Three boxes, B, C and A, contained 318 beads. Oliver added some beads into Box B and the number of beads in Box B tripled. He took out half of the number of beads from Box C and added another 62 beads into Box A. As a result, the ratio of the number of beads in Box B, Box C and Box A became 9 : 3 : 11. What was the ratio of the number of beads in Box C to the total number of beads in Box B and Box A at first? Give the answer in its lowest term.
| |
Box B |
Box C |
Box A |
Total |
| Before |
1x3 =
3 u |
2x3 =
6 u |
11 u - 62 |
318 |
| Change |
+ 2x3 =
+ 6 u |
- 1x3 =
- 3 u |
+ 62 |
|
| After |
3x3 =
9 u |
1x3 =
3 u |
|
|
Comparing the
3 boxes |
9 u |
3 u |
11 u |
|
The number of beads in Box B in the end is repeated. Make the number of beads in Box B in the end the same. LCM of 3 and 9 is 9.
The number of beads in Box C in the end is repeated. Make the number of beads in Box C in the end the same. LCM of 1 and 3 is 3.
Total number of beads at first
= 3 u + 6 u + 11 u - 62
= 20 u - 62
20 u - 62 = 318
20 u = 318 + 62
20 u = 380
1 u = 380 ÷ 20 = 19
Number of beads in Box C at first
= 6 u
= 6 x 19
= 114
Number of beads in Box B and Box A at first
= 318 - 114
= 204
Box C : Box B and Box A
114 : 204
(÷6)19 : 34
Answer(s): 19 : 34
