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On Monday, Lou drives his ford escort with 28inch tires
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Updated on: 29 Sep 2013, 10:38
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On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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.
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Originally posted by laxieqv on 14 Oct 2005, 20:17.
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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14 Oct 2005, 21:12
laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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 Louâ€™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 32inch tires < the avg # of revolutions of 28inch 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 = (78)/8 = 12.5%
Ans: B (OA pls)




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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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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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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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Joined: 21 Aug 2005
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AMBA, you are right.
I made a mistake in my 1st step.
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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No worries...making mistakes are part of the learning process.



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Re: On Monday, Lou drives his ford escort with 28inch tires
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31 Dec 2013, 06:33
laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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, 78/8 = 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 28inch tires
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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 28inch tires
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16 Jun 2014, 19:07
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 28inch tires
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19 Jun 2014, 04:37
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)
\(= \frac{78}{8} * 100\)
\(= \frac{1}{8} * 100\)
= 12.5
Answer = A



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Re: On Monday, Lou drives his ford escort with 28inch tires
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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 28inch tires
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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 28inch tires
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14 Mar 2015, 12:41
laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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 28inch tires
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16 Mar 2015, 16:56
santorasantu wrote: laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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 28inch tires
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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/28pix/32pi) / x/28pi = x(1/28pi1/32pi)*28pi / x = 17/8= 1/8 or 12.5%



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Re: On Monday, Lou drives his ford escort with 28inch tires
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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 28inch tires
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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 28inch tires
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20 Jul 2016, 23:22
laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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. Avg. speed old = avg. speed new total distance old/ total time old = total distance new/ total time new Now distance in terms of tire radius = 2*pi*r X no. of revolutions So now 2 * pi * rold * no.ofrevs old / total time = 2 * pi* rnew * no.ofrevs new/ total time no. of revs/ total time = no. of revs per second. canceling common terms rold * nrevsold = rnew * nrevsnew or nnew/ nold = 28/32 = 7/8 So has reduced by 1/8 = 12.25%



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Re: On Monday, Lou drives his ford escort with 28inch tires
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19 Sep 2017, 04:06
laxieqv wrote: On Monday, Lou drives his ford escort with 28inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32inch 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. 32 inch and 28 inch must clarify that they are circumference. They could be the width too.




Re: On Monday, Lou drives his ford escort with 28inch tires
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19 Sep 2017, 04:06



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