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The variables a and b are non-zero integers...

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The variables a and b are non-zero integers...

by BTGmoderatorLU » Fri Dec 22, 2017 2:19 pm
The variables a and b are non-zero integers. If a=2b^3/c, what happens to c when a is halved and b is doubled?

A. c is not changed
B. c is halved
C. c is doubled
D. c is multiplied by 4
E. c is multiplied by 16

The OA is E.

I'm confused with this PS question, what should I do to solve it? Isolate c? I'm not sure. Experts, any suggestion? Thanks in advance.
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by EconomistGMATTutor » Mon Dec 25, 2017 9:29 am
Hello LUANDATO.

Let's see your question.

We have that $$a=\frac{2b^3}{c},\ \ \ \text{then}\ \ c=\frac{2b^3}{a}.$$ Now, if we write "a/2" instead of "a" and "2b" instead of "b", we will get:
$$c=\frac{2\left(2b\right)^3}{\frac{a}{2}}=\frac{2\cdot8b^3}{\frac{a}{2}}=\frac{32b^3}{a}=16\cdot\frac{2b^3}{a}=16\cdot c.$$ So, the correct answer is E.

I hope this explanation may help you.

I'm available if you'd like a follow-up.

Regards.
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by [email protected] » Mon Dec 25, 2017 10:08 am
Hi Vincen,

To start, there's a slight 'typo' in the equation - I've noted the correct in my explanation:

We're told that the variables A and B are non-zero integers and that A = ((2B)^3)/C. We're asked what happens to A when A is halved and B is doubled. This question can be solved by TESTing VALUES.

IF...
B = 1 and C = 1
A = (8)/1 = 8

By halving A (A becomes 4) and doubling B (B becomes 2), we get....
4 = (64)/C
4C = 64
C = 16

Thus, C = 1 becomes C = 16, so C becomes multiplied by 16.

Final Answer: E

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