Three containers, C, A and B, contained 194 balls. Pierre 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 95 balls into Container B. As a result, the ratio of the number of balls in Container C, Container A and Container B became 6 : 4 : 7. 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 |
1x2 = 2 u |
2x4 = 8 u |
7 u - 95 |
194 |
Change |
+ 2x2 = + 4 u |
- 1x4 = - 4 u |
+ 95 |
|
After |
3x2 = 6 u |
1x4 = 4 u |
|
|
Comparing the 3 containers |
6 u |
4 u |
7 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 6 is 6.
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
= 2 u + 8 u + 7 u - 95
= 17 u - 95
17 u - 95 = 194
17 u = 194 + 95
17 u = 289
1 u = 289 ÷ 17 = 17
Number of balls in Container A at first
= 8 u
= 8 x 17
= 136
Number of balls in Container C and Container B at first
= 194 - 136
= 58
Container A : Container C and Container B
136 : 58
(÷2)68 : 29
Answer(s): 68 : 29