Three cartons, C, A and B, contained 302 balls. Cole added some balls into Carton C and the number of balls in Carton C tripled. He took out half of the number of balls from Carton A and removed 82 balls from Carton B. As a result, the ratio of the number of balls in Carton C, Carton A and Carton B became 6 : 4 : 1. What was the ratio of the number of balls in Carton B to the total number of balls 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 |
2x4 = 8 u |
1 u + 82 |
302 |
Change |
+ 2x2 = + 4 u |
- 1x4 = - 4 u |
- 82 |
|
After |
3x2 = 6 u |
1x4 = 4 u |
|
|
Comparing the 3 cartons |
6 u
|
4 u |
1 u |
|
The number of balls in Carton C in the end is the same. Make the number of balls in Carton C in the end the same. LCM of 3 and 6 is 6.
The number of balls in Carton A in the end is the same. Make the number of balls in Carton A in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 2 u + 8 u + 1 u + 82
= 11 u + 82
11 u + 82 = 302
11 u = 302 - 82
11 u = 220
1 u = 220 ÷ 11 = 20
Number of balls in Carton B at first
= 1 u + 82
= 1 x 20 + 82
= 20 + 82
= 102
Number of balls in Carton C and Carton A at first
= 302 - 102
= 200
Carton B : Carton C and Carton A
102 : 200
(÷2)51 : 100
Answer(s): 51 : 100