The useful life of a certain piece of equipment is...

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AAPL: Moderator; Posts: 2632; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Fri Feb 02, 2018 7:20 am

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%

The OA is D.

Is there a strategic approach to this PS question? Can any experts help me please? Thanks!

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Source: Beat The GMAT — Problem Solving | Fri Feb 02, 2018 7:20 am

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by GMATGuruNY » Fri Feb 02, 2018 8:22 am

AAPL wrote: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%

Original values:
Let d=1 and h=2, with the result that u = 8d/hÂ² = (8*1)/2Â² = 2.

New values:
When d is doubled to 2 and h is halved to 1, u = 8d/hÂ² = (8*2)/1Â² = 16.

Percent increase from 2 to 16 = Difference/Smaller * 100 = (16-2)/2 * 100 = 700.

The correct answer is D.

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by Scott@TargetTestPrep » Mon Feb 05, 2018 9:34 am

AAPL wrote: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%

We can let d = 1 and h = 2; thus, u = (8 x 1)/2^2 = 2.

Now, let's double d and make h half, so the new d = 2 and the new h = 1 and thus the new u = (8 x 2)/1^2 = 16.

So the percentage increase in u is:

(16 - 2)/2 x 100% = 7 x 100% = 700%

Answer: D

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