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# useful life

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Manager
Joined: 28 Mar 2009
Posts: 74

Kudos [?]: 368 [0], given: 0

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27 Jun 2009, 11:36
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100% (01:25) correct 0% (00:00) wrong based on 3 sessions

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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?

(A) 300%

(B) 400%

(C) 600%

(D) 700%

(E) 800%

Kudos [?]: 368 [0], given: 0

Senior Manager
Joined: 04 Jun 2008
Posts: 288

Kudos [?]: 154 [0], given: 15

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27 Jun 2009, 11:45
A = B/C2

after the changes given,

A = 2b/ (c/2)2

= 2b/ (c2/4)

= or 2*4 * b/c2

thus it will be 8 times the original value, or a 700% increase

Kudos [?]: 154 [0], given: 15

Re: useful life   [#permalink] 27 Jun 2009, 11:45
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