Question

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.

Answer 1

Answer:

$\% Change = \frac{|28-32|}{32} *100 = 12.5\%$

(B) Decrease by 12.5%

Step-by-step explanation:

For this case we know that the revolution is proportional to the circumference.

And we know that the average number of revolutions of 32 inch tires for Tuesday is higher than the original value of 28 inch tires for Monday.

We know that we have x mi/hr, so we can select a value fo x in order to find the average revolutions with the following formula:

$Avg = \frac{x}{mi}$

Let's say that we select a value for x for example x= 28*32 = 896, since this value is divisible by 32 and 28.

If we find the average revolutions per each case we got:

Tuesday:

$Avg = \frac{896}{32}=28$

Monday:

$Avg = \frac{896}{28}=32$

And then we can find the % of change like this:

$\% Change= \frac{|Final-Initial|}{Initial} *100$

And if we replace we got:

$\% Change = \frac{|28-32|}{32} *100 = 12.5\%$

Because we are assuming that the initial amount is the value for Monday and the final value for Tuesday.

So then the best answer for this case would be:

(B) Decrease by 12.5%