN PANTALLA — El tablero muestra el pasaje, en inglés, tal como sale en el examen: “Cities are measurably warmer than the countryside around them. Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The resulting difference is largest on clear, still nights and can approach seven degrees Celsius.” Es decir: las ciudades son mediblemente más cálidas que el campo que las rodea; el asfalto oscuro y la mampostería absorben radiación solar durante el día y la sueltan poco a poco después del atardecer, mientras que la escasez de vegetación elimina el enfriamiento que daría la evaporación de las hojas; y la diferencia resultante es mayor en noches despejadas y en calma, y puede acercarse a los siete grados Celsius. La pregunta debajo dice “It can be inferred from the passage that the temperature difference between a city and its surroundings is smallest when”, o sea: se puede inferir que la diferencia de temperatura entre la ciudad y sus alrededores es menor cuando… La instrucción de arriba dice “Select the answer the passage supports.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es wind and cloud cover are substantial, la B es the city contains a large amount of masonry, la C es the surrounding countryside has little vegetation y la D es solar radiation is at its daily maximum.
“Lee el pasaje completo antes de mirar una sola opción, y después busca la única oración de la que trata de verdad la pregunta. La pregunta te pide cuándo la diferencia es menor, y el pasaje nunca dice menor en ninguna parte: dice mayor. Ese hueco es todo el reactivo. Vuelve a leer esa oración: the resulting difference is largest on clear, still nights. Dos condiciones, y son las condiciones del máximo. Una pregunta de inferencia quiere lo que se desprende del texto, no lo que el texto dice con sus palabras, así que invierte las dos condiciones y di el resultado en voz alta. Lo contrario de clear, despejado, es nublado. Lo contrario de still, en calma, es con viento. Entonces la diferencia debería ser menor en una noche nublada y con viento, y esa inferencia la puedes defender señalando una sola oración. Fíjate en lo que no hiciste. No razonaste sobre el asfalto, ni sobre los árboles, ni sobre el campo. Las dos primeras oraciones explican por qué las ciudades son cálidas. Solo la tercera dice algo sobre el tamaño de la diferencia y bajo qué condiciones, así que solo la tercera puede contestar una pregunta sobre ese tamaño.”
Antes de que mire las opciones, pídale que subraye en el pasaje la única oración que habla del tamaño de la diferencia y que anote al margen sus dos condiciones, como pareja. Al lado de esa pareja, que escriba el contrario de cada palabra: una palabra cada uno, no una frase. Pídale que diga la inferencia terminada en voz alta, como oración completa. Contra esa oración es contra lo que debe comparar, no contra la lista de opciones.
El error que hay que vigilar es contestar desde el mecanismo en vez de desde la tercera oración, y eso lleva derecho a la B. La mampostería es una causa del calentamiento, así que más mampostería agranda la diferencia, no la reduce: la B tiene el tema correcto apuntando al revés. La D es la trampa más atractiva, porque “mayor de noche” parece autorizar “menor al mediodía”; el pasaje nunca compara el día con la noche por tamaño, y el mediodía no es lo contrario de despejado y en calma. La C es la inferencia sin respaldo: un campo con poca vegetación sí acortaría la diferencia en el mundo real, pero el pasaje asocia la vegetación únicamente con las ciudades y nunca liga el campo al tamaño de la diferencia. Pregúntele: ¿cuáles son las dos palabras del pasaje que nombran las condiciones de la diferencia mayor, y cuál es el contrario de cada una en una sola palabra? Si contesta clear y still, y luego nublado y con viento, la inferencia ya es suya.
“Wind and cloud cover are substantial”, es decir, cuando hay bastante viento y nubosidad. El pasaje afirma que la diferencia es mayor en noches despejadas y en calma, así que el complemento lógico es que sea menor cuando las condiciones son las contrarias: nublado y con viento. Las otras tres opciones o repiten una causa del calentamiento o meten una condición que el pasaje nunca liga al tamaño de la diferencia.
RONDA 2Propósito retórico
EN PANTALLA — El tablero muestra: “In the passage above, why does the author mention ‘the cooling that evaporation from leaves would otherwise provide’?” Es decir: ¿por qué menciona el autor “el enfriamiento que de otro modo daría la evaporación de las hojas”? La instrucción de arriba dice “Select the author's reason for including the detail.” El pasaje de la ronda uno sigue en pantalla. Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es To argue that planting trees is the only effective response to urban heat, la B es To identify a cooling process that is reduced when vegetation is scarce, la C es To question whether asphalt absorbs as much radiation as is claimed y la D es To explain why the difference is measured at night rather than at midday.
