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bmwhype2
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The useful life of a certain piece of equipment is determine Updated on: 27 Oct 2013, 05:21
Originally posted by bmwhype2 on 20 Feb 2008, 15:00.
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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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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maratikus
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The useful life of a certain piece of equipment is determine 20 Feb 2008, 15:09
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bmwhype2
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%
Show more

if it's h^2 then D is the answer
I really don't like units though :)
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neelesh
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The useful life of a certain piece of equipment is determine 20 Feb 2008, 15:09
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bmwhype2
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%
Show more

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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The useful life of a certain piece of equipment is determine 20 Feb 2008, 19:01
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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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The useful life of a certain piece of equipment is determine 21 Feb 2008, 02:42
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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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The useful life of a certain piece of equipment is determine 27 Oct 2013, 05:17
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hanumayamma
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%
Show more


Forgot to divide last h by 2 and then squared
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The useful life of a certain piece of equipment is determine 27 Oct 2013, 05:23
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OPEN DISCUSSION OF THIS QUESTION IS HERE: the-useful-life-of-a-certain-piece-of-equipment-is-determine-101496.html

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Bunuel
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