There are some white and red balls in a box. If 2 white balls are removed from the box, the total number of balls left will be 9 times the number of white balls left. If 39 red balls are removed from the box, the total number of balls left will be 4 times the number of white balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
White balls |
Red balls |
White balls |
Red balls |
Before |
1 u + 2 |
8 u |
1 p |
3 p + 39 |
Change |
- 2 |
|
|
- 39 |
After |
1 u |
8 u |
1 p |
3 p |
Number of red balls in the end for Case 1
= 9 u - 1 u
= 8 u
Number of white balls in the end for Case 2
= 4 p - 1 p
= 3 p
The number of white balls at first is the same for Case 1 and Case 2.
The number of red balls at first is also the same for Case 1 and Case 2.
1 u + 2 = 1 p --- (1)
8 u = 3 p + 39
8 u - 39 =
3 p --- (2)
Make p the same.
(1)
x3 3 u + 6 =
3 p --- (3)
(2) = (3)
8 u - 39 = 3 u + 6
8 u - 3 u = 6 + 39
5 u = 45
1 u = 45 ÷ 5 = 9
Number of balls in the box
= (1 u + 2) + 8 u
= 9 u + 2
= (9 x 9) + 2
= 81 + 2
= 83
Answer(s): 83