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The useful life of a certain piece of equipment is

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kairoshan
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The useful life of a certain piece of equipment is [#permalink] New post   Updated on: 19 Nov 2009, 06:45
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The useful life of a certain piece of equipment is determined by the following formula: u =(8d)/h^2, where u is the useful life of the equipment, in years, d is the density of the underlying material, in g/cm3, and h is the number of hours of daily usage of the equipment. If the density of the underlying material is doubled and the daily usage of the equipment is halved, what will be the percentage increase in the useful life of the equipment?

A. 300%
B. 400%
C. 600%
D. 700%
E. 800%

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-useful-life-of-a-certain-piece-of-equipment-is-determine-101496.html
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D

Originally posted by kairoshan on 18 Nov 2009, 21:47.
Last edited by kairoshan on 19 Nov 2009, 06:45, edited 1 time in total.
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Re: The useful life of a certain piece of equipment is determine [#permalink] New post   19 Nov 2009, 01:48
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Assuming you mean h^2 then we have:

Original useful life: \(u=8d/h^2\)
New useful life: \(u=\frac{8(2d)}{(h/2)^2}=\frac{64d}{h^2}\)
Amount Increase = new - old = \(\frac{64d}{h^2}-\frac{8d}{h^2}=\frac{56}{h^2}\)
Increase/Original = \(\frac{56d}{h^2}/\frac{8d}{h^2}\)
= 7.

To get percentage multiple by 100 = 700%
Ans = D
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diogoguitarrista
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Re: The useful life of a certain piece of equipment is determine [#permalink] New post   19 Nov 2009, 02:36
It is not a DS problem, but that's OK.

I understood by two ways:

1) U = 8d/hˆ2

U' = 8*2d/ (1/2*h)ˆ2 = 64d/hˆ2 --> U *64 = U ' --> 700% increase

Ans: D

2) U = 8d/(2h)

U '= 8*2d/(2*(1/2)*d) = 16d/h --> 300% increase

Ans A
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kairoshan
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Re: The useful life of a certain piece of equipment is determine [#permalink] New post   19 Nov 2009, 06:46
diogoguitarrista wrote:
It is not a DS problem, but that's OK.

I understood by two ways:

1) U = 8d/hˆ2

U' = 8*2d/ (1/2*h)ˆ2 = 64d/hˆ2 --> U *64 = U ' --> 700% increase

Ans: D

2) U = 8d/(2h)

U '= 8*2d/(2*(1/2)*d) = 16d/h --> 300% increase

Ans A


OA is D. sorry. I just edited the question.
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Re: The useful life of a certain piece of equipment is [#permalink] New post   04 May 2013, 17:03
In this problem why does it become \((64h^2)/h^2\) instead of \((32h^2)/h^2\)
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Re: The useful life of a certain piece of equipment is [#permalink] New post   05 May 2013, 04:21
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hfbamafan wrote:
In this problem why does it become \((64h^2)/h^2\) instead of \((32h^2)/h^2\)


Check here: the-useful-life-of-a-certain-piece-of-equipment-is-determine-101496.html#p786365

Hope it helps.
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Re: The useful life of a certain piece of equipment is [#permalink]
05 May 2013, 04:21
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Bunuel
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