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The useful life of a certain piece of equipment is determine
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22 Sep 2010, 06:25
Question Stats:
64% (01:31) correct 36% (01:44) wrong based on 210 sessions
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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% 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? 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.
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Re: The useful life of a certain piece of equipment is determine
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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
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 = \(\frac{NewOld}{Old} * 100\)
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Re: The useful life of a certain piece of equipment is determine
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22 Sep 2010, 06:32
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? 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: The useful life of a certain piece of equipment is determine
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22 Sep 2010, 06:41
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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Re: The useful life of a certain piece of equipment is determine
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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%.
Hope that helps.



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Re: The useful life of a certain piece of equipment is determine
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16 Jun 2015, 05:10
Why is the result different if I plug in 1 for both d and H? the result is 300% in which case, even by following the rule of (Increaseoriginal/100)



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Re: The useful life of a certain piece of equipment is determine
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16 Jun 2015, 05:43
RudeyboyZ wrote: Why is the result different if I plug in 1 for both d and H? the result is 300% in which case, even by following the rule of (Increaseoriginal/100) You have to show your work, the result should and IS the same.
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Re: The useful life of a certain piece of equipment is determine
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16 Jun 2015, 06:30
Dear Bunuel, You are right there isn't a way to get different answers, I forgot to square the h in the second part of the work. You're the best! Regards, Rudraksh.



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Re: The useful life of a certain piece of equipment is determine
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20 Jan 2016, 21:56
I'm confused because I'm getting a different answer. Where am I going wrong?
Original: u=8d/h(sqrd) With changes: u=8(2d)/[h(sqrd)/2]
I tried the plugging in approach. d=4 and h=4:
Original: 32/16=2 With changes: 64/8=8
Therefore the answer would be 300%, which is not correct, so I then tried d=6 and h=4:
Original: 48/16=4 With changes: 96/8=12
Therefore the answer would be 200%... What is going on here? Thanks



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Re: The useful life of a certain piece of equipment is determine
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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: \(u=\frac{2*8d}{(\frac{h}{2})^2}\) What you had was \(u=\frac{2*8d}{\frac{h^2}{2}}\) Do you see the difference? Cheers
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Re: The useful life of a certain piece of equipment is determine
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22 Jan 2016, 22:01
Hi davedekoos,
Ah! Yes I see my error now. Thanks for your help!



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Re: The useful life of a certain piece of equipment is determine
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09 May 2017, 11:43
I used the plug in approach for this question.
u=\(\frac{8d}{h^2}\)  eq 1
Case 1:
u = ? d = 10 h = 2
Substitute above values in eq 1
u = \(\frac{8 * 10}{2 * 2}\) u = 2*10 = 20
Case 2: If the density of the underlying material is doubled and the daily usage of the equipment is halved.
u = ? d= 2 * 10 = 20 h = 2/2 = 1
Substitute above values in eq 1
u = \(\frac{8 * 20}{1 * 1}\)= 160
% Increase =\(\frac{Change}{original Value}\) * 100
% Increase = \(\frac{160  20}{20 * 100}\) =\(\frac{140}{20}\)* 100 = 700
Ans:



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Re: The useful life of a certain piece of equipment is determine
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09 Dec 2018, 11:00
Quote: 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% My Solution :
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Re: The useful life of a certain piece of equipment is determine
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10 Dec 2018, 09:34
bruiz wrote: 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% \(u=\frac{(8d)}{h^2}\) Let initial values of \(d = 1\) and \(h = 1\) Then; \(u = \frac{8 * 1}{1^2} => 8\) Density is doubled; \(d = 2\) Daily usage of the equipment is halved; \(h = \frac{1}{2}\) New value of \(u = \frac{8*2}{(1/2)^2} = \frac{16}{(1/4)} =16*4 = 64\) Percentage Increase \(= \frac{648}{8}*100 = \frac{56}{8}*100 = 700\)% Answer D



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Re: The useful life of a certain piece of equipment is determine
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22 Apr 2019, 05:03
posting 2 approaches :
]original : \(\frac{8d}{h^2}\)
Now, d> 2d and h>\(\frac{h}{2}\)
substituting: \(\frac{8*2d}{h/2^2}\) > \(\frac{16d}{h^2/4}\)
on simplifying: \(\frac{64d}{h^2}\)
To calculate % increase: \(\frac{Newold}{old}\) X 100
new  old > \(\frac{64d}{h^2}\frac{8d}{h^2}\) = \(\frac{56d}{h^2}\)
now change / old X100 = \(\frac{56d}{h^2}\) X \(\frac{h^2}{8d}\) X100
which gives > 700%
by plugging values : original : \(U=\frac{8d}{h^2}\)
let, d=1 and h=2
\(U= \frac{8}{4}=2\)
After change: d=2 and h=1
\(U=\frac{8*2}{1}\) = 16
To calculate % increase: \(\frac{Newold}{old}\) X 100
change > 162 =14
% change> \(\frac{14}{2}\)X100 = 700%



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Re: The useful life of a certain piece of equipment is determine
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22 Apr 2019, 07:42
bruiz wrote: 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%
If the density of the underlying material is doubled and the daily usage of the equipment is halved, then the new useful life will be 8 times the existing useful life or there will be 700% percentage increase in the useful life of the equipment. Correct answer is D
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