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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  [#permalink]

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New post 24 Jul 2015, 00:51
2
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A
B
C
D
E

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  35% (medium)

Question Stats:

69% (01:39) correct 31% (01:43) wrong based on 146 sessions

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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 02:31
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Bunuel wrote:
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.

Kudos for a correct solution.


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.
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 03:47
Bunuel wrote:
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.

Kudos for a correct solution.


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.
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 04:06
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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.
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 04:15
Tornikea wrote:
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.


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.
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 05:12
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\(a = \frac{2b^3}{c}\), 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).
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 08:12
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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
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 10:48
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rearrange equation so that you have c=(2b^3)/a

a=2 b=1
c=1

a=1 b=2
c=16

Answer: E
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 24 Jul 2015, 10:56
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Bunuel wrote:
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.

Kudos for a correct solution.


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.
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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New post 26 Jul 2015, 12:20
Bunuel wrote:
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.

Kudos for a correct solution.


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).
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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe  [#permalink]

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Re: The variables a and b are non-zero integers. If a = 2b^3/c, what happe   [#permalink] 16 Jul 2018, 07:34
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