There are some silver and blue balls in a box. If 2 silver balls are removed from the box, the total number of balls left will be 6 times the number of silver balls left. If 32 blue balls are removed from the box, the total number of balls left will be 3 times the number of silver balls. How many balls are there in the box?

	Case 1		Case 2
	Silver balls	Blue balls	Silver balls	Blue balls
Before	1 u + 2	5 u	1 p	2 p + 32
Change	- 2			- 32
After	1 u	5 u	1 p	2 p

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

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

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

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

5 u = 2 p + 32
5 u - 32 = 2 p --- (2)

Make p the same.

(1)_x2
2 u + 4 = 2 p --- (3)

(2) = (3)
5 u - 32 = 2 u + 4
5 u - 2 u = 4 + 32
3 u = 36

1 u = 36 ÷ 3 = 12

Number of balls in the box
= (1 u + 2) + 5 u
= 6 u + 2
= (6 x 12) + 2
= 72 + 2
= 74

Answer(s): 74