Three boxes, C, A and B, contained 232 beads. George added some beads into Box C and the number of beads in Box C tripled. He took out half of the number of beads from Box A and added another 72 beads into Box B. As a result, the ratio of the number of beads in Box C, Box A and Box B became 6 : 4 : 9. What was the ratio of the number of beads in Box A to the total number of beads 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 |
2x4 =
8 u |
9 u - 72 |
232 |
| Change |
+ 2x2 =
+ 4 u |
- 1x4 =
- 4 u |
+ 72 |
|
| After |
3x2 =
6 u |
1x4 =
4 u |
|
|
Comparing the
3 boxes |
6 u |
4 u |
9 u |
|
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 3 and 6 is 6.
The number of beads in Box A in the end is repeated. Make the number of beads in Box A in the end the same. LCM of 1 and 4 is 4.
Total number of beads at first
= 2 u + 8 u + 9 u - 72
= 19 u - 72
19 u - 72 = 232
19 u = 232 + 72
19 u = 304
1 u = 304 ÷ 19 = 16
Number of beads in Box A at first
= 8 u
= 8 x 16
= 128
Number of beads in Box C and Box B at first
= 232 - 128
= 104
Box A : Box C and Box B
128 : 104
(÷8)16 : 13
Answer(s): 16 : 13
