Three cartons, C, A and B, contained 128 beads. Fred added some beads into Carton C and the number of beads in Carton C tripled. He took out half of the number of beads from Carton A and removed 18 beads from Carton B. As a result, the ratio of the number of beads in Carton C, Carton A and Carton B became 6 : 2 : 5. What was the ratio of the number of beads in Carton B to the total number of beads in Carton C and Carton A at first? Give the answer in its lowest term.
| |
Carton C |
Carton A |
Carton B |
Total |
| Before |
1x2 =
2 u |
2x2 =
4 u |
5 u + 18 |
128 |
| Change |
+ 2x2 =
+ 4 u |
- 1x2 =
- 2 u |
- 18 |
|
| After |
3x2 =
6 u |
1x2 =
2 u |
|
|
Comparing the
3 cartons |
6 u
|
2 u |
5 u |
|
The number of beads in Carton C in the end is the same. Make the number of beads in Carton C in the end the same. LCM of 3 and 6 is 6.
The number of beads in Carton A in the end is the same. Make the number of beads in Carton A in the end the same. LCM of 1 and 2 is 2.
Total number of beads at first
= 2 u + 4 u + 5 u + 18
= 11 u + 18
11 u + 18 = 128
11 u = 128 - 18
11 u = 110
1 u = 110 ÷ 11 = 10
Number of beads in Carton B at first
= 5 u + 18
= 5 x 10 + 18
= 50 + 18
= 68
Number of beads in Carton C and Carton A at first
= 128 - 68
= 60
Carton B : Carton C and Carton A
68 : 60
(÷4)17 : 15
Answer(s): 17 : 15