“Esta pregunta no te pide qué significa la frase. Te pide qué función cumple, así que contéstala primero por gramática y después por tema. Busca la frase en el pasaje y lee la cláusula completa a la que pertenece: while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. La frase depende del verbo removes, elimina. Algo se está quitando, y esta frase nombra qué. Entonces su función es nombrar el enfriamiento que las ciudades perdieron, y eso la convierte en la segunda causa del calentamiento, al lado del asfalto y la mampostería de la primera. Di la función en cuatro palabras antes de bajar la vista: nombra lo que falta. Ahora compara eso con el tono del pasaje. Aquí nadie recomienda nada, nadie duda de nada y nadie está discutiendo con un adversario. Explica un mecanismo, oración por oración. Un propósito que exige que el autor esté recomendando, dudando o discutiendo no puede ser correcto en un pasaje que solo explica.”
Pídale que localice la frase citada en el pasaje, que lea en voz alta la cláusula entera que la rodea y que diga de qué verbo depende. Que escriba el propósito con sus propias palabras en menos de diez antes de leer las cuatro opciones. Después lea las opciones en voz alta y pídale que etiquete cada una con un solo verbo — argumenta, identifica, cuestiona, explica — y que confronte ese verbo con lo que el pasaje está haciendo de verdad.
El error que hay que vigilar es elegir la A porque los árboles sí son un remedio contra el calor urbano, y entonces una opción sobre plantarlos suena a la idea central. El pasaje no recomienda nada, y la A además trae la palabra only, únicamente, un absoluto que el texto no respalda en ningún lugar. La C falla por postura, no por tema: el pasaje afirma que el asfalto y la mampostería absorben radiación, y afirmar es lo contrario de cuestionar. La D es la que pasa más cerca, porque el pasaje sí menciona la noche, pero ese dato de tiempo vive en la tercera oración y no en esta cláusula, así que la D cuelga la frase de la oración equivocada. Pregúntele: ¿de qué verbo del pasaje depende esta frase, y el autor está recomendando, dudando o explicando? Si contesta removes, y explicando, el propósito ya es suyo.
“To identify a cooling process that is reduced when vegetation is scarce”, es decir, identificar un proceso de enfriamiento que se reduce cuando la vegetación es escasa. Una pregunta de propósito retórico pregunta qué trabajo hace un detalle dentro del argumento. La cláusula está unida a la escasez de vegetación y nombra el mecanismo que esa escasez elimina, de modo que aporta una segunda causa del calentamiento. El pasaje no recomienda nada, no pone en duda la absorción y no usa la cláusula para justificar cuándo se mide la diferencia.
RONDA 3Vocabulario en contexto
EN PANTALLA — El tablero muestra: “In the sentence ‘The resulting difference is largest on clear, still nights,’ the word ‘still’ is closest in meaning to”. Es decir: en esa oración, la palabra “still” es la más cercana en significado a… La instrucción de arriba dice “Select the closest meaning as used in the passage.” El pasaje de la ronda uno sigue en pantalla. Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es silent, la B es windless, la C es continuing y la D es cloudless.
“Esta no la contestes por lo que la palabra suele significar. Un reactivo de vocabulario en contexto lo decide la oración que rodea la palabra, y el trozo que necesitas tiene solo tres palabras: clear, still nights. Pregúntate qué tipo de palabra es clear ahí. Es una palabra del clima: sin nubes. Entonces still está en una lista, junto a una palabra del clima, describiendo la misma noche, y eso te dice que still también se está usando sobre la atmósfera y no sobre el sonido. Ahora trae de vuelta el mecanismo. El pasaje dice que las superficies sueltan el calor poco a poco después del atardecer, y ese calor tiene que quedarse cerca del suelo para que la diferencia se acumule. ¿Qué se lleva el calor del suelo? El aire en movimiento. Así que la noche que produce la diferencia más grande es una noche sin viento. Di la sustitución en voz alta con la palabra puesta: the difference is largest on clear, windless nights. Encaja con la lista del clima y encaja con la física. Un último control antes de decidir: lo que elijas no puede repetir clear, porque quien enumera dos condiciones está nombrando dos cosas distintas.”
Pídale que tape las opciones con la mano y que escriba primero su propia sustitución de una palabra, solo a partir de la oración. Después que lea la oración en voz alta cuatro veces, una con cada opción metida en el hueco, y que descarte toda lectura que cambie el tema del clima o que repita una palabra que ya está en la lista. Pregúntele qué condición necesita el mecanismo de la segunda oración que cumpla el aire.
