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 Post subject: GMAT Algebra
PostPosted: Sat May 04, 2013 4:31 am 
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Joined: Tue Apr 13, 2010 8:48 am
Posts: 478
The variables a and b are non-zero integers. If a = 2b³/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.


(E) The easiest way to solve this problem is to plug in numbers for the variables and then backsolve for c. Use even numbers, because even numbers can be halved without leaving fractions.

Let b = 2 and c = 4. Solve the equation:
a = (2(2³))/4 = 16/4 = 4.
So, a = 4 when b = 2 and c = 4.

Now halve a and double b: a = 2, b = 4. Plug these values into the equation and see what happens to c:
2 = ((2(4³))/c
2 = (2(64))/c
2 = 128/c
2c = 128
c = 64.

Comparing this to the original value of c, which was 4, we see that the new value for c is 16 times the original value.

Another way to attack this problem is to solve for c to arrive at the same answer.

The correct answer is choice (E).
----------
Can you also show the solution for c? I get that c is increased by 4 and I think I'm missing something.

Thanks.


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 Post subject: Re: GMAT Algebra
PostPosted: Sat May 04, 2013 4:31 am 
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Joined: Fri Apr 09, 2010 2:11 pm
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Quote:
Can you also show the solution for c?
Here it is.

We start with a = 2b³/c
Solve for c:
c = 2b³/a

Now halve a and double b. In other words plug in a/2 instead of a and 2b instead of b.
?? × c = 2(2b)³/(a/2)

Then get all the new multipliers/dividers at the beginning of the right side. Note that we divide by (a/2), which is the same as multiplying by (2/a), so 2 goes to the numerator.
I've highlighted the new multipliers by the same colors to make transformations easier to spot.
?? × c = ( 2(8b³)/a ) × 2
?? × c = (8 × 2) × ( 2b³/a )

Thus ?? is 8 × 2 = 16


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