There are some silver and yellow balls in a container. If 3 silver balls are removed from the container, the total number of balls left will be 7 times the number of silver balls left. If 8 yellow balls are removed from the container, the total number of balls left will be 5 times the number of silver balls. How many balls are there in the container?
|
Case 1 |
Case 2 |
|
Silver balls |
Yellow balls |
Silver balls |
Yellow balls |
Before |
1 u + 3 |
6 u |
1 p |
4 p + 8 |
Change |
- 3 |
|
|
- 8 |
After |
1 u |
6 u |
1 p |
4 p |
Number of yellow balls in the end for Case 1
= 7 u - 1 u
= 6 u
Number of silver balls in the end for Case 2
= 5 p - 1 p
= 4 p
The number of silver balls at first is the same for Case 1 and Case 2.
The number of yellow balls at first is also the same for Case 1 and Case 2.
1 u + 3 = 1 p --- (1)
6 u = 4 p + 8
6 u - 8 =
4 p --- (2)
Make p the same.
(1)
x4 4 u + 12 =
4 p --- (3)
(2) = (3)
6 u - 8 = 4 u + 12
6 u - 4 u = 12 + 8
2 u = 20
1 u = 20 ÷ 2 = 10
Number of balls in the container
= (1 u + 3) + 6 u
= 7 u + 3
= (7 x 10) + 3
= 70 + 3
= 73
Answer(s): 73