It is currently 22 May 2019, 00:47
My Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Founder
Founder
User avatar
Joined: 18 Apr 2015
Posts: 6584
Followers: 107

Kudos [?]: 1258 [0], given: 5973

CAT Tests
Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 15 May 2019, 06:40
Expert's post
00:00

Question Stats:

100% (00:33) correct 0% (00:00) wrong based on 4 sessions
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

_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Manager
Manager
User avatar
Affiliations: Partner at MyGuru LLC.
Joined: 13 May 2019
Posts: 54
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE: Education (Education)
Followers: 1

Kudos [?]: 33 [1] , given: 0

CAT Tests
Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 15 May 2019, 10:25
1
This post received
KUDOS
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.
1 KUDOS received
Intern
Intern
Joined: 14 May 2019
Posts: 8
Followers: 0

Kudos [?]: 7 [1] , given: 0

Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 16 May 2019, 03:42
1
This post received
KUDOS
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).
1 KUDOS received
Manager
Manager
User avatar
Affiliations: Partner at MyGuru LLC.
Joined: 13 May 2019
Posts: 54
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE: Education (Education)
Followers: 1

Kudos [?]: 33 [1] , given: 0

CAT Tests
Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 17 May 2019, 07:47
1
This post received
KUDOS
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.
1 KUDOS received
Intern
Intern
Joined: 14 May 2019
Posts: 8
Followers: 0

Kudos [?]: 7 [1] , given: 0

Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 17 May 2019, 09:49
1
This post received
KUDOS
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
1 KUDOS received
Manager
Manager
User avatar
Affiliations: Partner at MyGuru LLC.
Joined: 13 May 2019
Posts: 54
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE: Education (Education)
Followers: 1

Kudos [?]: 33 [1] , given: 0

CAT Tests
Re: Jack, Karl, Marc, and Kate are friends. They collected just [#permalink] New post 17 May 2019, 09:57
1
This post received
KUDOS
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.
Re: Jack, Karl, Marc, and Kate are friends. They collected just   [#permalink] 17 May 2019, 09:57
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.