Three containers, B, C and A, contained 228 balls. Gabriel 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 removed 12 balls from Container A. As a result, the ratio of the number of balls in Container B, Container C and Container A became 9 : 4 : 1. What was the ratio of the number of balls in Container A to the total number of balls in Container B and Container C at first? Give the answer in its lowest term.
|
Container B |
Container C |
Container A |
Total |
Before |
1x3 = 3 u |
2x4 = 8 u |
1 u + 12 |
228 |
Change |
+ 2x3 = + 6 u |
- 1x4 = - 4 u |
- 12 |
|
After |
3x3 = 9 u |
1x4 = 4 u |
|
|
Comparing the 3 containers |
9 u
|
4 u |
1 u |
|
The number of balls in Container B in the end is the same. 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 the same. Make the number of balls in Container C in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 3 u + 8 u + 1 u + 12
= 12 u + 12
12 u + 12 = 228
12 u = 228 - 12
12 u = 216
1 u = 216 ÷ 12 = 18
Number of balls in Container A at first
= 1 u + 12
= 1 x 18 + 12
= 18 + 12
= 30
Number of balls in Container B and Container C at first
= 228 - 30
= 198
Container A : Container B and Container C
30 : 198
(÷6)5 : 33
Answer(s): 5 : 33