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j_shreyans: Legendary Member; Posts: 510; Joined: Thu Aug 07, 2014 2:24 am; Thanked: 3 times; Followed by:5 members

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by j_shreyans » Mon Oct 20, 2014 11:23 am

Ben is driving on the highway at x miles per hour ( one miles equals 5280 feet) Ben's tires have a circumference of y feet which of the following expressions gives the number of revolutions each wheel turns in one hour.

A)5280(x/y)

B)5280(y/x)

C)5280(xy)

D)5280/(xy)

E)(xy)/5280

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Source: Beat The GMAT — Problem Solving | Mon Oct 20, 2014 11:23 am

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by [email protected] » Mon Oct 20, 2014 11:57 am

HI J_shreyans,

This question is perfect for TESTing VALUES, but once you get to a certain point in your work, you won't have to check every answer - you should be able to recognize a pattern and save some time.

We have a car driving at X miles per hour that has tires with a CIRCUMFERENCE of Y feet each. The fact that the variable is for the circumference actually saves us a number of "math steps." Here's why.

Let's TEST....
X = 2
Y = 3

The car is traveling 2 miles per hour, which means it travels 2(5280) feet per hour. If you'd like to calculate that, then that's fine, but it's not really necessary. Each tire has a circumference of 3 feet, meaning that for each complete revolution/rotation, the tire has moved 3 feet.

So, the car traveled 2(5280) feet and each rotation is 3 feet. This gives us....

2(5280)/3

Visually, this means that the "X" is in the numerator and the "Y" is in the denominator. The correct answer is the only one that fits that pattern.

Final Answer: A

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Rich

Contact Rich at [email protected]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Oct 20, 2014 12:46 pm

j_shreyans wrote:Ben is driving on the highway at x miles per hour ( one miles equals 5280 feet) Ben's tires have a circumference of y feet which of the following expressions gives the number of revolutions each wheel turns in one hour.

A)5280(x/y)

B)5280(y/x)

C)5280(xy)

D)5280/(xy)

E)(xy)/5280

Let y = 10560 feet.
Since 10560 feet = 2 miles, the circumference of each tire = 2 miles.
Let x = 2 miles per hour.
Since Ben is traveling at 2 miles per hour, and the circumference of each tire = 2 miles, the number of revolutions per hour = 1. This is our target.
Now plug x=2 and y=10560 into the answers to see which yields our target of 1.
Only A works:
5280(x/y) = 5280(2/10560) = 1.

The correct answer is A.

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by Brent@GMATPrepNow » Mon Oct 27, 2014 5:21 pm

j_shreyans wrote:Ben is driving on the highway at x miles per hour ( one miles equals 5280 feet) Ben's tires have a circumference of y feet which of the following expressions gives the number of revolutions each wheel turns in one hour.

A)5280(x/y)
B)5280(y/x)
C)5280(xy)
D)5280/(xy)
E)(xy)/5280

Here's an algebraic approach:

Each tire has a circumference of y feet.
So, there will be 1 revolution every time the car travels y feet.
5280/y = the number of times y divides into 5280.
So, if the car travels 5280 feet (aka 1 mile) in one hour, then there will be 5280/y revolutions.
If the car travels (x)(5280) feet (aka x miles) in one hour, then there will be (x)(5280/y) revolutions.
(x)(5280/y) = 5280x/y = [spoiler]5280(x/y)[/spoiler]

Answer: A

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Brent

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by Scott@TargetTestPrep » Tue Dec 12, 2017 6:30 am

j_shreyans wrote:Ben is driving on the highway at x miles per hour ( one miles equals 5280 feet) Ben's tires have a circumference of y feet which of the following expressions gives the number of revolutions each wheel turns in one hour.

A)5280(x/y)

B)5280(y/x)

C)5280(xy)

D)5280/(xy)

E)(xy)/5280

We are given that Ben is driving at an average rate of x mph and we need to determine how many revolutions each wheel turns in one hour.

Since 1 mile = 5,280 feet, Ben drives 5280x feet in one hour.

Thus, the number of revolutions is distance/circumference = (5,280x)/y = 5280(x/y).

Answer: A

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