Three containers, C, A and B, contained 404 balls. Glen 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 36 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 : 11. 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 |
11 u - 36 |
404 |
Change |
+ 2x3 = + 6 u |
- 1x4 = - 4 u |
+ 36 |
|
After |
3x3 = 9 u |
1x4 = 4 u |
|
|
Comparing the 3 containers |
9 u |
4 u |
11 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 + 11 u - 36
= 22 u - 36
22 u - 36 = 404
22 u = 404 + 36
22 u = 440
1 u = 440 ÷ 22 = 20
Number of balls in Container A at first
= 8 u
= 8 x 20
= 160
Number of balls in Container C and Container B at first
= 404 - 160
= 244
Container A : Container C and Container B
160 : 244
(÷4)40 : 61
Answer(s): 40 : 61