There are some grey and brown balls in a box. If 6 grey balls are removed from the box, the total number of balls left will be 8 times the number of grey balls left. If 43 brown balls are removed from the box, the total number of balls left will be 3 times the number of grey balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Grey balls |
Brown balls |
Grey balls |
Brown balls |
Before |
1 u + 6 |
7 u |
1 p |
2 p + 43 |
Change |
- 6 |
|
|
- 43 |
After |
1 u |
7 u |
1 p |
2 p |
Number of brown balls in the end for Case 1
= 8 u - 1 u
= 7 u
Number of grey balls in the end for Case 2
= 3 p - 1 p
= 2 p
The number of grey 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 + 6 = 1 p --- (1)
7 u = 2 p + 43
7 u - 43 =
2 p --- (2)
Make p the same.
(1)
x2 2 u + 12 =
2 p --- (3)
(2) = (3)
7 u - 43 = 2 u + 12
7 u - 2 u = 12 + 43
5 u = 55
1 u = 55 ÷ 5 = 11
Number of balls in the box
= (1 u + 6) + 7 u
= 8 u + 6
= (8 x 11) + 6
= 88 + 6
= 94
Answer(s): 94