Three boxes, C, A and B, contained 225 beads. Vincent 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 38 beads from Box B. As a result, the ratio of the number of beads in Box C, Box A and Box B became 12 : 3 : 7. 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 |
1x4 = 4 u |
2x3 = 6 u |
7 u + 38 |
225 |
Change |
+ 2x4 = + 8 u |
- 1x3 = - 3 u |
- 38 |
|
After |
3x4 = 12 u |
1x3 = 3 u |
|
|
Comparing the 3 boxes |
12 u
|
3 u |
7 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 12 is 12.
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 3 is 3.
Total number of beads at first
= 4 u + 6 u + 7 u + 38
= 17 u + 38
17 u + 38 = 225
17 u = 225 - 38
17 u = 187
1 u = 187 ÷ 17 = 11
Number of beads in Box B at first
= 7 u + 38
= 7 x 11 + 38
= 77 + 38
= 115
Number of beads in Box C and Box A at first
= 225 - 115
= 110
Box B : Box C and Box A
115 : 110
(÷5)23 : 22
Answer(s): 23 : 22