Three containers, B, C and A, contained 168 balls. Rael added some balls into Container B and the number of balls in Container B tripled. He took out half of the number of balls from Container C and added another 22 balls into Container A. As a result, the ratio of the number of balls in Container B, Container C and Container A became 9 : 2 : 3. What was the ratio of the number of balls in Container C to the total number of balls in Container B and Container A at first? Give the answer in its lowest term.
|
Container B |
Container C |
Container A |
Total |
Before |
1x3 = 3 u |
2x2 = 4 u |
3 u - 22 |
168 |
Change |
+ 2x3 = + 6 u |
- 1x2 = - 2 u |
+ 22 |
|
After |
3x3 = 9 u |
1x2 = 2 u |
|
|
Comparing the 3 containers |
9 u |
2 u |
3 u |
|
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 3 and 9 is 9.
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 1 and 2 is 2.
Total number of balls at first
= 3 u + 4 u + 3 u - 22
= 10 u - 22
10 u - 22 = 168
10 u = 168 + 22
10 u = 190
1 u = 190 ÷ 10 = 19
Number of balls in Container C at first
= 4 u
= 4 x 19
= 76
Number of balls in Container B and Container A at first
= 168 - 76
= 92
Container C : Container B and Container A
76 : 92
(÷4)19 : 23
Answer(s): 19 : 23