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

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On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Sep 2013, 10:38, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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New post 14 Oct 2005, 20:29
A) Decrease of 14.3%

Revolution is proportional to the circumference which is proportional to radius R
so change from monday to tue = (Rm - Rt)/Rm * 100 = 14.3
Since it is a bigger tyre on tuesday, the number of revolution decreases.

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New post 14 Oct 2005, 20:33
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Answer is A

Total revolutions (mon) = x miles / 28 inches
Total revolutions (tue) = x miles / 32 inches

Inc/Dec = diff/orig amount * 100
= 4/28 * 100
= 14.28
= 14.3%

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New post 14 Oct 2005, 21:12
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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.


1) Revolution is proportional to the circumference
2) The avg # of revolutions of 32-inch tires < the avg # of revolutions of 28-inch tires =====> Decrease % change
3) Pick #: x = 224
4) avg # of revolutions on Tuesday ==> 224/32 = 7
5) avg # of revolutions on Monday ==> 224/28 = 8
6) Decrease % change = (7-8)/8 = 12.5%

Ans: B (OA pls)

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New post 14 Oct 2005, 21:19
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Ooops. AMBA is correct.

I went wrong in my calculation:
Inc/Dec = diff/orig amount

= (x/28 - x/32)/ (x/28)
= (4x/28*32) / (x/28)
= (x/28*8) / (x/28)
= 1/8 * 100 %
= 12.5

I am confident that OA should be B

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New post 14 Oct 2005, 21:23
AMBA, you are right.
I made a mistake in my 1st step. :oops:
Revolution is inversely proportional to Radius. Less the radius, more the number of revolutions.
So it is -> (1/Rm - 1/Rt)/(1/Rm) * 100
= ((1/14)-(1/16))/(1/14)
= (2/(14*16))/(1/14) = 1/8 = 0.125 or 12.5 % decrease

Ans B

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New post 14 Oct 2005, 21:34
No worries...making mistakes are part of the learning process. 8-)

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New post 15 Oct 2005, 19:42
OA is B and AMBA got it first! :)

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Re: On Monday, Lou drives his ford escort with 28-inch tires, [#permalink]

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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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.

Good question.

Let's tackle this one

We are told that we have 28 inch tires as well as 32 inch tires on two different days. Hence, they will have a circumference of 28 and 32 respectively
Now, one is also told that the average speed is the same, and so is the distance traveled to work.

So the number of revolutions = distance traveled/circumference = x/32 - x/28 = -4/32 = 12.5% decrease

Or one can also pick a smart number for the distance , say for example 56*4 = 224 the LCM

On Tuesday 224/32 =7

On Monday 224/28 = 8

Therefore, 7-8/8 = -1/8=-12.5% decrease

Same result

Answer is B

Hope it helps!
Cheers!

J :)

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 04 Feb 2014, 09:58
Let D denote Distance
Radius of 14" and 16" respectively:

( D/(2)(16)Pi - D/(2)(14)Pi ) / ( D/(2)(14)Pi )

= - D(2)(14)Pi / D(2)(16)(7)Pi

= -1/8 or a decrease of 12.5%

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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No. of revolutions is inversely proportional to radius. The greater the radius, the lesser the number of revolutions. Since the distance traveled is the same, if in the first case it is 32x, then in the second case it is 28x. The change is( 4x/32x) * 100 = 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 19 Jun 2014, 04:37
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Ratio of tire diameter

= 28:32

= 7:8

More the tire diameter, less are the revolutions & vice versa

So, calculation would be in inverse proportion

Percentage decrease (-ve sign shows decrease)

\(= \frac{7-8}{8} * 100\)

\(= \frac{1}{8} * 100\)

= 12.5

Answer = A
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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 21 Aug 2014, 10:26
Dear AMBA,

What is the significance of "make per second" in the question? Does that have any relevance? Or is it just for distraction? Please do reply. Thanks

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 25 Nov 2014, 13:09
I have a silly question. why are we dividing by 32 and not by 28? shouldn't we have the original in the den?

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 14 Mar 2015, 12:41
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.



OA is B:

I tried it the following way:

Perimeter of the 28 inch tire, = 28*pi
perimeter of the 32 inch tire = 32*pi
I take the perimeter so that, the circular tire is made as straight line and this is travelling the required distance. to be more clear. 28inch and 32 inch tires are shown as follows:

------------------------- 28 inch tire perimeter
-------------------------------- 32 inch tire perimeter

speed = x miles in both cases. so, 28*pi*x and 32*pi*x is the only distinct factor in the calculation of rev of tire/sec or min or hour

taking the ratio
(28*pi*x/32/pi*x)*100 = 0.875*100 = 87.5
= 12.5% decrease

so B

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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santorasantu wrote:
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.



OA is B:

I tried it the following way:

Perimeter of the 28 inch tire, = 28*pi
perimeter of the 32 inch tire = 32*pi
I take the perimeter so that, the circular tire is made as straight line and this is travelling the required distance. to be more clear. 28inch and 32 inch tires are shown as follows:

------------------------- 28 inch tire perimeter
-------------------------------- 32 inch tire perimeter

speed = x miles in both cases. so, 28*pi*x and 32*pi*x is the only distinct factor in the calculation of rev of tire/sec or min or hour

taking the ratio
(28*pi*x/32/pi*x)*100 = 0.875*100 = 87.5
= 12.5% decrease

so B


There is a short cut and easy method to problems related to inverse proportion.

Since, Speed(Constant)= Number of revolution x Perimeter of tire 2pi R.

This problem is of type a*b=Constant. So if A is increase by 1/x, then B should be decrease by 1/(x+1)
Coming to this problem, size increase by 4/28=1/7. Hence number of revolution will be decrease by 1/8 or 12.5%

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 21 Aug 2015, 06:08
for 28 inch tyres: circumference= 28pi , hence rev for 28inch tires= x/28pi
for 32 inch tyres: circumference= 32pi , hence rev for 32inch tires= x/32pi

% dec= (x/28pi-x/32pi) / x/28pi
= x(1/28pi-1/32pi)*28pi / x
= 1-7/8= 1/8 or 12.5%

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Re: On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 18 Jan 2016, 14:32
Revolutions on Monday = N = X/28
Revolutions on Tuesday = M = X/32
Percent change = ((M - N)*100)/N = ((X/32 - X/28)*100)/ (X/28) = (28/32 - 1)*100 = -12.5%

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On Monday, Lou drives his ford escort with 28-inch tires [#permalink]

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New post 19 Jan 2016, 12:18
28/32=7/8=.875=87.5%
100%-87.5%=12.5% decrease

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On Monday, Lou drives his ford escort with 28-inch tires   [#permalink] 19 Jan 2016, 12:18

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