The football fields at three different schools were measured, and it was found that their perimeters were 320 m, 360 m, and 400 m. if the sutdents at each school are to run the same distance, and this must be a whole number of laps, what is the shortest distance they need to run?

Let us say that,

A = perimeter of football field of the 1st school

B = perimeter of football field of the 2nd school

C = perimeter of football field of the 3rd school

So that:

A = 320 m

B = 360 m

C = 400 m

So the first thing we have to do to find for the minimum laps to make is to divide every perimeter by the smallest one, therefore divide all by 320 m, hence:

A’ = 1

B’ = 1.125

C’ = 1.25

Now multiply everything by a number such that all laps will be a whole number, in this case multiply everything by 8:

A’ = 8

B’ = 9

C’ = 10

Hence the laps to run are:

8, 9 and 10

Step-by-step explanation:

We have been given a sum . We are asked to rewrite this sum using distributive property and greatest common factor.

First of all, we will find greatest common factor of 72 and 60.

Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

We can see that the greatest common factor of 72 and 60 is 12.

We can rewrite our given sum as:

Upon factoring out 12, we will get:

Therefore, our required expression is .