There are some silver and white balls in a box. If 2 silver balls are removed from the box, the total number of balls left will be 8 times the number of silver balls left. If 2 white balls are removed from the box, the total number of balls left will be 4 times the number of silver balls. How many balls are there in the box?

	Case 1		Case 2
	Silver balls	White balls	Silver balls	White balls
Before	1 u + 2	7 u	1 p	3 p + 2
Change	- 2			- 2
After	1 u	7 u	1 p	3 p

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

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

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

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

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

Make p the same.

(1)_x3
3 u + 6 = 3 p --- (3)

(2) = (3)
7 u - 2 = 3 u + 6
7 u - 3 u = 6 + 2
4 u = 8

1 u = 8 ÷ 4 = 2

Number of balls in the box
= (1 u + 2) + 7 u
= 8 u + 2
= (8 x 2) + 2
= 16 + 2
= 18

Answer(s): 18