Three boxes, C, A and B, contained 225 beads. Sam 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 49 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 + 49 |
225 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
- 49 |
|
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 + 49
= 11 u + 49
11 u + 49 = 225
11 u = 225 - 49
11 u = 176
1 u = 176 ÷ 11 = 16
Number of beads in Box B at first
= 5 u + 49
= 5 x 16 + 49
= 80 + 49
= 129
Number of beads in Box C and Box A at first
= 225 - 129
= 96
Box B : Box C and Box A
129 : 96
(÷3)43 : 32
Answer(s): 43 : 32