Three boxes, A, B and C, contained 213 beads. Jenson 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 33 beads from Box C. As a result, the ratio of the number of beads in Box A, Box B and Box C became 12 : 3 : 5. 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 |
1x4 =
4 u |
2x3 =
6 u |
5 u + 33 |
213 |
| Change |
+ 2x4 =
+ 8 u |
- 1x3 =
- 3 u |
- 33 |
|
| After |
3x4 =
12 u |
1x3 =
3 u |
|
|
Comparing the
3 boxes |
12 u
|
3 u |
5 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 12 is 12.
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
= 4 u + 6 u + 5 u + 33
= 15 u + 33
15 u + 33 = 213
15 u = 213 - 33
15 u = 180
1 u = 180 ÷ 15 = 12
Number of beads in Box C at first
= 5 u + 33
= 5 x 12 + 33
= 60 + 33
= 93
Number of beads in Box A and Box B at first
= 213 - 93
= 120
Box C : Box A and Box B
93 : 120
(÷3)31 : 40
Answer(s): 31 : 40