El error que hay que vigilar es la A, silent, porque still night es la combinación que casi todo el mundo ha oído, y el significado familiar llega antes de que se haya leído la oración. Pregúntele qué tendría que ver el silencio con la temperatura de una ciudad y la trampa se hace visible sola. La C, continuing, es un sentido real de still — el adverbio de still raining, sigue lloviendo — y está en la lista justamente por eso; no funciona como adjetivo para describir una noche. La D es la fina de cazar: cloudless, sin nubes, es un significado real del vecino de still, no de still, y además duplica clear, que ya está en la oración. Pregúntele: ¿con qué palabra está emparejada still aquí, y qué necesita el mecanismo que esté haciendo el aire? Si contesta clear, que es del clima, y que no se mueva, el reactivo ya es suyo.
“Windless”, es decir, sin viento. Los reactivos de vocabulario se contestan desde la oración que rodea la palabra, no desde su significado más familiar. Still va emparejada con clear para describir condiciones atmosféricas, y el mecanismo del pasaje depende de que el aire no se mezcle, así que sin viento encaja. Continuing es otro sentido común de la palabra pero no describe una noche, y cloudless duplica clear.
RONDA 4Simplificación de oraciones
EN PANTALLA — El tablero muestra: “Which of the following best expresses the essential information in this sentence: ‘Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide.’?” Es decir: ¿cuál de las siguientes expresa mejor la información esencial de esa oración? La instrucción de arriba dice “Select the choice that best expresses the essential information.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling; la B es Asphalt absorbs more solar radiation than masonry does, so cities with more paving are warmer at night; la C es Vegetation cools cities by evaporation, which is why urban surfaces release heat slowly after sunset; y la D es Solar radiation is absorbed by cities during the day but has little effect on nighttime temperatures.
“La simplificación es un problema de contabilidad, no un concurso de estilo. Antes de leer una opción, cuenta las ideas independientes de la oración original. Tiene dos, unidas por la palabra while, mientras que. Idea uno: las superficies oscuras absorben radiación todo el día y la sueltan poco a poco después del atardecer. Idea dos: como hay poca vegetación, el enfriamiento que daría la evaporación de las hojas desaparece. Escríbelas como dos renglones cortos. Ahora la regla. Una simplificación correcta conserva todas las ideas principales y no agrega ninguna. Dos ideas que entran son dos ideas que salen, así que cualquier opción que traiga una sola idea perdió algo, y cualquier opción que traiga un dato que la original nunca dijo agregó algo. Dos formas de fallar, y con tus dos renglones puedes probar las dos. Lee cada opción contra el renglón uno, luego contra el renglón dos, y después pregúntate si dice algo que no está en ninguno de los dos. La respuesta correcta va a ser la aburrida. Va a sonar como tus dos renglones pegados, porque es exactamente lo que debe ser.”
Pídale que escriba las dos ideas como dos renglones numerados antes de leer una sola opción, y que tape las opciones con la mano mientras lo hace. Después que califique cada opción en voz alta con tres marcas: ¿conserva el renglón uno, conserva el renglón dos, agrega algo que no está en ninguno? Pídale que diga con una palabra qué falla comete cada opción descartada, perdió o inventó.
El error que hay que vigilar es elegir una opción porque todas sus palabras aparecen en la original, y eso es lo que hace tan eficaz a la C: trae vegetation, evaporation, cooling y release heat slowly, y de todos modos invierte la causalidad, porque convierte el enfriamiento por evaporación en la razón por la que las superficies sueltan el calor despacio, en vez de un mecanismo aparte que se perdió. La B inventa una comparación que la original nunca hace — la original agrupa asfalto y mampostería y no las ordena — y agrega una afirmación sobre cuánto pavimento tiene la ciudad. La D contradice la oración de frente, porque todo el sentido de release it slowly after sunset es un efecto sobre la noche. Pregúntele: ¿cuántas ideas había en la original, y esta opción tiene las dos y nada más? Si logra nombrar la idea perdida o la inventada en cada una de las tres opciones incorrectas, ya domina el tipo de pregunta.
“Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling”, es decir, las superficies urbanas almacenan el calor del día y lo sueltan de noche, y la falta de plantas elimina una fuente de enfriamiento. Una simplificación correcta conserva las dos ideas principales y no agrega ninguna. La original tiene dos cláusulas — las superficies almacenan y vuelven a soltar calor, y la vegetación ausente elimina el enfriamiento por evaporación — y solo esta opción conserva ambas. La segunda inventa una comparación entre asfalto y mampostería, la tercera invierte la relación causal y la cuarta contradice la oración.
ROUND 1Inference
ON SCREEN — Your board says, showing the passage: “Cities are measurably warmer than the countryside around them. Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The resulting difference is largest on clear, still nights and can approach seven degrees Celsius.” The question beneath it reads “It can be inferred from the passage that the temperature difference between a city and its surroundings is smallest when”, and the instruction above it reads “Select the answer the passage supports.” This round has no manipulative. Four answer choices sit below the item — A is wind and cloud cover are substantial, B is the city contains a large amount of masonry, C is the surrounding countryside has little vegetation, D is solar radiation is at its daily maximum.
