There are some black and silver balls in a container. If 2 black balls are removed from the container, the total number of balls left will be 8 times the number of black balls left. If 22 silver balls are removed from the container, the total number of balls left will be 5 times the number of black balls. How many balls are there in the container?

	Case 1		Case 2
	Black balls	Silver balls	Black balls	Silver balls
Before	1 u + 2	7 u	1 p	4 p + 22
Change	- 2			- 22
After	1 u	7 u	1 p	4 p

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

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

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

1 u + 2 = 1 p --- (1)

7 u = 4 p + 22
7 u - 22 = 4 p --- (2)

Make p the same.

(1)_x4
4 u + 8 = 4 p --- (3)

(2) = (3)
7 u - 22 = 4 u + 8
7 u - 4 u = 8 + 22
3 u = 30

1 u = 30 ÷ 3 = 10

Number of balls in the container
= (1 u + 2) + 7 u
= 8 u + 2
= (8 x 10) + 2
= 80 + 2
= 82

Answer(s): 82