There are some red and grey balls in a container. If 3 red balls are removed from the container, the total number of balls left will be 9 times the number of red balls left. If 18 grey balls are removed from the container, the total number of balls left will be 6 times the number of red balls. How many balls are there in the container?

	Case 1		Case 2
	Red balls	Grey balls	Red balls	Grey balls
Before	1 u + 3	8 u	1 p	5 p + 18
Change	- 3			- 18
After	1 u	8 u	1 p	5 p

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

Number of red balls in the end for Case 2
= 6 p - 1 p
= 5 p

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

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

8 u = 5 p + 18
8 u - 18 = 5 p --- (2)

Make p the same.

(1)_x5
5 u + 15 = 5 p --- (3)

(2) = (3)
8 u - 18 = 5 u + 15
8 u - 5 u = 15 + 18
3 u = 33

1 u = 33 ÷ 3 = 11

Number of balls in the container
= (1 u + 3) + 8 u
= 9 u + 3
= (9 x 11) + 3
= 99 + 3
= 102

Answer(s): 102