Three boxes, A, B and C, contained 378 balls. Tim added some balls into Box A and the number of balls in Box A tripled. He took out half of the number of balls from Box B and removed 56 balls from Box C. As a result, the ratio of the number of balls in Box A, Box B and Box C became 12 : 4 : 11. What was the ratio of the number of balls in Box C to the total number of balls in Box A and Box B at first? Give the answer in its lowest term.
| |
Box A |
Box B |
Box C |
Total |
| Before |
1x4 =
4 u |
2x4 =
8 u |
11 u + 56 |
378 |
| Change |
+ 2x4 =
+ 8 u |
- 1x4 =
- 4 u |
- 56 |
|
| After |
3x4 =
12 u |
1x4 =
4 u |
|
|
Comparing the
3 boxes |
12 u
|
4 u |
11 u |
|
The number of balls in Box A in the end is the same. Make the number of balls in Box A in the end the same. LCM of 3 and 12 is 12.
The number of balls in Box B in the end is the same. Make the number of balls in Box B in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 4 u + 8 u + 11 u + 56
= 23 u + 56
23 u + 56 = 378
23 u = 378 - 56
23 u = 322
1 u = 322 ÷ 23 = 14
Number of balls in Box C at first
= 11 u + 56
= 11 x 14 + 56
= 154 + 56
= 210
Number of balls in Box A and Box B at first
= 378 - 210
= 168
Box C : Box A and Box B
210 : 168
(÷42)5 : 4
Answer(s): 5 : 4
