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Speed & Distance Problem 5

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by aditiniyer » Thu Jan 26, 2017 6:07 am
On Monday Luo drives her ford escort with 28 inch tires, averaging x miles per hour. On tuesday Lou switches the tire of her car to 32 inches tire yet drives to work at the same average speed as on monday. What is the percent change from monday to tuesday in the average number of revolutions that lou's tires makes per second?
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by GMATGuruNY » Thu Jan 26, 2017 6:26 am
aditiniyer wrote:On Monday Luo drives her ford escort with 28 inch tires, averaging x miles per hour. On tuesday Lou switches the tire of her car to 32 inches tire yet drives to work at the same average speed as on monday. What is the percent change from monday to tuesday in the average number of revolutions that lou's tires makes per second?
The GMAT would not expect us to know how tires are measured.
The given measurements represent the DIAMETERS of the two tires.

Circumference of each 28-inch tire = πd ≈ 3*4*7 inches.
Circumference of each 32-inch tire = πd ≈ 3*4*8 inches.

Let the distance = 3*4*7*8 inches.
Number of revolutions made by each 28-inch tire = (distance)/(circumference) = (3*4*7*8)/(3*4*7) = 8.
Number of revolutions made by each 32-inch tire = (distance)/(circumference) = (3*4*7*8)/(3*4*8) = 7.
Percent change from 8 to 7 = (difference)/(original) * 100 = (8-7)/8 * 100 = 25/2 = [spoiler]12.5%[/spoiler].
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