Three boxes, A, B and C, contained 322 beads. Luke added some beads into Box A and the number of beads in Box A tripled. He took out half of the number of beads from Box B and removed 56 beads from Box C. As a result, the ratio of the number of beads in Box A, Box B and Box C became 9 : 2 : 7. What was the ratio of the number of beads in Box C to the total number of beads in Box A and Box B at first? Give the answer in its lowest term.
|
Box A |
Box B |
Box C |
Total |
Before |
1x3 = 3 u |
2x2 = 4 u |
7 u + 56 |
322 |
Change |
+ 2x3 = + 6 u |
- 1x2 = - 2 u |
- 56 |
|
After |
3x3 = 9 u |
1x2 = 2 u |
|
|
Comparing the 3 boxes |
9 u
|
2 u |
7 u |
|
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 3 and 9 is 9.
The number of beads in Box B in the end is the same. Make the number of beads in Box B in the end the same. LCM of 1 and 2 is 2.
Total number of beads at first
= 3 u + 4 u + 7 u + 56
= 14 u + 56
14 u + 56 = 322
14 u = 322 - 56
14 u = 266
1 u = 266 ÷ 14 = 19
Number of beads in Box C at first
= 7 u + 56
= 7 x 19 + 56
= 133 + 56
= 189
Number of beads in Box A and Box B at first
= 322 - 189
= 133
Box C : Box A and Box B
189 : 133
(÷7)27 : 19
Answer(s): 27 : 19