There are some red and black balls in a container. If 4 red balls are removed from the container, the total number of balls left will be 7 times the number of red balls left. If 8 black balls are removed from the container, the total number of balls left will be 5 times the number of red balls. How many balls are there in the container?

	Case 1		Case 2
	Red balls	Black balls	Red balls	Black balls
Before	1 u + 4	6 u	1 p	4 p + 8
Change	- 4			- 8
After	1 u	6 u	1 p	4 p

Number of black balls in the end for Case 1
= 7 u - 1 u
= 6 u

Number of red balls in the end for Case 2
= 5 p - 1 p
= 4 p

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

1 u + 4 = 1 p --- (1)

6 u = 4 p + 8
6 u - 8 = 4 p --- (2)

Make p the same.

(1)_x4
4 u + 16 = 4 p --- (3)

(2) = (3)
6 u - 8 = 4 u + 16
6 u - 4 u = 16 + 8
2 u = 24

1 u = 24 ÷ 2 = 12

Number of balls in the container
= (1 u + 4) + 6 u
= 7 u + 4
= (7 x 12) + 4
= 84 + 4
= 88

Answer(s): 88