There are some purple and silver balls in a box. If 4 purple balls are removed from the box, the total number of balls left will be 7 times the number of purple balls left. If 12 silver balls are removed from the box, the total number of balls left will be 3 times the number of purple balls. How many balls are there in the box?
|
Case 1 |
Case 2 |
|
Purple balls |
Silver balls |
Purple balls |
Silver balls |
Before |
1 u + 4 |
6 u |
1 p |
2 p + 12 |
Change |
- 4 |
|
|
- 12 |
After |
1 u |
6 u |
1 p |
2 p |
Number of silver balls in the end for Case 1
= 7 u - 1 u
= 6 u
Number of purple balls in the end for Case 2
= 3 p - 1 p
= 2 p
The number of purple balls at first is the same for Case 1 and Case 2.
The number of silver balls at first is also the same for Case 1 and Case 2.
1 u + 4 = 1 p --- (1)
6 u = 2 p + 12
6 u - 12 =
2 p --- (2)
Make p the same.
(1)
x2 2 u + 8 =
2 p --- (3)
(2) = (3)
6 u - 12 = 2 u + 8
6 u - 2 u = 8 + 12
4 u = 20
1 u = 20 ÷ 4 = 5
Number of balls in the box
= (1 u + 4) + 6 u
= 7 u + 4
= (7 x 5) + 4
= 35 + 4
= 39
Answer(s): 39