There are some green and silver balls in a box. If 4 green balls are removed from the box, the total number of balls left will be 6 times the number of green balls left. If 2 silver balls are removed from the box, the total number of balls left will be 4 times the number of green balls. How many balls are there in the box?

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

Number of silver balls in the end for Case 1
= 6 u - 1 u
= 5 u

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

The number of green 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 + 4 = 1 p --- (1)

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

Make p the same.

(1)_x3
3 u + 12 = 3 p --- (3)

(2) = (3)
5 u - 2 = 3 u + 12
5 u - 3 u = 12 + 2
2 u = 14

1 u = 14 ÷ 2 = 7

Number of balls in the box
= (1 u + 4) + 5 u
= 6 u + 4
= (6 x 7) + 4
= 42 + 4
= 46

Answer(s): 46