Three cartons, C, A and B, contained 219 balls. Charlie 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 added another 51 balls into Carton B. As a result, the ratio of the number of balls in Carton C, Carton A and Carton B became 6 : 3 : 7. What was the ratio of the number of balls in Carton A to the total number of balls in Carton C and Carton B at first? Give the answer in its lowest term.
|
Carton C |
Carton A |
Carton B |
Total |
Before |
1x2 = 2 u |
2x3 = 6 u |
7 u - 51 |
219 |
Change |
+ 2x2 = + 4 u |
- 1x3 = - 3 u |
+ 51 |
|
After |
3x2 = 6 u |
1x3 = 3 u |
|
|
Comparing the 3 cartons |
6 u |
3 u |
7 u |
|
The number of balls in Carton C in the end is repeated. 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 repeated. Make the number of balls in Carton A in the end the same. LCM of 1 and 3 is 3.
Total number of balls at first
= 2 u + 6 u + 7 u - 51
= 15 u - 51
15 u - 51 = 219
15 u = 219 + 51
15 u = 270
1 u = 270 ÷ 15 = 18
Number of balls in Carton A at first
= 6 u
= 6 x 18
= 108
Number of balls in Carton C and Carton B at first
= 219 - 108
= 111
Carton A : Carton C and Carton B
108 : 111
(÷3)36 : 37
Answer(s): 36 : 37