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The useful life of a certain piece of equipment is determine [#permalink]

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22 Sep 2010, 06:25

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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?

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%

Re: The useful life of a certain piece of equipment is determine [#permalink]

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22 Sep 2010, 06:31

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

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%

Re: The useful life of a certain piece of equipment is determine [#permalink]

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22 Sep 2010, 06:32

2

This post received KUDOS

Expert's post

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

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.

Hi, and welcome do Gmat Club.

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

Re: The useful life of a certain piece of equipment is determine [#permalink]

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22 Sep 2010, 06:41

1

This post received KUDOS

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%.

Re: The useful life of a certain piece of equipment is determine [#permalink]

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22 Sep 2010, 07:13

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%.

Re: The useful life of a certain piece of equipment is determine [#permalink]

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17 Nov 2014, 10:54

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Re: The useful life of a certain piece of equipment is determine [#permalink]

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21 Jan 2016, 13:23

Hi Anonamy,

The original formula is \(u=\frac{8d}{h^2}\), so when the daily usage (h) is halved, you must replace \(h\) with \(h/2\). All of \(h/2\) is then squared.

So the new formula looks like this: \(u=\frac{2*8d}{(\frac{h}{2})^2}\)

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