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The useful life of a certain piece of equipment is determine

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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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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.
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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
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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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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%
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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%
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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
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Re: The useful life of a certain piece of equipment is determine [#permalink]

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27 Oct 2013, 05:23
Expert's post
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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