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22 Sep 2010, 06:25
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:40) correct
32%
(01:48)
wrong
based on 816
sessions
History
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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%
A. 300%
B. 400%
C. 600%
D. 700%
E. 800%
I plugged the following numbers:
d=4
h=4
Formula:
u=\frac{(8d)}{h^2}
First part
u=\frac{(8*4)}{4^2}
u=\frac{32}{16}
u=\frac{2}{1}
u=2
Second part
Density doubled
d=4*2
Usage halved
h=4/2
d=8
h=2
u=\frac{(8*8)}{2^2}
u=\frac{64}{4}
u=\frac{16}{1}
u=[fraction]16[/fraction]
Third Part
So my initial value is 2 and my value after the change is 16.
16/2=8, so there is an 8 fold increase, which is 800%
This is wrong. In the spoiler is the explanation, which uses the exact same method but with other numbers and its result is different. What am I missing here?
d=4
h=4
Formula:
u=\frac{(8d)}{h^2}
First part
u=\frac{(8*4)}{4^2}
u=\frac{32}{16}
u=\frac{2}{1}
u=2
Second part
Density doubled
d=4*2
Usage halved
h=4/2
d=8
h=2
u=\frac{(8*8)}{2^2}
u=\frac{64}{4}
u=\frac{16}{1}
u=[fraction]16[/fraction]
Third Part
So my initial value is 2 and my value after the change is 16.
16/2=8, so there is an 8 fold increase, which is 800%
This is wrong. In the spoiler is the explanation, which uses the exact same method but with other numbers and its result is different. What am I missing here?
One of the most effective ways to solve problems involving formulas is to pick numbers. Note that since we are not given actual values but are asked to compute only the relative change in the useful life, we can select easy numbers and plug them into the formula to compute the percentage increase. Let’s pick d = 3 and h = 2 to simplify our computations:
Before the change: d = 3, h = 2; u = (8)(3)/22 = 24/4 = 6
After the change: d = (2)(3)= 6, h =2/2 =1; u = (8)(6)/12 = 48
Finally, percent increase is found by first calculating the change in value divided by the original value and then multiplying by 100:
(48 – 6)/6 = (42/6) = 7
(7)(100) = 700%
The correct answer is D.
Before the change: d = 3, h = 2; u = (8)(3)/22 = 24/4 = 6
After the change: d = (2)(3)= 6, h =2/2 =1; u = (8)(6)/12 = 48
Finally, percent increase is found by first calculating the change in value divided by the original value and then multiplying by 100:
(48 – 6)/6 = (42/6) = 7
(7)(100) = 700%
The correct answer is D.
22 Sep 2010, 06:41
Kudos
Bookmarks
bruiz
Algebraic approach:
\(u_1=\frac{8d}{h^2}\)
Density of the underlying material is doubled and the daily usage of the equipment is halved --> \(u_2=\frac{8(2d)}{(\frac{h}{2})^2}=8*\frac{8d}{h^2}\) --> \(u_2=8u_1\) --> 8 times more = increase by 700%.
Answer: D.
General Discussion
22 Sep 2010, 06:31
Kudos
Bookmarks
bruiz
You are on the right track, its just the last bit of the calculation.
The useful life goes from x to 8x (8 times). Now the increase is therefore 7x. And percent increase is 7x/x * 100 or 700%
Remember : %-age increase = \(\frac{New-Old}{Old} * 100\)
