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# If a = b × c^2 and c decreases by 20% while a remains

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If a = b × c^2 and c decreases by 20% while a remains [#permalink]  21 Jun 2017, 13:52
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If a = b × c^2 and c decreases by 20% while a remains constant, by what percent does b increase?

[Reveal] Spoiler: OA
56.3

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Re: If a = b × c^2 and c decreases by 20% while a remains [#permalink]  22 Jun 2017, 10:27
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Carcass wrote:

If a = b × c^2 and c decreases by 20% while a remains constant, by what percent does b increase?

[Reveal] Spoiler: OA
56.3

a= bxc^2

Now c is reduced by 20% but a remain constant i.e

a= bx(0.8c)^2 = bx0.64c

therefore b= 1/0.64 = 1.5625

% increase = (1.5625 - 1 )/ 1 x100% = 56.25% ~ 56.3%
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Re: If a = b × c^2 and c decreases by 20% while a remains [#permalink]  11 Jul 2017, 04:09
Expert's post
Carcass wrote:

If a = b × c^2 and c decreases by 20% while a remains constant, by what percent does b increase?

[Reveal] Spoiler: OA
56.3

Another approach is to PLUG IN some numbers that satisfy the given conditions.

Given: a = b x c²
So, how about a = 2500, b = 100 and c = 5
We get: 2500 = 100 x 5²

c decreases by 20% while a remains constant
c becomes 4 and a stays at 2500
So, we get: 2500 = b x 4²
Simplify: 2500 = b x 16
Solve: 2500/16 = b
Simplify: 156.25 = b

So b INCREASES from 100 to 156.25
This represents an increase of 56.25%
Since the question asks us to round our answer to the nearest TENTH, the correct response is 56.3

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Re: If a = b × c^2 and c decreases by 20% while a remains   [#permalink] 11 Jul 2017, 04:09
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