Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

22 Sep 2010, 05:25

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

64% (01:49) correct
36% (01:03) wrong based on 165 sessions

HideShow timer Statistics

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]

Show Tags

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

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%

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%.

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]

Show Tags

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%.

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

Show Tags

17 Nov 2014, 09:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

21 Jan 2016, 12:23

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: \(u=\frac{2*8d}{(\frac{h}{2})^2}\)

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...