There are some black and silver balls in a box. If 8 black balls are removed from the box, the total number of balls left will be 8 times the number of black balls left. If 12 silver balls are removed from the box, the total number of balls left will be 4 times the number of black balls. How many balls are there in the box?

	Case 1		Case 2
	Black balls	Silver balls	Black balls	Silver balls
Before	1 u + 8	7 u	1 p	3 p + 12
Change	- 8			- 12
After	1 u	7 u	1 p	3 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
= 4 p - 1 p
= 3 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 + 8 = 1 p --- (1)

7 u = 3 p + 12
7 u - 12 = 3 p --- (2)

Make p the same.

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

(2) = (3)
7 u - 12 = 3 u + 24
7 u - 3 u = 24 + 12
4 u = 36

1 u = 36 ÷ 4 = 9

Number of balls in the box
= (1 u + 8) + 7 u
= 8 u + 8
= (8 x 9) + 8
= 72 + 8
= 80

Answer(s): 80