Three containers, B, C and A, contained 305 beads. Sean added some beads into Container B and the number of beads in Container B tripled. He took out half of the number of beads from Container C and removed 45 beads from Container A. As a result, the ratio of the number of beads in Container B, Container C and Container A became 6 : 2 : 7. What was the ratio of the number of beads in Container A to the total number of beads in Container B and Container C at first? Give the answer in its lowest term.
| |
Container B |
Container C |
Container A |
Total |
| Before |
1x2 =
2 u |
2x2 =
4 u |
7 u + 45 |
305 |
| Change |
+ 2x2 =
+ 4 u |
- 1x2 =
- 2 u |
- 45 |
|
| After |
3x2 =
6 u |
1x2 =
2 u |
|
|
Comparing the
3 containers |
6 u
|
2 u |
7 u |
|
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 3 and 6 is 6.
The number of beads in Container C in the end is the same. Make the number of beads in Container C in the end the same. LCM of 1 and 2 is 2.
Total number of beads at first
= 2 u + 4 u + 7 u + 45
= 13 u + 45
13 u + 45 = 305
13 u = 305 - 45
13 u = 260
1 u = 260 ÷ 13 = 20
Number of beads in Container A at first
= 7 u + 45
= 7 x 20 + 45
= 140 + 45
= 185
Number of beads in Container B and Container C at first
= 305 - 185
= 120
Container A : Container B and Container C
185 : 120
(÷5)37 : 24
Answer(s): 37 : 24
