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# Jack, Karl, Marc, and Kate are friends. They collected just

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Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  15 May 2019, 06:40
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Jack, Karl, Marc, and Kate are friends. They collected just enough money to buy a car. Jack contributed 1/3 of what his three friends contributed together. Karl contributed 1/4 of what his three friends contributed together. Marc contributed 2/5 of what his three friends contributed together.

 Quantity A Quantity B The amount paid by Jack The amount paid by Marc

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  15 May 2019, 10:25
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Quote:
Jack, Karl, Marc, and Kate are friends. They collected just enough money to buy a car. Jack contributed 1/3 of what his three friends contributed together. Karl contributed 1/4 of what his three friends contributed together. Marc contributed 2/5 of what his three friends contributed together.

 Quantity A Quantity B The amount paid by Jack The amount paid by Marc

To most efficiently evaluate Quantitative Comparisons first attempt to eliminate values shared by both quantities to precisely determine the relationship between the quantities.

In this scenario, no matter what total amount of money the friends received, it will be the same impact for both Quantity A and Quantity B, so conceptually, we can eliminate the total from consideration in both.

Then, we are just comparing 2/5 of three friends vs 1/3 of three friends. Since the values must be non-negative to add to a real total, we can confidently compare 1/3 vs 2/5 alone without worrying about negative values affecting any products.

2/5 > 1/3, so the correct answer is choice B - Quantity B is always greater.
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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  16 May 2019, 03:42
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Hi,

The right answer here is (D), since no solution can be determined.

Please be cautious while answering this question, so as to not miss that there are 4 different friends who paid different values, and not 3 friends who paid the same amount. It may be simple to assume that Jack, paying just 1/3rd of a given amount, is paying less than Marc, paying 2/5th of the amount. Therefore, one may be tempted to say 2/5 (40%) is greater than 1/3 (34%). But this is not the case.

This method would only be applicable if the value of the amounts paid by each of the friends was known, or was equal in a 1:1:1:1 ratio (which would make no sense). The real equations that come out are:

Jack = 1/3 (Karl+Kate+Marc)
Marc = 2/5 (Jack+Kate+Karl)
Karl = 1/4 (Kate+Marc+Jack)

None of the values are known, and therefore it is impossible to answer this question based on the information given. Hence, the answer is (D).
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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  17 May 2019, 07:47
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Expert's post
Since there seem to be some competing thoughts on this problem, let's take a look at an algebraic solution

Quote:
Jack, Karl, Marc, and Kate are friends. They collected just enough money to buy a car. Jack contributed 1/3 of what his three friends contributed together. Karl contributed 1/4 of what his three friends contributed together. Marc contributed 2/5 of what his three friends contributed together.

 Quantity A Quantity B The amount paid by Jack The amount paid by Marc

Because the values in the quantities are all related to the same shared total, we can set that total = x. Therefore, using the following variables (Jack = j | Kate = k | Karl = l | Marc = m) we know that x = j + k + l + m.

Now, for Quantity A we know algebraically that j = 1/3 (x - j) because x - j is the total without Jack. Multiply the equation by 3 to eliminate the fraction to determine that 3j = x - j . Then, add j to each side to find x = 4j as our new value relating to Quantity A.

Now, for Quantity B we know that algebraically m = 2/5 (x - m) because x - m is the total without Marc. Multiply the equation by 5 to eliminate the fraction to determine that 5m = 2x - 2m . Then, add 2m to each side and divide the full equation by 2 to find 7m/2 = x as our new value relating to Quantity B.

Finally, we can relate the values for m and j directly to each other as 7m/2 = x = 4j. Simplified, 7m/2 = 4j. Multiply each side of the equation by 2 to find that 7m = 8j. Divide the equation by 7 to find that no matter what values are used, m = 8j/7 and is therefore always the greater value. So, select choice B.
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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  17 May 2019, 09:49
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StefanMaisnier wrote:
Since there seem to be some competing thoughts on this problem, let's take a look at an algebraic solution

Quote:
Jack, Karl, Marc, and Kate are friends. They collected just enough money to buy a car. Jack contributed 1/3 of what his three friends contributed together. Karl contributed 1/4 of what his three friends contributed together. Marc contributed 2/5 of what his three friends contributed together.

 Quantity A Quantity B The amount paid by Jack The amount paid by Marc

Because the values in the quantities are all related to the same shared total, we can set that total = x. Therefore, using the following variables (Jack = j | Kate = k | Karl = l | Marc = m) we know that x = j + k + l + m.

Now, for Quantity A we know algebraically that j = 1/3 (x - j) because x - j is the total without Jack. Multiply the equation by 3 to eliminate the fraction to determine that 3j = x - j . Then, add j to each side to find x = 4j as our new value relating to Quantity A.

Now, for Quantity B we know that algebraically m = 2/5 (x - m) because x - m is the total without Marc. Multiply the equation by 5 to eliminate the fraction to determine that 5m = 2x - 2m . Then, add 2m to each side and divide the full equation by 2 to find 7m/2 = x as our new value relating to Quantity B.

Finally, we can relate the values for m and j directly to each other as 7m/2 = x = 4j. Simplified, 7m/2 = 4j. Multiply each side of the equation by 2 to find that 7m = 8j. Divide the equation by 7 to find that no matter what values are used, m = 8j/7 and is therefore always the greater value. So, select choice B.

Thanks! I stand corrected. I jumped to a conclusion too quickly :D
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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  17 May 2019, 09:57
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Expert's post
grewhiz wrote:
Thanks! I stand corrected. I jumped to a conclusion too quickly :D

It's how those tricky GRE test makers work! Happens to all of us, but I'm glad that I had a reason to do both explanations.

Remember that most frequently there is more than one method to solving any particular GRE problem - especially Quantitative Comparisons.
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Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink]  28 May 2019, 03:53
StefanMaisnier wrote:
Quote:
Jack, Karl, Marc, and Kate are friends. They collected just enough money to buy a car. Jack contributed 1/3 of what his three friends contributed together. Karl contributed 1/4 of what his three friends contributed together. Marc contributed 2/5 of what his three friends contributed together.

 Quantity A Quantity B The amount paid by Jack The amount paid by Marc

To most efficiently evaluate Quantitative Comparisons first attempt to eliminate values shared by both quantities to precisely determine the relationship between the quantities.

In this scenario, no matter what total amount of money the friends received, it will be the same impact for both Quantity A and Quantity B, so conceptually, we can eliminate the total from consideration in both.

Then, we are just comparing 2/5 of three friends vs 1/3 of three friends. Since the values must be non-negative to add to a real total, we can confidently compare 1/3 vs 2/5 alone without worrying about negative values affecting any products.

2/5 > 1/3, so the correct answer is choice B - Quantity B is always greater.

Why will it have same impact on both quantity A & B? How did you deduce that?
Re: Jack, Karl, Marc, and Kate are friends. They collected just   [#permalink] 28 May 2019, 03:53
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