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# The useful life of a certain piece of equipment is determine

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CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City
The useful life of a certain piece of equipment is determine [#permalink]

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20 Feb 2008, 15:00
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84% (01:59) correct 16% (00:38) wrong based on 73 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
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Last edited by Bunuel on 27 Oct 2013, 05:21, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Director
Joined: 01 Jan 2008
Posts: 622

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20 Feb 2008, 15:09
bmwhype2 wrote:
The useful life of a certain piece of equipment is determined by the following formula: u =(8d)/h2, 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?

300%
400%
600%
700%
800%

if it's h^2 then D is the answer
I really don't like units though
Senior Manager
Joined: 20 Dec 2004
Posts: 251

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20 Feb 2008, 15:09
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bmwhype2 wrote:
The useful life of a certain piece of equipment is determined by the following formula: u =(8d)/h2, 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?

300%
400%
600%
700%
800%

I am assuming your question is u =(8d)/h^2,

Hence new equation would be
$$u_1 = \frac{8 * (2d)}{(\frac{h}{2})^2}$$ --> $$u_1 = 8 * \frac{8 * d} {({h})^2}$$ --> $$u_1 = 8*u$$

Hence 700% increase....
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Joined: 28 Dec 2005
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20 Feb 2008, 19:01
agreed. assuming question is h^2, we have:

percentage change = u2-u1/u2 * 100

u1=8d/h^2

u2=[8*2d]/[(h/2)^2] -->64d/h^2

64d/h^2 - 8d/h^2 / 8d/h^2 = 7*100 = 700%
Director
Joined: 30 Jun 2007
Posts: 786

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21 Feb 2008, 02:42
U = (8d) / h2 without any percentage change
D = 1 h = 1
Uold = 4
With change
Unew = 16d / h
D = 1 h = 1
Unew = 16

400%
Intern
Joined: 09 Jul 2013
Posts: 7
GMAT 1: 700 Q47 V39

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27 Oct 2013, 05:17
hanumayamma wrote:
U = (8d) / h2 without any percentage change
D = 1 h = 1
Uold = 4
With change
Unew = 16d / h
D = 1 h = 1
Unew = 16

400%

Forgot to divide last h by 2 and then squared
Math Expert
Joined: 02 Sep 2009
Posts: 39672
Re: The useful life of a certain piece of equipment is determine [#permalink]

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27 Oct 2013, 05:23
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-useful-life-of-a-certain-piece-of-equipment-is-determine-101496.html
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Re: The useful life of a certain piece of equipment is determine   [#permalink] 27 Oct 2013, 05:23
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