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

	Case 1		Case 2
	Black balls	Green balls	Black balls	Green balls
Before	1 u + 5	5 u	1 p	3 p + 5
Change	- 5			- 5
After	1 u	5 u	1 p	3 p

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

Number of black balls in the end for Case 2
= 4 p - 1 p
= 3 p

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

1 u + 5 = 1 p --- (1)

5 u = 3 p + 5
5 u - 5 = 3 p --- (2)

Make p the same.

(1)_x3
3 u + 15 = 3 p --- (3)

(2) = (3)
5 u - 5 = 3 u + 15
5 u - 3 u = 15 + 5
2 u = 20

1 u = 20 ÷ 2 = 10

Number of balls in the container
= (1 u + 5) + 5 u
= 6 u + 5
= (6 x 10) + 5
= 60 + 5
= 65

Answer(s): 65