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If a = b × c^2 and c decreases by 20% while a remains
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21 Jun 2017, 13:52
21 Jun 2017, 13:52
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Question Stats:
73% (01:39) correct
26% (01:59) wrong based on 192 sessions
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If a = b × \(c^2\) and c decreases by 20% while a remains constant, by what percent does b increase?
Round your answer to the nearest 0.1%
enter your value as %
Show: :: OA
56.3
Re: If a = b × c^2 and c decreases by 20% while a remains
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22 Jun 2017, 10:27
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?
Round your answer to the nearest 0.1%
enter your value as %
Show: :: 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
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11 Jul 2017, 04:09
11 Jul 2017, 04:09
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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?
Round your answer to the nearest 0.1%
enter your value as %
Show: :: 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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