There are some silver and grey balls in a box. If 6 silver balls are removed from the box, the total number of balls left will be 9 times the number of silver balls left. If 8 grey balls are removed from the box, the total number of balls left will be 5 times the number of silver balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Silver balls |
Grey balls |
Silver balls |
Grey balls |
Before |
1 u + 6 |
8 u |
1 p |
4 p + 8 |
Change |
- 6 |
|
|
- 8 |
After |
1 u |
8 u |
1 p |
4 p |
Number of grey balls in the end for Case 1
= 9 u - 1 u
= 8 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 grey balls at first is also the same for Case 1 and Case 2.
1 u + 6 = 1 p --- (1)
8 u = 4 p + 8
8 u - 8 =
4 p --- (2)
Make p the same.
(1)
x4 4 u + 24 =
4 p --- (3)
(2) = (3)
8 u - 8 = 4 u + 24
8 u - 4 u = 24 + 8
4 u = 32
1 u = 32 ÷ 4 = 8
Number of balls in the box
= (1 u + 6) + 8 u
= 9 u + 6
= (9 x 8) + 6
= 72 + 6
= 78
Answer(s): 78