There are some red and silver balls in a box. If 3 red balls are removed from the box, the total number of balls left will be 9 times the number of red balls left. If 4 silver balls are removed from the box, the total number of balls left will be 5 times the number of red balls. How many balls are there in the box?

	Case 1		Case 2
	Red balls	Silver balls	Red balls	Silver balls
Before	1 u + 3	8 u	1 p	4 p + 4
Change	- 3			- 4
After	1 u	8 u	1 p	4 p

Number of silver balls in the end for Case 1
= 9 u - 1 u
= 8 u

Number of red balls in the end for Case 2
= 5 p - 1 p
= 4 p

The number of red 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 + 3 = 1 p --- (1)

8 u = 4 p + 4
8 u - 4 = 4 p --- (2)

Make p the same.

(1)_x4
4 u + 12 = 4 p --- (3)

(2) = (3)
8 u - 4 = 4 u + 12
8 u - 4 u = 12 + 4
4 u = 16

1 u = 16 ÷ 4 = 4

Number of balls in the box
= (1 u + 3) + 8 u
= 9 u + 3
= (9 x 4) + 3
= 36 + 3
= 39

Answer(s): 39