There are some grey and red balls in a container. If 8 grey balls are removed from the container, the total number of balls left will be 8 times the number of grey balls left. If 20 red balls are removed from the container, the total number of balls left will be 4 times the number of grey balls. How many balls are there in the container?

	Case 1		Case 2
	Grey balls	Red balls	Grey balls	Red balls
Before	1 u + 8	7 u	1 p	3 p + 20
Change	- 8			- 20
After	1 u	7 u	1 p	3 p

Number of red balls in the end for Case 1
= 8 u - 1 u
= 7 u

Number of grey balls in the end for Case 2
= 4 p - 1 p
= 3 p

The number of grey balls at first is the same for Case 1 and Case 2.
The number of red balls at first is also the same for Case 1 and Case 2.

1 u + 8 = 1 p --- (1)

7 u = 3 p + 20
7 u - 20 = 3 p --- (2)

Make p the same.

(1)_x3
3 u + 24 = 3 p --- (3)

(2) = (3)
7 u - 20 = 3 u + 24
7 u - 3 u = 24 + 20
4 u = 44

1 u = 44 ÷ 4 = 11

Number of balls in the container
= (1 u + 8) + 7 u
= 8 u + 8
= (8 x 11) + 8
= 88 + 8
= 96

Answer(s): 96