Three boxes, A, B and C, contained 460 beads. Michael 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 60 beads from Box C. As a result, the ratio of the number of beads in Box A, Box B and Box C became 9 : 3 : 11. 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 |
2x3 = 6 u |
11 u + 60 |
460 |
Change |
+ 2x3 = + 6 u |
- 1x3 = - 3 u |
- 60 |
|
After |
3x3 = 9 u |
1x3 = 3 u |
|
|
Comparing the 3 boxes |
9 u
|
3 u |
11 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 3 is 3.
Total number of beads at first
= 3 u + 6 u + 11 u + 60
= 20 u + 60
20 u + 60 = 460
20 u = 460 - 60
20 u = 400
1 u = 400 ÷ 20 = 20
Number of beads in Box C at first
= 11 u + 60
= 11 x 20 + 60
= 220 + 60
= 280
Number of beads in Box A and Box B at first
= 460 - 280
= 180
Box C : Box A and Box B
280 : 180
(÷20)14 : 9
Answer(s): 14 : 9