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Tough word problem - dividing $

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by leekaru14 » Wed Sep 25, 2013 1:38 am
Bob gets: $4 + 1/3 remaining
Chloe gets: 2/3 remaining = 32
==> Bob gets: $4 + $16 = $20
I expected a trap in your question. :)
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by ProGMAT » Tue Mar 11, 2014 4:26 am
kamu wrote:Bob took $4 + 1/3 of leftover..

that implies 2/3rds is left for Chloe..

if 2/3 of leftover = $32

1/3 of leftover = $16

$16+$4 = $20.
Got stuck in calculation and spent around 4 mins to solve. This is the simplest one I found here. Thanks for solution kamu. Good one.
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by susahntgupta402 » Fri Apr 18, 2014 2:15 pm
Looks like I am the only lateral thinker in the group :)

Bob + Chloe = X

Bob = 4 + x/3
Chloe = (2X/3) - 4 = 32 ; X= 54

Bob = 4+ 54/3 = 22
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by Brent@GMATPrepNow » Fri Apr 18, 2014 2:33 pm
susahntgupta402 wrote:Looks like I am the only lateral thinker in the group :)

Bob + Chloe = X

Bob = 4 + x/3
Chloe = (2X/3) - 4 = 32 ; X= 54

Bob = 4+ 54/3 = 22
Sorry, but Bob receives $20

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Fri Apr 18, 2014 7:14 pm
Tricky problem Brent!

Brent@GMATPrepNow wrote:Nice work, Logitech
My solution is below. Note that we don�t need to consider Ann�s portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bob�s portion, we have $32 for Chloe.
So, we get K � Bob�s $ = 32
The equation is K-4 � (K-4)/3 = 32
Solve for K (K=52) and then determine Bob�s portion ($20).
The answer is, indeed, A
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by Kapstera1 » Fri Oct 24, 2014 3:19 am
A: 4 + 1/2(X-4)
B: 4 + 1/3(X/2 + 2)

C: 2/3 (X/2 + 2) = 32
Solve equation C to get X = 92
Substitute X=92 in equation B to get 20
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by akash singhal » Sun May 03, 2015 12:56 am
answer is A 20

since we know last remaining part is 32
we can say that
Let the remaining money after first Ann took her share be x
So bob will have x/3 part
and we know
x-x/3=32
Solving we get x = 48
So, bob gets (x/3)+4 = 20$

I guess i understood the question correct
Thanks...
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by nikhilgmat31 » Sun May 31, 2015 9:58 pm
What about doing the solution in following way

lets total amount be x

1. A = 4 + (x-4)/2 = (x + 4)/2
2. B = 4 + 1/3 of remaining = 4 + (x - (x+4)/2)/3 = (x+20)/6
3. C = 32

C = x-A-B

32 = x- (x+4)/2 - (x+20)/6

x= 112

A = (x+4)/2 = (112+4)/2 = 58
B = (x+20)/6 = (112+20)/6 = 22
C = 32

To confirm 58+22+32 = 112

So Answer is 22

Please review & confirm.
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by nikhilgmat31 » Mon Jun 01, 2015 12:55 am
You calculated Bob's value wrongly.
Bob receives $4 plus one third of what remains -> 4 + 44/3 = 56/3

It should be 4+ 48/3 = 20 :)
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by deepak4mba » Thu Mar 01, 2018 1:44 pm
A
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