Tough word problem - dividing $

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by immaculatesahai » Wed Nov 09, 2011 12:39 am
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
I think the crux of the problem is managing the time and not really the solution part of it. I am sure most ppl here would arrive at the answer (which is A btw). But hats off to you Brent, a very nice solution indeed.

Unfortunately, I went the conventional route: Anyways, here is what I did:

Let total money be x. (i have omitted several steps)

Amount Anna got: (x+4)/2
Amount Bob got: (x+12)/6
Equate that to amount received by C and you get x= 108.

Therefore Bob got $20.

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by nskandan » Thu Dec 01, 2011 2:39 am
Rastis wrote:Anyone have any explanations that are easier to understand?
For me, these types of problems are easier to understand and solve if a diagram is drawn (e: Pie diagram).

A 's pay = 4 + (A+B+C-4)/2
now that A is paid A pie portion can be taken out
B's Pay = 4 + 1/3(B+C-4)
C's pay = 32

We are concerned only about B's Pay
So, just try to solve the 2nd equation forget the rest.
NSK

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by Mr.Hollywood » Mon Dec 12, 2011 6:35 pm
logitech wrote:A) 4 + (x-4)/2

B) 4 + 1/3 [(X-12)/2]

C) 2/3 [(X-12)/2] = 32 ; (x-12) = 3 x 32

SO 4 + 1/3 [(X-12)/2] = 4+ 1/3 ( 3x32/2) = 20
Good job! I used the same approach as you, i was wondering how did you get the "12" in 4 + 1/3 [(X-12)/2] ?

thank you!

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by him1985 » Thu Jan 26, 2012 3:52 am
Correct Answer: A
I made it correct in one go But Thanks Brantt for a different & shorter approach.
:)
Himanshu Chauhan

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by bachvudao » Thu Feb 02, 2012 5:23 pm
We can back solve this one to get the answer. As Chloe received the remaining amount after Bobb, 32 is 2/3 of the remaining after Bob took 4 dollars so Bob took:

4 + 32/2 = 20.

Answer is A

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by [email protected] » Fri Feb 03, 2012 11:56 pm
Thanx Logitech for a wonderful explantion!!!
A) 4 + (x-4)/2

B) 4 + 1/3 [(X-12)/2]

C) 2/3 [(X-12)/2] = 32 ; (x-12) = 3 x 32

SO 4 + 1/3 [(X-12)/2] = 4+ 1/3 ( 3x32/2) = 20

I was stuck at step B but eventually got hint....

ThankkSSS
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by ronnie1985 » Sat Feb 04, 2012 2:18 am
Assume the amount to be x
A: gets x/2 + 2
Rem: x/2 - 2
B: 4+ 1/3*(x/2-6) = x/6 + 2
Rem: x/2 - 2- (x/6+2) = x/3 - 4 = 32 = > x = 108
B = x/6 + 2 = 20
(A) is answer
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by fxsunny » Sun Feb 26, 2012 3:44 pm
badpoem wrote:I started with $100.

Ann receives a $4 plus one half of what remains -> 4 + 96/2 = 52.
Remaining = 100-52 = 48.
Bob receives $4 plus one third of what remains -> 4 + 44/3 = 56/3

Chloe receives the remaining $32. In this case Chloe receives 48 - 4 - 44/3 = 44*2/3.

Therefore, 44*2/3 equivalent to 32
and 56/3 (Bob's share) equivalent to (32*(56/3)*3)/(44*2) ~= 20.

Hence A.
I like the technique, but the result is also approximate...in that (32*(56/3)*3)/(44*2) ends up being 20.36 - which is pretty close to 20. But, I'm curious what's your success rate with this technique?

Is this a pretty popular technique? It certainly does make sense...but I'm not too comfortable with the "approximate" result especially when the other solutions lead to 20 on the dot.

Thoughts?

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by ka_t_rin » Fri Mar 16, 2012 11:44 pm
Although I started with X - as a whole and then made all the computations the solution is much easier...
Suppose that 32 = 2/3 of the remainder, then Bob has 1/3*R + 4 => 16+4 = 20

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by kannan007 » Mon Mar 26, 2012 8:33 am
my calculations goes as below..
of the remaining money chole receives 2 times as bob..
so bob get (32/2=16 of the remaning money )and he get a 4 as well..so 16+4=20!!

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by asax » Thu Jun 21, 2012 11:53 pm
wow! It took me full 4 minutes to do the algebra! thanks for the shortcuts folks!
Looking forward to 2013 MBA admissions!

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by Brent@GMATPrepNow » Fri Jun 22, 2012 6:53 am
asax wrote:wow! It took me full 4 minutes to do the algebra! thanks for the shortcuts folks!
A good thing to keep in mind is that most (if not all) GMAT math questions can be solved in 2 or more ways. So, if you find yourself headed down the path of a very long solution, you should stop and see if there's a shorter approach. Of course, that doesn't necessarily mean that you'll necessarily spot that faster approach. However, just knowing that the GMAT would never ask a math question that can be solved using only one, very time-consuming approach can help you judge the practicality of some solutions.

Cheers,
Brent
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by sureshcpa » Fri Aug 31, 2012 12:40 pm
Well i did it this way.

The remaining for Chole is 32 which is 2/3rd after the 1/3rd share of Bob

So, before the share of Bob the amount was 32 x 3/2 = 48

Bob's share = $4 + (1/3 x 48) = 16 = 20

No need of solving the equation!

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by rajeshsinghgmat » Sun Jan 20, 2013 6:09 pm
A for answer.

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by glensuchitha » Thu Aug 01, 2013 5:34 am
Hey I'm not a math whiz so I look for logic. my logic for the following question was as follows,

If Ann gets 1/2 (plus 4$) and Bob gets 1/3(plus 4$),

Their fractions are 1/2 + 1/3 = 5/6ths of the total, hence, chloe gets 32 or 1/6th.


so therefore, if 1/6 = 32, bob (1/3) = ?

so we get (cross multiply), (32X3)/6 = 16 + 4$(as mentioned) = 20$ hence, A