There are some grey and white balls in a container. If 4 grey balls are removed from the container, the total number of balls left will be 7 times the number of grey balls left. If 12 white 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	White balls	Grey balls	White balls
Before	1 u + 4	6 u	1 p	2 p + 12
Change	- 4			- 12
After	1 u	6 u	1 p	2 p

Number of white balls in the end for Case 1
= 7 u - 1 u
= 6 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 white balls at first is also the same for Case 1 and Case 2.

1 u + 4 = 1 p --- (1)

6 u = 2 p + 12
6 u - 12 = 2 p --- (2)

Make p the same.

(1)_x2
2 u + 8 = 2 p --- (3)

(2) = (3)
6 u - 12 = 2 u + 8
6 u - 2 u = 8 + 12
4 u = 20

1 u = 20 ÷ 4 = 5

Number of balls in the container
= (1 u + 4) + 6 u
= 7 u + 4
= (7 x 5) + 4
= 35 + 4
= 39

Answer(s): 39