Three containers, B, C and A, contained 158 balls. Howard 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 51 balls into Container A. As a result, the ratio of the number of balls in Container B, Container C and Container A became 6 : 2 : 5. 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 |
1x2 = 2 u |
2x2 = 4 u |
5 u - 51 |
158 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
+ 51 |
|
After |
3x2 = 6 u |
1x2 = 2 u |
|
|
Comparing the 3 containers |
6 u |
2 u |
5 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 6 is 6.
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
= 2 u + 4 u + 5 u - 51
= 11 u - 51
11 u - 51 = 158
11 u = 158 + 51
11 u = 209
1 u = 209 ÷ 11 = 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
= 158 - 76
= 82
Container C : Container B and Container A
76 : 82
(÷2)38 : 41
Answer(s): 38 : 41