Three boxes, C, A and B, contained 124 balls. Jeremy 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 71 balls into Box B. As a result, the ratio of the number of balls in Box C, Box A and Box B became 6 : 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 |
1x2 =
2 u |
2x3 =
6 u |
5 u - 71 |
124 |
| Change |
+ 2x2 =
+ 4 u |
- 1x3 =
- 3 u |
+ 71 |
|
| After |
3x2 =
6 u |
1x3 =
3 u |
|
|
Comparing the
3 boxes |
6 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 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 3 is 3.
Total number of balls at first
= 2 u + 6 u + 5 u - 71
= 13 u - 71
13 u - 71 = 124
13 u = 124 + 71
13 u = 195
1 u = 195 ÷ 13 = 15
Number of balls in Box A at first
= 6 u
= 6 x 15
= 90
Number of balls in Box C and Box B at first
= 124 - 90
= 34
Box A : Box C and Box B
90 : 34
(÷2)45 : 17
Answer(s): 45 : 17
