Three boxes, B, C and A, contained 188 beads. Pierre added some beads into Box B and the number of beads in Box B tripled. He took out half of the number of beads from Box C and added another 78 beads into Box A. As a result, the ratio of the number of beads in Box B, Box C and Box A became 9 : 3 : 5. What was the ratio of the number of beads in Box C to the total number of beads in Box B and Box A at first? Give the answer in its lowest term.
|
Box B |
Box C |
Box A |
Total |
Before |
1x3 = 3 u |
2x3 = 6 u |
5 u - 78 |
188 |
Change |
+ 2x3 = + 6 u |
- 1x3 = - 3 u |
+ 78 |
|
After |
3x3 = 9 u |
1x3 = 3 u |
|
|
Comparing the 3 boxes |
9 u |
3 u |
5 u |
|
The number of beads in Box B in the end is repeated. Make the number of beads in Box B in the end the same. LCM of 3 and 9 is 9.
The number of beads in Box C in the end is repeated. Make the number of beads in Box C in the end the same. LCM of 1 and 3 is 3.
Total number of beads at first
= 3 u + 6 u + 5 u - 78
= 14 u - 78
14 u - 78 = 188
14 u = 188 + 78
14 u = 266
1 u = 266 ÷ 14 = 19
Number of beads in Box C at first
= 6 u
= 6 x 19
= 114
Number of beads in Box B and Box A at first
= 188 - 114
= 74
Box C : Box B and Box A
114 : 74
(÷2)57 : 37
Answer(s): 57 : 37