problem solving ques

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Tarun Khanna: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Wed Sep 14, 2011 1:48 amFollowed by: 1 member; Thanked: 1 times; GMAT Score:630

by Tarun Khanna » Fri Sep 23, 2011 9:32 am

brent had posted this long ago and the ques is

A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money did Bob receive?
A) 20
B) 22
C) 24
D) 26
E) 52

Answer: Option A i.e.20
my explanation is as below but my answer is wrong.

A gets= 4+(x-4)/2 which is (x+4)/2
B gets= 4+1/3[x-(x+4)/2] which is (x+20)/6
C gets 32 which is x-[(x+4)/2 + (x+20)/6] solving which x=112 which means B gets 22 but answer is 20...can someone help explain?

Thanks

Source: Beat The GMAT — Problem Solving | Fri Sep 23, 2011 9:32 am

cavaliar: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Wed Oct 07, 2009 4:29 am

by cavaliar » Fri Sep 23, 2011 9:52 am

ur ans is correct

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Fri Sep 23, 2011 10:35 am

A: 4 + (x-4)/2 = (x+4)/2
remaining = (x-4)/2
Bob: 4+ 1/3 [(x-4)/2 -4] = 4 + (x-12)/6
remaining = (x-12)/3 = 32
thus x-12 = 32*3
or Bob: 4 + 32*3/6 = 20

Last edited by cans on Sat Sep 24, 2011 4:59 am, edited 1 time in total.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Sep 23, 2011 11:06 am

Tarun Khanna wrote:brent had posted this long ago and the ques is

A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money did Bob receive?
A) 20
B) 22
C) 24
D) 26
E) 52

Let R = what remains after Ann takes her share and Bob takes $4.

Since, Bob receives (1/3)R, Chloe receives (2/3)R.
Since Chloe receives 32:
(2/3)R = 32
R = 48.

Since Bob receives 1/3 of what remains plus $4:
B = (1/3)*48 + 4 = 20.

The correct answer is A.

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Tarun Khanna: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Wed Sep 14, 2011 1:48 amFollowed by: 1 member; Thanked: 1 times; GMAT Score:630

by Tarun Khanna » Fri Sep 23, 2011 11:25 pm

cans wrote:A: 4 + (x-4)/2 = (x+4)/2
remaining = (x-4)/2
Bob: 4+ 1/3 [(x-4)/2 -4] = 4 + (x-12)/6
remaining = (x-12)/3 = 32
thus x-12 = 32*3
or Bob: 4 + 32*3/6 = 22

hey thnx cans...you subtracted 4 here what i have highlighted in red Bob:4+ 1/3 [(x-4)/2 -4] because B also got 4 dollars first and we need to calculate the remaining amount after B got his 4 dollars right??

Tarun Khanna: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Wed Sep 14, 2011 1:48 amFollowed by: 1 member; Thanked: 1 times; GMAT Score:630

by Tarun Khanna » Fri Sep 23, 2011 11:27 pm

cavaliar wrote:ur ans is correct

Hi cavaliar...my ans is wrong...my ans is 22 but the correct ans is 20...have a look at post by cans and then my reply to it

cheers

Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Tue Dec 19, 2017 6:21 am

A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money did Bob receive?
A) 20
B) 22
C) 24
D) 26
E) 52

We can let the total amount of money = n.

Thus, Ann receives:

4 + (1/2)(n - 4)

4 + (1/2)n - 2

(1/2)n + 2

Bob receives:

4 + (1/3)[n - 4 - Ann]

4 + (1/3)[n - 4 - [(1/2)n + 2)]

4 + (1/3)[n - 4 - (1/2)n - 2]

4 + (1/3)[(1/2)n - 6] = 4 + (1/6)n - 2

(1/6)n + 2

Since we are given that Chloe receives the remaining $32, we can create the following equation, which combines Ann’s money, Bob’s money, and Chloe’s money:

(1/2)n + 2 + (1/6)n + 2 + 32 = n

Multiplying both sides by 6, we have:

3n + 12 + n + 12 + 192 = 6n

4n + 216 = 6n

2n = 216

n = 108

Thus, Bob receives:

1/6 x (108) + 2 = 18 + 2 = $20

Answer: A

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