Three cartons, C, A and B, contained 161 beads. Xavier 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 added another 77 beads into Carton B. As a result, the ratio of the number of beads in Carton C, Carton A and Carton B became 12 : 2 : 9. What was the ratio of the number of beads in Carton A to the total number of beads in Carton C and Carton B at first? Give the answer in its lowest term.
| |
Carton C |
Carton A |
Carton B |
Total |
| Before |
1x4 =
4 u |
2x2 =
4 u |
9 u - 77 |
161 |
| Change |
+ 2x4 =
+ 8 u |
- 1x2 =
- 2 u |
+ 77 |
|
| After |
3x4 =
12 u |
1x2 =
2 u |
|
|
Comparing the
3 cartons |
12 u |
2 u |
9 u |
|
The number of beads in Carton C in the end is repeated. Make the number of beads in Carton C in the end the same. LCM of 3 and 12 is 12.
The number of beads in Carton A in the end is repeated. 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
= 4 u + 4 u + 9 u - 77
= 17 u - 77
17 u - 77 = 161
17 u = 161 + 77
17 u = 238
1 u = 238 ÷ 17 = 14
Number of beads in Carton A at first
= 4 u
= 4 x 14
= 56
Number of beads in Carton C and Carton B at first
= 161 - 56
= 105
Carton A : Carton C and Carton B
56 : 105
(÷7)8 : 15
Answer(s): 8 : 15
