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On Monday, Lou drives his ford escort with 28-inch tires

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Status: Target 760
Joined: 20 Aug 2014
Posts: 63
Location: India
Concentration: Strategy, Economics
GMAT 1: 670 Q50 V30
GPA: 3.25
WE: Corporate Finance (Investment Banking)
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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 04 Nov 2017, 06:36
laxieqv wrote:
On Monday, Lou drives his ford escort with 28-inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32-inch tires 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 make per second?

(A) Decrease by 14.3%
(B) Decrease by 12.5%
(C) Increase by 14.3%
(D) Increase by 12.5%
(E) Cannot be determined with the given information.



Basic common sense pertaining to numbers will make you find this Q's answer in a minute.

Tire's 1 revolution is equal its circumference,isnt it? speed is constant so you can ignore. Now, lets keep a distance same- 224 (multiple of 28 and 32)

28 inch tire will roll for 8 times in order to travel a distance of 224; 32 inch will take 7 revolutions.

DIfference- 8-7=1

therefore, percentage change is 1/8= 12.5% decrease
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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 06 Nov 2017, 10:46
the number of revolutions is INVERSELY proportional to the circumference:

Number of revolutions ~ 1/circumference
circumference is directly proportional to the radius

so if circumference increases by 14.28% , the number of revolutions decreases and becomes 1/1.1428 which when deducted from 1: 1 - 1/1.1428 = almost 1/8= 12.5% decrease
Re: On Monday, Lou drives his ford escort with 28-inch tires   [#permalink] 06 Nov 2017, 10:46

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On Monday, Lou drives his ford escort with 28-inch tires

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