There are some red and white balls in a container. If 10 red balls are removed from the container, the total number of balls left will be 8 times the number of red balls left. If 6 white balls are removed from the container, the total number of balls left will be 4 times the number of red balls. How many balls are there in the container?
|
Case 1 |
Case 2 |
|
Red balls |
White balls |
Red balls |
White balls |
Before |
1 u + 10 |
7 u |
1 p |
3 p + 6 |
Change |
- 10 |
|
|
- 6 |
After |
1 u |
7 u |
1 p |
3 p |
Number of white balls in the end for Case 1
= 8 u - 1 u
= 7 u
Number of red balls in the end for Case 2
= 4 p - 1 p
= 3 p
The number of red balls at first is the same for Case 1 and Case 2.
The number of white balls at first is also the same for Case 1 and Case 2.
1 u + 10 = 1 p --- (1)
7 u = 3 p + 6
7 u - 6 =
3 p --- (2)
Make p the same.
(1)
x3 3 u + 30 =
3 p --- (3)
(2) = (3)
7 u - 6 = 3 u + 30
7 u - 3 u = 30 + 6
4 u = 36
1 u = 36 ÷ 4 = 9
Number of balls in the container
= (1 u + 10) + 7 u
= 8 u + 10
= (8 x 9) + 10
= 72 + 10
= 82
Answer(s): 82