Three boxes, C, A and B, contained 112 balls. Bryan 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 57 balls into Box B. As a result, the ratio of the number of balls in Box C, Box A and Box B became 6 : 2 : 7. 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 |
1x2 = 2 u |
2x2 = 4 u |
7 u - 57 |
112 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
+ 57 |
|
After |
3x2 = 6 u |
1x2 = 2 u |
|
|
Comparing the 3 boxes |
6 u |
2 u |
7 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 6 is 6.
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 2 is 2.
Total number of balls at first
= 2 u + 4 u + 7 u - 57
= 13 u - 57
13 u - 57 = 112
13 u = 112 + 57
13 u = 169
1 u = 169 ÷ 13 = 13
Number of balls in Box A at first
= 4 u
= 4 x 13
= 52
Number of balls in Box C and Box B at first
= 112 - 52
= 60
Box A : Box C and Box B
52 : 60
(÷4)13 : 15
Answer(s): 13 : 15