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The useful life of a certain piece of equipment is determine

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The useful life of a certain piece of equipment is determine [#permalink] New post 22 Sep 2010, 05:25
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A
B
C
D
E

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Question Stats:

61% (01:19) correct 39% (00:31) wrong based on 36 sessions
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%

[Reveal] Spoiler:
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?


[Reveal] Spoiler:
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.
[Reveal] Spoiler: OA
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Re: Useful Life [#permalink] New post 22 Sep 2010, 05:31
bruiz wrote:
Hi,

First post! I'm having problems with the following question.

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%

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?


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
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Re: Useful Life [#permalink] New post 22 Sep 2010, 05:32
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bruiz wrote:
Hi,

First post! I'm having problems with the following question.

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%

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?

[Reveal] Spoiler:
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.


Hi, and welcome do Gmat Club.

8 times more means increase by 700% the same way as twice as much means increase by 100%.

x(1+p/100)=x(1+700/100)=x(1+7)=8x

Hope it's clear.
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Re: Useful Life [#permalink] New post 22 Sep 2010, 05:41
Expert's post
bruiz wrote:
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%


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.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Useful Life [#permalink] New post 22 Sep 2010, 06:13
As its asking for an increase you have to remember to subtract 100% from the original answer. For example, lets say you had $2 and you invest it and get back $2. Your return isn't 100% as would be found if we had taken $2/$2 = 100%, you need to subtract the original value or this can be accomplished by just subtracting 100%. So the equation would read ($2/$2)-1 = 0%. Taken one step further if we invested $2 and got back $4 our return would be ($4/$2)-1 = 100%.

Hope that helps.
Re: Useful Life   [#permalink] 22 Sep 2010, 06:13
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