The variables a and b are nonzero 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.
