There are some silver and brown balls in a container. If 7 silver balls are removed from the container, the total number of balls left will be 8 times the number of silver balls left. If 11 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 + 7 |
7 u |
1 p |
3 p + 11 |
Change |
- 7 |
|
|
- 11 |
After |
1 u |
7 u |
1 p |
3 p |
Number of brown balls in the end for Case 1
= 8 u - 1 u
= 7 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 + 7 = 1 p --- (1)
7 u = 3 p + 11
7 u - 11 =
3 p --- (2)
Make p the same.
(1)
x3 3 u + 21 =
3 p --- (3)
(2) = (3)
7 u - 11 = 3 u + 21
7 u - 3 u = 21 + 11
4 u = 32
1 u = 32 ÷ 4 = 8
Number of balls in the container
= (1 u + 7) + 7 u
= 8 u + 7
= (8 x 8) + 7
= 64 + 7
= 71
Answer(s): 71