Three containers, A, B and C, contained 215 balls. Liam added some balls into Container A and the number of balls in Container A tripled. He took out half of the number of balls from Container B and removed 50 balls from Container C. As a result, the ratio of the number of balls in Container A, Container B and Container C became 6 : 4 : 5. What was the ratio of the number of balls in Container C to the total number of balls in Container A and Container B at first? Give the answer in its lowest term.
| |
Container A |
Container B |
Container C |
Total |
| Before |
1x2 =
2 u |
2x4 =
8 u |
5 u + 50 |
215 |
| Change |
+ 2x2 =
+ 4 u |
- 1x4 =
- 4 u |
- 50 |
|
| After |
3x2 =
6 u |
1x4 =
4 u |
|
|
Comparing the
3 containers |
6 u
|
4 u |
5 u |
|
The number of balls in Container A in the end is the same. Make the number of balls in Container A in the end the same. LCM of 3 and 6 is 6.
The number of balls in Container B in the end is the same. Make the number of balls in Container B in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 2 u + 8 u + 5 u + 50
= 15 u + 50
15 u + 50 = 215
15 u = 215 - 50
15 u = 165
1 u = 165 ÷ 15 = 11
Number of balls in Container C at first
= 5 u + 50
= 5 x 11 + 50
= 55 + 50
= 105
Number of balls in Container A and Container B at first
= 215 - 105
= 110
Container C : Container A and Container B
105 : 110
(÷5)21 : 22
Answer(s): 21 : 22
