There are some white and brown balls in a container. If 6 white balls are removed from the container, the total number of balls left will be 7 times the number of white balls left. If 28 brown balls are removed from the container, the total number of balls left will be 3 times the number of white balls. How many balls are there in the container?
|
Case 1 |
Case 2 |
|
White balls |
Brown balls |
White balls |
Brown balls |
Before |
1 u + 6 |
6 u |
1 p |
2 p + 28 |
Change |
- 6 |
|
|
- 28 |
After |
1 u |
6 u |
1 p |
2 p |
Number of brown balls in the end for Case 1
= 7 u - 1 u
= 6 u
Number of white balls in the end for Case 2
= 3 p - 1 p
= 2 p
The number of white 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)
6 u = 2 p + 28
6 u - 28 =
2 p --- (2)
Make p the same.
(1)
x2 2 u + 12 =
2 p --- (3)
(2) = (3)
6 u - 28 = 2 u + 12
6 u - 2 u = 12 + 28
4 u = 40
1 u = 40 ÷ 4 = 10
Number of balls in the container
= (1 u + 6) + 6 u
= 7 u + 6
= (7 x 10) + 6
= 70 + 6
= 76
Answer(s): 76