There are some silver and gold balls in a container. If 2 silver balls are removed from the container, the total number of balls left will be 6 times the number of silver balls left. If 4 gold balls are removed from the container, the total number of balls left will be 5 times the number of silver balls. How many balls are there in the container?

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

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

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

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

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

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

Make p the same.

(1)_x4
4 u + 8 = 4 p --- (3)

(2) = (3)
5 u - 4 = 4 u + 8
5 u - 4 u = 8 + 4
1 u = 12

1 u = 12 ÷ 1 = 12

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

Answer(s): 74