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Difficulty:
35%
(medium)
Question Stats:
69% (01:28) correct
31%
(01:49)
wrong
based on 269
sessions
History
Date
Time
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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
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
19 Nov 2009, 01:48
Kudos
Bookmarks
Assuming you mean h^2 then we have:
Original useful life: \(u=8d/h^2\)
New useful life: \(u=\frac{8(2d)}{(h/2)^2}=\frac{64d}{h^2}\)
Amount Increase = new - old = \(\frac{64d}{h^2}-\frac{8d}{h^2}=\frac{56}{h^2}\)
Increase/Original = \(\frac{56d}{h^2}/\frac{8d}{h^2}\)
= 7.
To get percentage multiple by 100 = 700%
Ans = D
Original useful life: \(u=8d/h^2\)
New useful life: \(u=\frac{8(2d)}{(h/2)^2}=\frac{64d}{h^2}\)
Amount Increase = new - old = \(\frac{64d}{h^2}-\frac{8d}{h^2}=\frac{56}{h^2}\)
Increase/Original = \(\frac{56d}{h^2}/\frac{8d}{h^2}\)
= 7.
To get percentage multiple by 100 = 700%
Ans = D
19 Nov 2009, 02:36
Kudos
Bookmarks
It is not a DS problem, but that's OK.
I understood by two ways:
1) U = 8d/hˆ2
U' = 8*2d/ (1/2*h)ˆ2 = 64d/hˆ2 --> U *64 = U ' --> 700% increase
Ans: D
2) U = 8d/(2h)
U '= 8*2d/(2*(1/2)*d) = 16d/h --> 300% increase
Ans A
I understood by two ways:
1) U = 8d/hˆ2
U' = 8*2d/ (1/2*h)ˆ2 = 64d/hˆ2 --> U *64 = U ' --> 700% increase
Ans: D
2) U = 8d/(2h)
U '= 8*2d/(2*(1/2)*d) = 16d/h --> 300% increase
Ans A