Read the passage all the way through before you read a single choice, and then find the one sentence the question is really about. The question asks when the difference is smallest, and the passage never says smallest anywhere — it says largest. That gap is the whole item. Read that sentence again: the resulting difference is largest on clear, still nights. Two conditions, and they are the conditions for the maximum. An inference question wants what follows from the text rather than what the text states, so flip both conditions and say the result out loud. The opposite of clear is cloudy. The opposite of still is windy. So the difference should be smallest on a cloudy, windy night, and that is an inference you can defend by pointing at one sentence. Notice what you did not do. You did not reason about asphalt, or about trees, or about the countryside. The first two sentences explain why cities are warm at all. Only the third sentence says anything about how big the gap gets and under what conditions, so only the third sentence can answer a question about the size of the gap.
Before you look at the choices, underline the one sentence in the passage that mentions the size of the difference, then write its two conditions in the margin as a pair. Next to that pair, write the opposite of each word — one word each, not a phrase. Say the finished inference as a full sentence out loud. That sentence, not the list of choices, is what you should be matching against.
Don't answer from the mechanism instead of from the third sentence, which pulls you straight to B. Masonry is a cause of the warming, so more of it makes the difference larger, not smaller — B is the right topic pointing the wrong way. D is the more attractive trap, because “largest at night” feels like it licenses “smallest at noon”; the passage never compares day with night by size, and midday is not the opposite of clear and still. C is the unsupported inference: countryside with little vegetation would plausibly narrow the gap in the real world, but the passage attaches vegetation only to cities and never links the countryside to the size of the difference. Ask yourself: which two words in the passage name the conditions for the largest difference, and what is the one-word opposite of each? If you answer clear and still, then cloudy and windy, the inference is yours.
Wind and cloud cover are substantial. The passage states the difference is largest on clear, still nights, so the logical complement is that it is smallest when conditions are the opposite of clear and still — cloudy and windy. The other three options either restate a cause of the warming or introduce a condition the passage never links to the size of the difference.
ROUND 2Rhetorical Purpose
ON SCREEN — Your board says: “In the passage above, why does the author mention ‘the cooling that evaporation from leaves would otherwise provide’?” The instruction above it reads “Select the author's reason for including the detail.” The passage from round one stays on screen. This round has no manipulative. Four answer choices sit below the item — A is To argue that planting trees is the only effective response to urban heat, B is To identify a cooling process that is reduced when vegetation is scarce, C is To question whether asphalt absorbs as much radiation as is claimed, D is To explain why the difference is measured at night rather than at midday.
This question type is not asking what the phrase means. It is asking what job the phrase is doing, so answer it grammatically before you answer it thematically. Find the phrase in the passage and read the whole clause it belongs to: while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The phrase is attached to the verb removes. Something is being taken away, and this phrase names what. So its job is to name the cooling that cities have lost, which makes it the second cause of the warming, sitting alongside the asphalt and masonry in the first cause. Say the function in four words before you look down: it names what's missing. Now hold that against the passage's tone. Nothing here recommends anything, nothing here doubts anything, and nothing here is arguing with an opponent. It explains a mechanism, sentence by sentence. A purpose that requires the author to be recommending, doubting or arguing cannot be right in a passage that is only explaining.
Find the quoted phrase in the passage and read out the full clause around it, then say which verb the phrase depends on. Write the purpose in your own words in under ten words before you read the four choices. Then read the choices aloud and label each with one verb — argues, identifies, questions, explains — and check that verb against what the passage is actually doing.
Don't pick A because trees really are a remedy for urban heat, so a choice about planting them sounds like the point of the sentence. The passage makes no recommendation at all, and A also carries the word only, an absolute the text nowhere supports. C fails on stance rather than on topic: the passage asserts that asphalt and masonry absorb radiation, and asserting is the opposite of questioning. D is the closest miss, because the passage does mention night — but the timing lives in the third sentence, not in this clause, so D attaches the phrase to the wrong sentence. Ask yourself: which verb in the passage does this phrase depend on, and is the author recommending, doubting or explaining? If you answer removes, and explaining, the purpose is yours.
To identify a cooling process that is reduced when vegetation is scarce. A rhetorical purpose question asks what work a detail does in the argument. The clause is attached to the scarcity of vegetation and names the mechanism that scarcity removes, so it supplies a second cause of the warming. The passage makes no recommendation, casts no doubt on absorption, and does not use the clause to justify when the difference is measured.
