There are some green and yellow balls in a container. If 6 green balls are removed from the container, the total number of balls left will be 8 times the number of green balls left. If 2 yellow balls are removed from the container, the total number of balls left will be 4 times the number of green balls. How many balls are there in the container?
|
Case 1 |
Case 2 |
|
Green balls |
Yellow balls |
Green balls |
Yellow balls |
Before |
1 u + 6 |
7 u |
1 p |
3 p + 2 |
Change |
- 6 |
|
|
- 2 |
After |
1 u |
7 u |
1 p |
3 p |
Number of yellow balls in the end for Case 1
= 8 u - 1 u
= 7 u
Number of green balls in the end for Case 2
= 4 p - 1 p
= 3 p
The number of green balls at first is the same for Case 1 and Case 2.
The number of yellow balls at first is also the same for Case 1 and Case 2.
1 u + 6 = 1 p --- (1)
7 u = 3 p + 2
7 u - 2 =
3 p --- (2)
Make p the same.
(1)
x3 3 u + 18 =
3 p --- (3)
(2) = (3)
7 u - 2 = 3 u + 18
7 u - 3 u = 18 + 2
4 u = 20
1 u = 20 ÷ 4 = 5
Number of balls in the container
= (1 u + 6) + 7 u
= 8 u + 6
= (8 x 5) + 6
= 40 + 6
= 46
Answer(s): 46