Three boxes, A, B and C, contained 228 beads. Lee added some beads into Box A and the number of beads in Box A tripled. He took out half of the number of beads from Box B and added another 52 beads into Box C. As a result, the ratio of the number of beads in Box A, Box B and Box C became 9 : 2 : 7. What was the ratio of the number of beads in Box B to the total number of beads in Box A and Box C at first? Give the answer in its lowest term.
| |
Box A |
Box B |
Box C |
Total |
| Before |
1x3 =
3 u |
2x2 =
4 u |
7 u - 52 |
228 |
| Change |
+ 2x3 =
+ 6 u |
- 1x2 =
- 2 u |
+ 52 |
|
| After |
3x3 =
9 u |
1x2 =
2 u |
|
|
Comparing the
3 boxes |
9 u |
2 u |
7 u |
|
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 3 and 9 is 9.
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 1 and 2 is 2.
Total number of beads at first
= 3 u + 4 u + 7 u - 52
= 14 u - 52
14 u - 52 = 228
14 u = 228 + 52
14 u = 280
1 u = 280 ÷ 14 = 20
Number of beads in Box B at first
= 4 u
= 4 x 20
= 80
Number of beads in Box A and Box C at first
= 228 - 80
= 148
Box B : Box A and Box C
80 : 148
(÷4)20 : 37
Answer(s): 20 : 37