ROUND 3Vocabulary in Context
ON SCREEN — Your board says: “In the sentence ‘The resulting difference is largest on clear, still nights,’ the word ‘still’ is closest in meaning to”. The instruction above it reads “Select the closest meaning as used in the passage.” The passage from round one stays on screen. This round has no manipulative. Four answer choices sit below the item — A is silent, B is windless, C is continuing, D is cloudless.
Do not answer this one from what the word usually means. A vocabulary-in-context item is decided by the sentence around the word, and the piece you need is only three words long: clear, still nights. Ask what kind of word clear is there. It is a weather word — no cloud. So still is sitting in a list beside a weather word, describing the same night, which tells you still is also being used about the atmosphere and not about sound. Now bring the mechanism back. The passage says the surfaces release heat slowly after sunset, and that heat has to stay near the ground for the difference to build up. What carries heat away from the ground? Moving air. So the night that produces the biggest difference is a night with no wind. Say the substitution out loud with the word in place: the difference is largest on clear, windless nights. It fits the weather list and it fits the physics. One more check before you commit — whatever you pick must not repeat clear, because a writer listing two conditions is naming two different things.
Cover the choices with a hand and write your own one-word substitution first, from the sentence alone. Then read the sentence aloud four times, once with each choice slotted in, and reject any reading that either changes the topic from weather or repeats a word already in the list. Say which condition the mechanism in the second sentence actually needs the air to satisfy.
Don't take A, silent, because still night is the collocation almost everyone has heard, and the familiar meaning arrives before the sentence has been read. Ask what silence would have to do with the temperature of a city and the trap becomes visible on its own. C, continuing, is a genuine sense of still — the adverb in still raining — and that is exactly why it is on the list; it cannot work as an adjective describing a night. D is the neat one to catch: cloudless is a real meaning of still's neighbour, not of still, and it duplicates clear, which is already in the sentence. Ask yourself: what is still paired with here, and what does the mechanism need the air to be doing? If you answer clear, a weather word, and not moving, the item is yours.
Windless. Vocabulary items are answered from the surrounding sentence, not from the most familiar meaning of the word. Still is paired with clear to describe atmospheric conditions, and the passage's mechanism depends on air that does not mix, so windless fits. Continuing is a common alternative sense of the word but does not describe a night, and cloudless duplicates clear.
ROUND 4Sentence Simplification
ON SCREEN — Your board says: “Which of the following best expresses the essential information in this sentence: ‘Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide.’?” The instruction above it reads “Select the choice that best expresses the essential information.” This round has no manipulative. Four answer choices sit below the item — A is Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling, B is Asphalt absorbs more solar radiation than masonry does, so cities with more paving are warmer at night, C is Vegetation cools cities by evaporation, which is why urban surfaces release heat slowly after sunset, D is Solar radiation is absorbed by cities during the day but has little effect on nighttime temperatures.
Simplification is an accounting problem, not a style contest. Before you read a choice, count the independent ideas in the original sentence. It has two, joined by the word while. Idea one: dark surfaces take in radiation all day and let it out slowly after sunset. Idea two: because there is little vegetation, the cooling that evaporation from leaves would give is gone. Write those as two short lines. Now the rule. A correct simplification keeps every main idea and adds nothing. Two ideas in means two ideas out, so any choice carrying one idea has dropped something, and any choice carrying a fact the original never stated has added something. Two failure modes, and you can test for both with your two lines. Read each choice against line one, then against line two, then ask whether it says anything that is in neither line. The right answer will be the dull one. It will read like your two lines pushed together, because that is exactly what it is supposed to be.
Write the two ideas as two short numbered lines before you read a single choice — cover the choices with a hand while you do it. Then score each choice out loud with three marks: does it keep line one, does it keep line two, does it add anything that is in neither. Say which failure each rejected choice commits, dropped or invented, in one word.
Don't pick a choice because every word in it appears in the original, which is what makes C so effective: it contains vegetation, evaporation, cooling and release heat slowly, and it still reverses the causation, turning evaporative cooling into the reason surfaces release heat slowly rather than a separate lost mechanism. B invents a comparison the original never draws — the original groups asphalt and masonry together and never ranks them — and adds a claim about how much paving a city has. D contradicts the sentence outright, because the whole point of release it slowly after sunset is an effect on the night. Ask yourself: how many ideas were in the original, and does this choice have both of them and nothing extra? If you can name the dropped idea or the invented one for each of the three wrong choices, the question type is yours.
Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling. A correct simplification keeps both main ideas and adds nothing. The original has two clauses — surfaces store and re-release heat, and missing vegetation removes evaporative cooling — and only this choice preserves both. The second invents a comparison between asphalt and masonry, the third reverses the causal relationship, and the fourth contradicts the sentence.
