Three boxes, C, A and B, contained 306 beads. Pierre 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 50 beads from Box B. As a result, the ratio of the number of beads in Box C, Box A and Box B became 9 : 2 : 9. 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 |
1x3 = 3 u |
2x2 = 4 u |
9 u + 50 |
306 |
Change |
+ 2x3 = + 6 u |
- 1x2 = - 2 u |
- 50 |
|
After |
3x3 = 9 u |
1x2 = 2 u |
|
|
Comparing the 3 boxes |
9 u
|
2 u |
9 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 9 is 9.
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
= 3 u + 4 u + 9 u + 50
= 16 u + 50
16 u + 50 = 306
16 u = 306 - 50
16 u = 256
1 u = 256 ÷ 16 = 16
Number of beads in Box B at first
= 9 u + 50
= 9 x 16 + 50
= 144 + 50
= 194
Number of beads in Box C and Box A at first
= 306 - 194
= 112
Box B : Box C and Box A
194 : 112
(÷2)97 : 56
Answer(s): 97 : 56