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by pradeepkaushal9518 » Fri Jul 16, 2010 4:06 am
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%
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by kvcpk » Fri Jul 16, 2010 4:21 am
u = 8d/h^2

newU=8*2d/(1/2h)^2 = 64d/h^2
newU= U *8

hence increase is 700%
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by ryantherocket » Fri Jul 16, 2010 4:56 am
plug in numbers.

Initial: d=1, h=2, u=(8*1/2^2)=2

d doubles and h is halved. what happens to u?

Final: d=2, h=1, u=(8*2/1^2)=16

From 2 to 16 is an increase of 14, or 700% of 2. Pick D
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by Rahul@gurome » Fri Jul 16, 2010 6:07 am
u =(8d)/h^2
New u = [8(2d)]/(h/2)^2 = 64d/(h^2)
Increase = 64d/(h^2) - (8d)/h^2 = 56d/(h^2)
Percentage increase = [56d/(h^2) / (8d)/h^2] * 100 = 700%

The correct answer is (D).
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by pradeepkaushal9518 » Fri Jul 16, 2010 9:10 am
thaks rahul i have choosen 800% by mistake
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by selango » Fri Jul 16, 2010 9:24 am
Pradeep,

When I am working on this prob in MGMAT test,in a hurry I also thought the same 800%.

Simply 64/8=8%

But I thought some trick ll be there and used the formula.Then 700%

Whenever % prob come use the formula

Change/Original %
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