Three boxes, C, A and B, contained 134 balls. Eric added some balls into Box C and the number of balls in Box C tripled. He took out half of the number of balls from Box A and added another 31 balls into Box B. As a result, the ratio of the number of balls in Box C, Box A and Box B became 12 : 3 : 5. What was the ratio of the number of balls in Box A to the total number of balls in Box C and Box B at first? Give the answer in its lowest term.
|
Box C |
Box A |
Box B |
Total |
Before |
1x4 = 4 u |
2x3 = 6 u |
5 u - 31 |
134 |
Change |
+ 2x4 = + 8 u |
- 1x3 = - 3 u |
+ 31 |
|
After |
3x4 = 12 u |
1x3 = 3 u |
|
|
Comparing the 3 boxes |
12 u |
3 u |
5 u |
|
The number of balls in Box C in the end is repeated. Make the number of balls in Box C in the end the same. LCM of 3 and 12 is 12.
The number of balls in Box A in the end is repeated. Make the number of balls in Box A in the end the same. LCM of 1 and 3 is 3.
Total number of balls at first
= 4 u + 6 u + 5 u - 31
= 15 u - 31
15 u - 31 = 134
15 u = 134 + 31
15 u = 165
1 u = 165 ÷ 15 = 11
Number of balls in Box A at first
= 6 u
= 6 x 11
= 66
Number of balls in Box C and Box B at first
= 134 - 66
= 68
Box A : Box C and Box B
66 : 68
(÷2)33 : 34
Answer(s): 33 : 34