Three boxes, C, A and B, contained 144 beads. Luis 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 removed 39 beads from Box B. As a result, the ratio of the number of beads in Box C, Box A and Box B became 6 : 2 : 1. What was the ratio of the number of beads in Box B to the total number of beads in Box C and Box A at first? Give the answer in its lowest term.
| |
Box C |
Box A |
Box B |
Total |
| Before |
1x2 =
2 u |
2x2 =
4 u |
1 u + 39 |
144 |
| Change |
+ 2x2 =
+ 4 u |
- 1x2 =
- 2 u |
- 39 |
|
| After |
3x2 =
6 u |
1x2 =
2 u |
|
|
Comparing the
3 boxes |
6 u
|
2 u |
1 u |
|
The number of beads in Box C in the end is the same. 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 the same. Make the number of beads in Box A in the end the same. LCM of 1 and 2 is 2.
Total number of beads at first
= 2 u + 4 u + 1 u + 39
= 7 u + 39
7 u + 39 = 144
7 u = 144 - 39
7 u = 105
1 u = 105 ÷ 7 = 15
Number of beads in Box B at first
= 1 u + 39
= 1 x 15 + 39
= 15 + 39
= 54
Number of beads in Box C and Box A at first
= 144 - 54
= 90
Box B : Box C and Box A
54 : 90
(÷18)3 : 5
Answer(s): 3 : 5
