Three boxes, C, A and B, contained 222 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 68 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 : 5. 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 |
5 u + 68 |
222 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
- 68 |
|
After |
3x2 = 6 u |
1x2 = 2 u |
|
|
Comparing the 3 boxes |
6 u
|
2 u |
5 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 + 5 u + 68
= 11 u + 68
11 u + 68 = 222
11 u = 222 - 68
11 u = 154
1 u = 154 ÷ 11 = 14
Number of beads in Box B at first
= 5 u + 68
= 5 x 14 + 68
= 70 + 68
= 138
Number of beads in Box C and Box A at first
= 222 - 138
= 84
Box B : Box C and Box A
138 : 84
(÷6)23 : 14
Answer(s): 23 : 14