Question

Alice bought 18 toys, 48 books and 54 crayons to make boxes for a charity. She wants to use all of them in such a way that each box contains the same number of each item. What is the greatest number of boxes she can make? How many toys, books and crayons would be in each box?

Answer 1

There would be 6 boxes, each with 3 toys, 8 books and 9 crayons.

Explanation:

Find the greatest common factor of 18, 48 and 54 (the toys, books and crayons).

18 = 2*3*3

48 = 2*2*2*2*3

54 = 2*3*3*3

All of them have at least one 2 and one 3.

The greatest common factor is 6 because 2*3 is 6.

There will be 6 boxes.

The number of each item is the total items divided by number of boxes.

18/6 = 3

48/6 = 8

54/6 = 9

There would be 3 toys, 8 books and 9 crayons in each box.

Answer 2

There would be 6 boxes, each with 3 toys, 8 books and 9 crayons.

Explanation:

Find the greatest common factor of 18, 48 and 54 (the toys, books and crayons).

18 = 2*3*3

48 = 2*2*2*2*3

54 = 2*3*3*3

All of them have at least one 2 and one 3.

The greatest common factor is 6 because 2*3 is 6.

There will be 6 boxes.

The number of each item is the total items divided by number of boxes.

18/6 = 3

48/6 = 8

54/6 = 9

There would be 3 toys, 8 books and 9 crayons in each box.