There are some grey and blue balls in a container. If 5 grey balls are removed from the container, the total number of balls left will be 8 times the number of grey balls left. If 5 blue balls are removed from the container, the total number of balls left will be 3 times the number of grey balls. How many balls are there in the container?

	Case 1		Case 2
	Grey balls	Blue balls	Grey balls	Blue balls
Before	1 u + 5	7 u	1 p	2 p + 5
Change	- 5			- 5
After	1 u	7 u	1 p	2 p

Number of blue balls in the end for Case 1
= 8 u - 1 u
= 7 u

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

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

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

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

Make p the same.

(1)_x2
2 u + 10 = 2 p --- (3)

(2) = (3)
7 u - 5 = 2 u + 10
7 u - 2 u = 10 + 5
5 u = 15

1 u = 15 ÷ 5 = 3

Number of balls in the container
= (1 u + 5) + 7 u
= 8 u + 5
= (8 x 3) + 5
= 24 + 5
= 29

Answer(s): 29