There are some silver and blue balls in a container. If 2 silver balls are removed from the container, the total number of balls left will be 7 times the number of silver balls left. If 14 blue 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 |
Blue balls |
Silver balls |
Blue balls |
Before |
1 u + 2 |
6 u |
1 p |
4 p + 14 |
Change |
- 2 |
|
|
- 14 |
After |
1 u |
6 u |
1 p |
4 p |
Number of blue 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 blue balls at first is also the same for Case 1 and Case 2.
1 u + 2 = 1 p --- (1)
6 u = 4 p + 14
6 u - 14 =
4 p --- (2)
Make p the same.
(1)
x4 4 u + 8 =
4 p --- (3)
(2) = (3)
6 u - 14 = 4 u + 8
6 u - 4 u = 8 + 14
2 u = 22
1 u = 22 ÷ 2 = 11
Number of balls in the container
= (1 u + 2) + 6 u
= 7 u + 2
= (7 x 11) + 2
= 77 + 2
= 79
Answer(s): 79