Problem solving

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Problem solving

by BTGmoderatorRO » Fri Sep 15, 2017 1:35 pm
15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?
A:37.5
B:37
C:45
D:25
E:35

OA is a...
what is the best approach to solving this question. Did you also get 45 as your answer

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by Jay@ManhattanReview » Tue Sep 19, 2017 1:05 am
Roland2rule wrote:15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?
A:37.5
B:37
C:45
D:25
E:35

OA is a...
what is the best approach to solving this question. Did you also get 45 as your answer
Hi Roland2rule,

You have been posting many questions. It would be nice if you write the source of questions. Is it a GMAT source?

The correct answer for this one is A, but it involves a lot of calculation which is not expected in the GMAT.

Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide

-Jay
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by GMATGuruNY » Tue Sep 19, 2017 2:56 am
Roland2rule wrote:15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?
A:37.5
B:37
C:45
D:25
E:35
I agree that this problem is beyond the scope of the GMAT.
That said, here's an efficient approach:

Use the formula for REPEATED FRACTIONAL CHANGE.
If amount x DECREASES by fraction a/b exactly n times:

Final amount = x * (1 - a/b)^n.

In the problem at hand:

Original amount:
Let the capacity of the container = x.
Since the container is full of A, the original amount of A = x.

Final amount:
After two replacements, A:B = 9:16.
Since 9+16 = 25, A will constitute 9/25 of the container.
Thus, the final amount of A = (9/25)x.

Fractional decrease:
Since 15 liters of x are removed with each replacement, the fractional decrease = 15/x.
Since the volume decreases TWICE, n = 2.

Plugging these values into the formula, we get:

(9/25)x = x * (1 - 15/x)²

9/25 = (1 - 15/x)²

3/5 = 1 - 15/x

15/x = 2/5

2x = 75

x = 37.5.

Note that if amount x INCREASES by fraction a/b exactly n times, the formula is as follows:
Final amount = x * (1 + a/b)^n.
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