There are some gold and silver balls in a container. If 2 gold balls are removed from the container, the total number of balls left will be 9 times the number of gold balls left. If 38 silver balls are removed from the container, the total number of balls left will be 3 times the number of gold balls. How many balls are there in the container?

	Case 1		Case 2
	Gold balls	Silver balls	Gold balls	Silver balls
Before	1 u + 2	8 u	1 p	2 p + 38
Change	- 2			- 38
After	1 u	8 u	1 p	2 p

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

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

The number of gold 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)

8 u = 2 p + 38
8 u - 38 = 2 p --- (2)

Make p the same.

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

(2) = (3)
8 u - 38 = 2 u + 4
8 u - 2 u = 4 + 38
6 u = 42

1 u = 42 ÷ 6 = 7

Number of balls in the container
= (1 u + 2) + 8 u
= 9 u + 2
= (9 x 7) + 2
= 63 + 2
= 65

Answer(s): 65