RONDA 1Inferencia
EN PANTALLA — Tu tablero muestra el pasaje, en inglés, tal como sale en el examen: “Cities are measurably warmer than the countryside around them. Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. The resulting difference is largest on clear, still nights and can approach seven degrees Celsius.” Es decir: las ciudades son mediblemente más cálidas que el campo que las rodea; el asfalto oscuro y la mampostería absorben radiación solar durante el día y la sueltan poco a poco después del atardecer, mientras que la escasez de vegetación elimina el enfriamiento que daría la evaporación de las hojas; y la diferencia resultante es mayor en noches despejadas y en calma, y puede acercarse a los siete grados Celsius. La pregunta debajo dice “It can be inferred from the passage that the temperature difference between a city and its surroundings is smallest when”, o sea: se puede inferir que la diferencia de temperatura entre la ciudad y sus alrededores es menor cuando… La instrucción de arriba dice “Select the answer the passage supports.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es wind and cloud cover are substantial, la B es the city contains a large amount of masonry, la C es the surrounding countryside has little vegetation y la D es solar radiation is at its daily maximum.
Lee el pasaje completo antes de mirar una sola opción, y después busca la única oración de la que trata de verdad la pregunta. La pregunta te pide cuándo la diferencia es menor, y el pasaje nunca dice menor en ninguna parte: dice mayor. Ese hueco es todo el reactivo. Vuelve a leer esa oración: the resulting difference is largest on clear, still nights. Dos condiciones, y son las condiciones del máximo. Una pregunta de inferencia quiere lo que se desprende del texto, no lo que el texto dice con sus palabras, así que invierte las dos condiciones y di el resultado en voz alta. Lo contrario de clear, despejado, es nublado. Lo contrario de still, en calma, es con viento. Entonces la diferencia debería ser menor en una noche nublada y con viento, y esa inferencia la puedes defender señalando una sola oración. Fíjate en lo que no hiciste. No razonaste sobre el asfalto, ni sobre los árboles, ni sobre el campo. Las dos primeras oraciones explican por qué las ciudades son cálidas. Solo la tercera dice algo sobre el tamaño de la diferencia y bajo qué condiciones, así que solo la tercera puede contestar una pregunta sobre ese tamaño.
Antes de mirar las opciones, subraya en el pasaje la única oración que habla del tamaño de la diferencia y anota al margen sus dos condiciones, como pareja. Al lado de esa pareja, escribe el contrario de cada palabra: una palabra cada uno, no una frase. Di la inferencia terminada en voz alta, como oración completa. Contra esa oración es contra lo que debes comparar, no contra la lista de opciones.
No contestes desde el mecanismo en vez de desde la tercera oración, porque eso lleva derecho a la B. La mampostería es una causa del calentamiento, así que más mampostería agranda la diferencia, no la reduce: la B tiene el tema correcto apuntando al revés. La D es la trampa más atractiva, porque “mayor de noche” parece autorizar “menor al mediodía”; el pasaje nunca compara el día con la noche por tamaño, y el mediodía no es lo contrario de despejado y en calma. La C es la inferencia sin respaldo: un campo con poca vegetación sí acortaría la diferencia en el mundo real, pero el pasaje asocia la vegetación únicamente con las ciudades y nunca liga el campo al tamaño de la diferencia. Pregúntate: ¿cuáles son las dos palabras del pasaje que nombran las condiciones de la diferencia mayor, y cuál es el contrario de cada una en una sola palabra? Si contestas clear y still, y luego nublado y con viento, la inferencia ya es tuya.
“Wind and cloud cover are substantial”, es decir, cuando hay bastante viento y nubosidad. El pasaje afirma que la diferencia es mayor en noches despejadas y en calma, así que el complemento lógico es que sea menor cuando las condiciones son las contrarias: nublado y con viento. Las otras tres opciones o repiten una causa del calentamiento o meten una condición que el pasaje nunca liga al tamaño de la diferencia.
RONDA 2Propósito retórico
EN PANTALLA — Tu tablero muestra: “In the passage above, why does the author mention ‘the cooling that evaporation from leaves would otherwise provide’?” Es decir: ¿por qué menciona el autor “el enfriamiento que de otro modo daría la evaporación de las hojas”? La instrucción de arriba dice “Select the author's reason for including the detail.” El pasaje de la ronda uno sigue en pantalla. Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es To argue that planting trees is the only effective response to urban heat, la B es To identify a cooling process that is reduced when vegetation is scarce, la C es To question whether asphalt absorbs as much radiation as is claimed y la D es To explain why the difference is measured at night rather than at midday.
