There are some brown and green balls in a box. If 6 brown balls are removed from the box, the total number of balls left will be 6 times the number of brown balls left. If 2 green balls are removed from the box, the total number of balls left will be 4 times the number of brown balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Brown balls |
Green balls |
Brown balls |
Green balls |
Before |
1 u + 6 |
5 u |
1 p |
3 p + 2 |
Change |
- 6 |
|
|
- 2 |
After |
1 u |
5 u |
1 p |
3 p |
Number of green balls in the end for Case 1
= 6 u - 1 u
= 5 u
Number of brown balls in the end for Case 2
= 4 p - 1 p
= 3 p
The number of brown balls at first is the same for Case 1 and Case 2.
The number of green balls at first is also the same for Case 1 and Case 2.
1 u + 6 = 1 p --- (1)
5 u = 3 p + 2
5 u - 2 =
3 p --- (2)
Make p the same.
(1)
x3 3 u + 18 =
3 p --- (3)
(2) = (3)
5 u - 2 = 3 u + 18
5 u - 3 u = 18 + 2
2 u = 20
1 u = 20 ÷ 2 = 10
Number of balls in the box
= (1 u + 6) + 5 u
= 6 u + 6
= (6 x 10) + 6
= 60 + 6
= 66
Answer(s): 66