There are some black and grey balls in a box. If 9 black balls are removed from the box, the total number of balls left will be 7 times the number of black balls left. If 26 grey balls are removed from the box, the total number of balls left will be 3 times the number of black balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Black balls |
Grey balls |
Black balls |
Grey balls |
Before |
1 u + 9 |
6 u |
1 p |
2 p + 26 |
Change |
- 9 |
|
|
- 26 |
After |
1 u |
6 u |
1 p |
2 p |
Number of grey balls in the end for Case 1
= 7 u - 1 u
= 6 u
Number of black balls in the end for Case 2
= 3 p - 1 p
= 2 p
The number of black 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 + 9 = 1 p --- (1)
6 u = 2 p + 26
6 u - 26 =
2 p --- (2)
Make p the same.
(1)
x2 2 u + 18 =
2 p --- (3)
(2) = (3)
6 u - 26 = 2 u + 18
6 u - 2 u = 18 + 26
4 u = 44
1 u = 44 ÷ 4 = 11
Number of balls in the box
= (1 u + 9) + 6 u
= 7 u + 9
= (7 x 11) + 9
= 77 + 9
= 86
Answer(s): 86