There are some red and gold balls in a container. If 7 red balls are removed from the container, the total number of balls left will be 8 times the number of red balls left. If 5 gold balls are removed from the container, the total number of balls left will be 5 times the number of red balls. How many balls are there in the container?

	Case 1		Case 2
	Red balls	Gold balls	Red balls	Gold balls
Before	1 u + 7	7 u	1 p	4 p + 5
Change	- 7			- 5
After	1 u	7 u	1 p	4 p

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

Number of red balls in the end for Case 2
= 5 p - 1 p
= 4 p

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

1 u + 7 = 1 p --- (1)

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

Make p the same.

(1)_x4
4 u + 28 = 4 p --- (3)

(2) = (3)
7 u - 5 = 4 u + 28
7 u - 4 u = 28 + 5
3 u = 33

1 u = 33 ÷ 3 = 11

Number of balls in the container
= (1 u + 7) + 7 u
= 8 u + 7
= (8 x 11) + 7
= 88 + 7
= 95

Answer(s): 95