Esta pregunta no te pide qué significa la frase. Te pide qué función cumple, así que contéstala primero por gramática y después por tema. Busca la frase en el pasaje y lee la cláusula completa a la que pertenece: while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide. La frase depende del verbo removes, elimina. Algo se está quitando, y esta frase nombra qué. Entonces su función es nombrar el enfriamiento que las ciudades perdieron, y eso la convierte en la segunda causa del calentamiento, al lado del asfalto y la mampostería de la primera. Di la función en cuatro palabras antes de bajar la vista: nombra lo que falta. Ahora compara eso con el tono del pasaje. Aquí nadie recomienda nada, nadie duda de nada y nadie está discutiendo con un adversario. Explica un mecanismo, oración por oración. Un propósito que exige que el autor esté recomendando, dudando o discutiendo no puede ser correcto en un pasaje que solo explica.
Localiza la frase citada en el pasaje, lee en voz alta la cláusula entera que la rodea y di de qué verbo depende. Escribe el propósito con tus propias palabras en menos de diez antes de leer las cuatro opciones. Después lee las opciones en voz alta y etiqueta cada una con un solo verbo — argumenta, identifica, cuestiona, explica — y confronta ese verbo con lo que el pasaje está haciendo de verdad.
No elijas la A porque los árboles sí son un remedio contra el calor urbano y entonces una opción sobre plantarlos suena a la idea central. El pasaje no recomienda nada, y la A además trae la palabra only, únicamente, un absoluto que el texto no respalda en ningún lugar. La C falla por postura, no por tema: el pasaje afirma que el asfalto y la mampostería absorben radiación, y afirmar es lo contrario de cuestionar. La D es la que pasa más cerca, porque el pasaje sí menciona la noche, pero ese dato de tiempo vive en la tercera oración y no en esta cláusula, así que la D cuelga la frase de la oración equivocada. Pregúntate: ¿de qué verbo del pasaje depende esta frase, y el autor está recomendando, dudando o explicando? Si contestas removes, y explicando, el propósito ya es tuyo.
“To identify a cooling process that is reduced when vegetation is scarce”, es decir, identificar un proceso de enfriamiento que se reduce cuando la vegetación es escasa. Una pregunta de propósito retórico pregunta qué trabajo hace un detalle dentro del argumento. La cláusula está unida a la escasez de vegetación y nombra el mecanismo que esa escasez elimina, de modo que aporta una segunda causa del calentamiento. El pasaje no recomienda nada, no pone en duda la absorción y no usa la cláusula para justificar cuándo se mide la diferencia.
RONDA 3Vocabulario en contexto
EN PANTALLA — Tu tablero muestra: “In the sentence ‘The resulting difference is largest on clear, still nights,’ the word ‘still’ is closest in meaning to”. Es decir: en esa oración, la palabra “still” es la más cercana en significado a… La instrucción de arriba dice “Select the closest meaning as used in the passage.” El pasaje de la ronda uno sigue en pantalla. Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es silent, la B es windless, la C es continuing y la D es cloudless.
Esta no la contestes por lo que la palabra suele significar. Un reactivo de vocabulario en contexto lo decide la oración que rodea la palabra, y el trozo que necesitas tiene solo tres palabras: clear, still nights. Pregúntate qué tipo de palabra es clear ahí. Es una palabra del clima: sin nubes. Entonces still está en una lista, junto a una palabra del clima, describiendo la misma noche, y eso te dice que still también se está usando sobre la atmósfera y no sobre el sonido. Ahora trae de vuelta el mecanismo. El pasaje dice que las superficies sueltan el calor poco a poco después del atardecer, y ese calor tiene que quedarse cerca del suelo para que la diferencia se acumule. ¿Qué se lleva el calor del suelo? El aire en movimiento. Así que la noche que produce la diferencia más grande es una noche sin viento. Di la sustitución en voz alta con la palabra puesta: the difference is largest on clear, windless nights. Encaja con la lista del clima y encaja con la física. Un último control antes de decidir: lo que elijas no puede repetir clear, porque quien enumera dos condiciones está nombrando dos cosas distintas.
Tapa las opciones con la mano y escribe primero tu propia sustitución de una palabra, solo a partir de la oración. Después lee la oración en voz alta cuatro veces, una con cada opción metida en el hueco, y descarta toda lectura que cambie el tema del clima o que repita una palabra que ya está en la lista. Pregúntate qué condición necesita el mecanismo de la segunda oración que cumpla el aire.
