There are some black and white balls in a box. If 6 black balls are removed from the box, the total number of balls left will be 8 times the number of black balls left. If 38 white balls are removed from the box, the total number of balls left will be 3 times the number of black balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Black balls |
White balls |
Black balls |
White balls |
Before |
1 u + 6 |
7 u |
1 p |
2 p + 38 |
Change |
- 6 |
|
|
- 38 |
After |
1 u |
7 u |
1 p |
2 p |
Number of white balls in the end for Case 1
= 8 u - 1 u
= 7 u
Number of black balls in the end for Case 2
= 3 p - 1 p
= 2 p
The number of black 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 + 6 = 1 p --- (1)
7 u = 2 p + 38
7 u - 38 =
2 p --- (2)
Make p the same.
(1)
x2 2 u + 12 =
2 p --- (3)
(2) = (3)
7 u - 38 = 2 u + 12
7 u - 2 u = 12 + 38
5 u = 50
1 u = 50 ÷ 5 = 10
Number of balls in the box
= (1 u + 6) + 7 u
= 8 u + 6
= (8 x 10) + 6
= 80 + 6
= 86
Answer(s): 86