There are some silver and brown balls in a container. If 4 silver balls are removed from the container, the total number of balls left will be 7 times the number of silver balls left. If 15 brown balls are removed from the container, the total number of balls left will be 4 times the number of silver balls. How many balls are there in the container?

	Case 1		Case 2
	Silver balls	Brown balls	Silver balls	Brown balls
Before	1 u + 4	6 u	1 p	3 p + 15
Change	- 4			- 15
After	1 u	6 u	1 p	3 p

Number of brown balls in the end for Case 1
= 7 u - 1 u
= 6 u

Number of silver balls in the end for Case 2
= 4 p - 1 p
= 3 p

The number of silver 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 + 4 = 1 p --- (1)

6 u = 3 p + 15
6 u - 15 = 3 p --- (2)

Make p the same.

(1)_x3
3 u + 12 = 3 p --- (3)

(2) = (3)
6 u - 15 = 3 u + 12
6 u - 3 u = 12 + 15
3 u = 27

1 u = 27 ÷ 3 = 9

Number of balls in the container
= (1 u + 4) + 6 u
= 7 u + 4
= (7 x 9) + 4
= 63 + 4
= 67

Answer(s): 67