Three containers, A, B and C, contained 201 beads. Oscar added some beads into Container A and the number of beads in Container A tripled. He took out half of the number of beads from Container B and removed 69 beads from Container C. As a result, the ratio of the number of beads in Container A, Container B and Container C became 6 : 4 : 1. What was the ratio of the number of beads in Container C to the total number of beads in Container A and Container B at first? Give the answer in its lowest term.
| |
Container A |
Container B |
Container C |
Total |
| Before |
1x2 =
2 u |
2x4 =
8 u |
1 u + 69 |
201 |
| Change |
+ 2x2 =
+ 4 u |
- 1x4 =
- 4 u |
- 69 |
|
| After |
3x2 =
6 u |
1x4 =
4 u |
|
|
Comparing the
3 containers |
6 u
|
4 u |
1 u |
|
The number of beads in Container A in the end is the same. Make the number of beads in Container A in the end the same. LCM of 3 and 6 is 6.
The number of beads in Container B in the end is the same. Make the number of beads in Container B in the end the same. LCM of 1 and 4 is 4.
Total number of beads at first
= 2 u + 8 u + 1 u + 69
= 11 u + 69
11 u + 69 = 201
11 u = 201 - 69
11 u = 132
1 u = 132 ÷ 11 = 12
Number of beads in Container C at first
= 1 u + 69
= 1 x 12 + 69
= 12 + 69
= 81
Number of beads in Container A and Container B at first
= 201 - 81
= 120
Container C : Container A and Container B
81 : 120
(÷3)27 : 40
Answer(s): 27 : 40
