There are some white and silver balls in a box. If 2 white balls are removed from the box, the total number of balls left will be 5 times the number of white balls left. If 5 silver balls are removed from the box, the total number of balls left will be 4 times the number of white balls. How many balls are there in the box?

	Case 1		Case 2
	White balls	Silver balls	White balls	Silver balls
Before	1 u + 2	4 u	1 p	3 p + 5
Change	- 2			- 5
After	1 u	4 u	1 p	3 p

Number of silver balls in the end for Case 1
= 5 u - 1 u
= 4 u

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

The number of white 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)

4 u = 3 p + 5
4 u - 5 = 3 p --- (2)

Make p the same.

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

(2) = (3)
4 u - 5 = 3 u + 6
4 u - 3 u = 6 + 5
1 u = 11

1 u = 11 ÷ 1 = 11

Number of balls in the box
= (1 u + 2) + 4 u
= 5 u + 2
= (5 x 11) + 2
= 55 + 2
= 57

Answer(s): 57