The variables a and b are non-zero integers. If a = 2b^3/c, what happe

Question

The variables a and b are non-zero integers. If  a = \frac{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.

balamoon · Answer

Solution -

Given that a = 2*b^3/c --> c = 2*b^3/a

What is the value of c when a->a/2 and b->2b.

c = 2*b^3/a = 2*(2b)^3/(a/2) = 2*2*8*b^3/a =  16 *(2*b^3/a)

The value of c is 16 times the original value. ANS E.

ENGRTOMBA2018 · Answer

Assume a=2, b =1 000----> c = 2(1)/2 =1

Final value of c = with b =2 and a=1 ---> c = 2(2)^3/1 ---> c = 16. Thus C is multiplied by 16. E is the correct answer.

Tornikea · Answer

if a is divided by 2, the whole equation is divided by two - a/2=(2b^3/c)/2 => a/2=2b^3/2c - it means that c has to be halved
if b is doubled - b^3 is multiplied by 8, which means that c also has to be multiplied by 8 to keep the equation.
hence, C is halved and than multiplied by 8, so correct answer is D - multiplied by 4.

ENGRTOMBA2018 · Answer

The text in red above is not correct. When you half 'a' , c gets multiplied by 2.

c=(2b^3)/(a/2) =  2*  (2b^3)/a

Thus, the final effect will be to multiply c by 2 and then by 8 in effect multiplying by 16.

TimeTraveller · Answer

a = \frac{2b^3}{c} , so  c = \frac{2b^3}{a} .

when  a  is halved and  b  is doubled, c =  \frac{2*2(2b)^3}{a}  =  \frac{32b^3}{a} .

so,  c  is multiplied by 16 when  a  is halved and  b  is doubled. Ans (E).

twilliamson88 · Answer

When a = 2 and b = 2 ; C = 8
a*.5 = 1
b *2 = 4

When a = 1 and b = 4 ; C = 128

8x = 128
x = 16

kasinowr · Answer

rearrange equation so that you have c=(2b^3)/a

a=2 b=1
c=1

a=1 b=2
c=16

Answer: E

JJo · Answer

a=2b^3/c

c=2b^3/a

when a is halved and b is doubled

c=2(2b)^3/(a/2)
c=16*2b^3/a

Ans. E c is multiplied by 16.

Bunuel · Answer

800score Official Solution:

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).

rak08 · Answer

I feel the question should have been written better,
i felt its b^(3/c)

Bunuel · Answer

2b^3/c can  only  mean  \frac{2b^3}{c} . If it were 2b^(3/c), it would have been written that way. Formatted to avoid this kind of confusion for others.

The variables a and b are non-zero integers. If a = 2b^3/c, what happe

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The variables a and b are non-zero integers. If a = 2b^3/c, what happe

Prep Toolkit

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