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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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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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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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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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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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laxieqv
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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laxieqv
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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28/32=7/8=.875=87.5%
100%-87.5%=12.5% decrease
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laxieqv
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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First let us understand the concept behind the question.
For example, If 100 is increased by 25% (ie.,\(\frac{1}{4}\)) followed by a decrease of 20% (ie.,\(\frac{1}{5}\)) we get 100. In general if any number is increased by a factor of \(\frac{1}{N}\) then it should be decreased by a factor of \(\frac{1}{N+1}\). This is the relationship between the increase factor and the decrease factor.

Coming to the question, we need to know a formula that is, DISTANCE TRAVELED BY CIRCULAR WHEEL (D) = CIRCUMFERENCE OF WHEEL (C) * NUMBER OF REVOLUTIONS (N).

ie., N = D/C

which means N and C are inversely proportional. As N increases, C decreases to maintain the same distance.

Given in the question, Circumference increase from 28 to 32, which represent an increase of \(\frac{1}{7}\) (ie., \(\frac{32-28}{28}\)). So there should be a decrease of \(\frac{1}{8}\) in the number of revolutions to keep the distance constant.

\(\frac{1}{8}\) = 12.5%

Answer Option: B
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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.

The circumference of 28-inch tire = \(28\pi r\).
The circumference of 32-inch tire = \(32\pi r\).

The number of revolutions made by 28-inch tire = \(\frac{d}{28\pi r}\).
The number of revolutions made by 32-inch tire = \(\frac{d}{32\pi r}\).

The percent change from Monday to Tuesday in the average number of revolutions =

\(\frac{change}{original} = \)

\(= \frac{\frac{d}{28\pi r} - \frac{d}{32\pi r}}{\frac{d}{28\pi r}} = \)

\(= \frac{\frac{1}{28\pi r} - \frac{1}{32\pi r}}{\frac{1}{28\pi r}} =\)

\(= (\frac{1}{28\pi r} - \frac{1}{32\pi r})*28\pi r =\)

\(=1-\frac{7}{8}=\)

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

\(=0.125\)

Answer: B.

Hope it helps.
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laxieqv
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.

Monday - he drives with 28 inch tires
Tuesday - he drives with 32 inch tires
and yet we're told that average speed is the same.
So percentage change has to be a decrease - so we're only looking at A and B
Pick LCM of 28 and 32 ie 224
Monday : 224/28 = 8
Tuesday : 224/32 = 7

Percentage change: 8-7/8 % = 1/8% = 12.5%
Therefore B
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Hi KarishmaB , could you please help in understanding what's the significance of the speed x miles per hour here and the fact that he has the same average speed on Monday. Also in all the above solutions, why didn't we take into account the fact that we needed the percent change in the average number of revolutions ''per second''?
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laxieqv
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.

Miles per hour is same on both days.

Tire circumference * Number of revolutions on Monday = Tire circumference * Number of revolutions on Tuesday

\(\pi * 28 * \text{Number of revolutions on Monday} = \pi * 32 * \text{Number of revolutions on Tuesday}\)

\(\text{Number of revolutions on Tuesday}/ \text{Number of revolutions on Monday} = \frac{7}{8}\)

So number of revolutions on Tuesday are 1/8th fewer i.e. 12.5% less

Answer (B)
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Natansha
Hi KarishmaB , could you please help in understanding what's the significance of the speed x miles per hour here and the fact that he has the same average speed on Monday. Also in all the above solutions, why didn't we take into account the fact that we needed the percent change in the average number of revolutions ''per second''?

x miles/hour is given to clarify that miles driven in an hour are the same - to clarify what they mean by average speed. Say another measure could be number of revolutions per hour could be another measure of speed in this context.

The units on both sides of the equation are the same so the units all get cancelled off. Hence, they become irrelevant.
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