No tomes la A, silent, porque still night es la combinación que casi todo el mundo ha oído y el significado familiar llega antes de que se haya leído la oración. Pregúntate qué tendría que ver el silencio con la temperatura de una ciudad y la trampa se hace visible sola. La C, continuing, es un sentido real de still — el adverbio de still raining, sigue lloviendo — y está en la lista justamente por eso; no funciona como adjetivo para describir una noche. La D es la fina de cazar: cloudless, sin nubes, es un significado real del vecino de still, no de still, y además duplica clear, que ya está en la oración. Pregúntate: ¿con qué palabra está emparejada still aquí, y qué necesita el mecanismo que esté haciendo el aire? Si contestas clear, que es del clima, y que no se mueva, el reactivo ya es tuyo.
“Windless”, es decir, sin viento. Los reactivos de vocabulario se contestan desde la oración que rodea la palabra, no desde su significado más familiar. Still va emparejada con clear para describir condiciones atmosféricas, y el mecanismo del pasaje depende de que el aire no se mezcle, así que sin viento encaja. Continuing es otro sentido común de la palabra pero no describe una noche, y cloudless duplica clear.
RONDA 4Simplificación de oraciones
EN PANTALLA — Tu tablero muestra: “Which of the following best expresses the essential information in this sentence: ‘Dark asphalt and masonry absorb solar radiation through the day and release it slowly after sunset, while the scarcity of vegetation removes the cooling that evaporation from leaves would otherwise provide.’?” Es decir: ¿cuál de las siguientes expresa mejor la información esencial de esa oración? La instrucción de arriba dice “Select the choice that best expresses the essential information.” Esta ronda no lleva material para tocar. Debajo hay cuatro opciones: la A es Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling; la B es Asphalt absorbs more solar radiation than masonry does, so cities with more paving are warmer at night; la C es Vegetation cools cities by evaporation, which is why urban surfaces release heat slowly after sunset; y la D es Solar radiation is absorbed by cities during the day but has little effect on nighttime temperatures.
La simplificación es un problema de contabilidad, no un concurso de estilo. Antes de leer una opción, cuenta las ideas independientes de la oración original. Tiene dos, unidas por la palabra while, mientras que. Idea uno: las superficies oscuras absorben radiación todo el día y la sueltan poco a poco después del atardecer. Idea dos: como hay poca vegetación, el enfriamiento que daría la evaporación de las hojas desaparece. Escríbelas como dos renglones cortos. Ahora la regla. Una simplificación correcta conserva todas las ideas principales y no agrega ninguna. Dos ideas que entran son dos ideas que salen, así que cualquier opción que traiga una sola idea perdió algo, y cualquier opción que traiga un dato que la original nunca dijo agregó algo. Dos formas de fallar, y con tus dos renglones puedes probar las dos. Lee cada opción contra el renglón uno, luego contra el renglón dos, y después pregúntate si dice algo que no está en ninguno de los dos. La respuesta correcta va a ser la aburrida. Va a sonar como tus dos renglones pegados, porque es exactamente lo que debe ser.
Escribe las dos ideas como dos renglones numerados antes de leer una sola opción, y tapa las opciones con la mano mientras lo haces. Después califica cada opción en voz alta con tres marcas: ¿conserva el renglón uno, conserva el renglón dos, agrega algo que no está en ninguno? Di con una palabra qué falla comete cada opción descartada, perdió o inventó.
No elijas una opción porque todas sus palabras aparecen en la original, y eso es lo que hace tan eficaz a la C: trae vegetation, evaporation, cooling y release heat slowly, y de todos modos invierte la causalidad, porque convierte el enfriamiento por evaporación en la razón por la que las superficies sueltan el calor despacio, en vez de un mecanismo aparte que se perdió. La B inventa una comparación que la original nunca hace — la original agrupa asfalto y mampostería y no las ordena — y agrega una afirmación sobre cuánto pavimento tiene la ciudad. La D contradice la oración de frente, porque todo el sentido de release it slowly after sunset es un efecto sobre la noche. Pregúntate: ¿cuántas ideas había en la original, y esta opción tiene las dos y nada más? Si logras nombrar la idea perdida o la inventada en cada una de las tres opciones incorrectas, ya dominas el tipo de pregunta.
“Urban surfaces store daytime heat and give it off at night, and the lack of plants removes a source of cooling”, es decir, las superficies urbanas almacenan el calor del día y lo sueltan de noche, y la falta de plantas elimina una fuente de enfriamiento. Una simplificación correcta conserva las dos ideas principales y no agrega ninguna. La original tiene dos cláusulas — las superficies almacenan y vuelven a soltar calor, y la vegetación ausente elimina el enfriamiento por evaporación — y solo esta opción conserva ambas. La segunda inventa una comparación entre asfalto y mampostería, la tercera invierte la relación causal y la cuarta contradice la oración.
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