There are some red and blue balls in a container. If 3 red balls are removed from the container, the total number of balls left will be 8 times the number of red balls left. If 5 blue balls are removed from the container, the total number of balls left will be 6 times the number of red balls. How many balls are there in the container?
|
Case 1 |
Case 2 |
|
Red balls |
Blue balls |
Red balls |
Blue balls |
Before |
1 u + 3 |
7 u |
1 p |
5 p + 5 |
Change |
- 3 |
|
|
- 5 |
After |
1 u |
7 u |
1 p |
5 p |
Number of blue balls in the end for Case 1
= 8 u - 1 u
= 7 u
Number of red balls in the end for Case 2
= 6 p - 1 p
= 5 p
The number of red balls at first is the same for Case 1 and Case 2.
The number of blue balls at first is also the same for Case 1 and Case 2.
1 u + 3 = 1 p --- (1)
7 u = 5 p + 5
7 u - 5 =
5 p --- (2)
Make p the same.
(1)
x5 5 u + 15 =
5 p --- (3)
(2) = (3)
7 u - 5 = 5 u + 15
7 u - 5 u = 15 + 5
2 u = 20
1 u = 20 ÷ 2 = 10
Number of balls in the container
= (1 u + 3) + 7 u
= 8 u + 3
= (8 x 10) + 3
= 80 + 3
= 83
Answer(s): 83