There are some silver and red balls in a container. If 9 silver balls are removed from the container, the total number of balls left will be 9 times the number of silver balls left. If 18 red balls are removed from the container, the total number of balls left will be 3 times the number of silver balls. How many balls are there in the container?

	Case 1		Case 2
	Silver balls	Red balls	Silver balls	Red balls
Before	1 u + 9	8 u	1 p	2 p + 18
Change	- 9			- 18
After	1 u	8 u	1 p	2 p

Number of red balls in the end for Case 1
= 9 u - 1 u
= 8 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 red balls at first is also the same for Case 1 and Case 2.

1 u + 9 = 1 p --- (1)

8 u = 2 p + 18
8 u - 18 = 2 p --- (2)

Make p the same.

(1)_x2
2 u + 18 = 2 p --- (3)

(2) = (3)
8 u - 18 = 2 u + 18
8 u - 2 u = 18 + 18
6 u = 36

1 u = 36 ÷ 6 = 6

Number of balls in the container
= (1 u + 9) + 8 u
= 9 u + 9
= (9 x 6) + 9
= 54 + 9
= 63

Answer(s): 63