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

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)

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

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%

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

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

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%

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
_________________

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

I have a silly question. why are we dividing by 32 and not by 28? shouldn't we have the original in the den?

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%

28/32=7/8=.875=87.5%
100%-87.5%=12.5% decrease

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

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

same question. Please help

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.
_________________

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

Mathematically this is giving +ve sign and not -12.5, shouldn't it be increase?

We aren’t dividing by 32.

No of revolutions made on Monday is m= D/(2*Pi*28)


No of revolutions made on Tuesday is t= D/(2*Pi*32)

Hence % diff in revolutions made = (t-m)/m *100%

= t/m -1 = 28/32-1 = -4/32 = -1/8 = decrease of 12.5%

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