Three containers, C, A and B, contained 206 balls. Luke added some balls into Container C and the number of balls in Container C tripled. He took out half of the number of balls from Container A and added another 54 balls into Container B. As a result, the ratio of the number of balls in Container C, Container A and Container B became 9 : 4 : 9. What was the ratio of the number of balls in Container A to the total number of balls in Container C and Container B at first? Give the answer in its lowest term.
| |
Container C |
Container A |
Container B |
Total |
| Before |
1x3 =
3 u |
2x4 =
8 u |
9 u - 54 |
206 |
| Change |
+ 2x3 =
+ 6 u |
- 1x4 =
- 4 u |
+ 54 |
|
| After |
3x3 =
9 u |
1x4 =
4 u |
|
|
Comparing the
3 containers |
9 u |
4 u |
9 u |
|
The number of balls in Container C in the end is repeated. Make the number of balls in Container C in the end the same. LCM of 3 and 9 is 9.
The number of balls in Container A in the end is repeated. Make the number of balls in Container A in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 3 u + 8 u + 9 u - 54
= 20 u - 54
20 u - 54 = 206
20 u = 206 + 54
20 u = 260
1 u = 260 ÷ 20 = 13
Number of balls in Container A at first
= 8 u
= 8 x 13
= 104
Number of balls in Container C and Container B at first
= 206 - 104
= 102
Container A : Container C and Container B
104 : 102
(÷2)52 : 51
Answer(s): 52 : 51
