Three containers, A, B and C, contained 169 balls. Riordan 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 added another 39 balls into Container C. As a result, the ratio of the number of balls in Container A, Container B and Container C became 9 : 2 : 9. What was the ratio of the number of balls in Container B to the total number of balls in Container A and Container C at first? Give the answer in its lowest term.
| |
Container A |
Container B |
Container C |
Total |
| Before |
1x3 =
3 u |
2x2 =
4 u |
9 u - 39 |
169 |
| Change |
+ 2x3 =
+ 6 u |
- 1x2 =
- 2 u |
+ 39 |
|
| After |
3x3 =
9 u |
1x2 =
2 u |
|
|
Comparing the
3 containers |
9 u |
2 u |
9 u |
|
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 3 and 9 is 9.
The number of balls in Container B in the end is repeated. Make the number of balls in Container B in the end the same. LCM of 1 and 2 is 2.
Total number of balls at first
= 3 u + 4 u + 9 u - 39
= 16 u - 39
16 u - 39 = 169
16 u = 169 + 39
16 u = 208
1 u = 208 ÷ 16 = 13
Number of balls in Container B at first
= 4 u
= 4 x 13
= 52
Number of balls in Container A and Container C at first
= 169 - 52
= 117
Container B : Container A and Container C
52 : 117
(÷13)4 : 9
Answer(s): 4 : 9
