There are some red and brown balls in a container. If 2 red balls are removed from the container, the total number of balls left will be 8 times the number of red balls left. If 4 brown 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	Brown balls	Red balls	Brown balls
Before	1 u + 2	7 u	1 p	5 p + 4
Change	- 2			- 4
After	1 u	7 u	1 p	5 p

Number of brown balls in the end for Case 1
= 8 u - 1 u
= 7 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 brown balls at first is also the same for Case 1 and Case 2.

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

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

Make p the same.

(1)_x5
5 u + 10 = 5 p --- (3)

(2) = (3)
7 u - 4 = 5 u + 10
7 u - 5 u = 10 + 4
2 u = 14

1 u = 14 ÷ 2 = 7

Number of balls in the container
= (1 u + 2) + 7 u
= 8 u + 2
= (8 x 7) + 2
= 56 + 2
= 58

Answer(s): 58