The useful life of a certain piece of equipment is : GMAT Data Sufficiency (DS)
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# The useful life of a certain piece of equipment is

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The useful life of a certain piece of equipment is [#permalink]

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18 Nov 2009, 20:47
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Question Stats:

61% (01:16) correct 39% (00:20) wrong based on 133 sessions

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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
[Reveal] Spoiler: OA

Last edited by kairoshan on 19 Nov 2009, 05:45, edited 1 time in total.
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Re: The useful life of a certain piece of equipment is determine [#permalink]

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19 Nov 2009, 00:48
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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Re: The useful life of a certain piece of equipment is determine [#permalink]

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19 Nov 2009, 01:36
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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Re: The useful life of a certain piece of equipment is determine [#permalink]

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19 Nov 2009, 05:46
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]

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04 May 2013, 16:03
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]

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05 May 2013, 03:21
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, 03:21
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