Three cartons, B, C and A, contained 232 balls. Daniel added some balls into Carton B and the number of balls in Carton B tripled. He took out half of the number of balls from Carton C and removed 37 balls from Carton A. As a result, the ratio of the number of balls in Carton B, Carton C and Carton A became 6 : 2 : 9. What was the ratio of the number of balls in Carton A to the total number of balls in Carton B and Carton C at first? Give the answer in its lowest term.
| |
Carton B |
Carton C |
Carton A |
Total |
| Before |
1x2 =
2 u |
2x2 =
4 u |
9 u + 37 |
232 |
| Change |
+ 2x2 =
+ 4 u |
- 1x2 =
- 2 u |
- 37 |
|
| After |
3x2 =
6 u |
1x2 =
2 u |
|
|
Comparing the
3 cartons |
6 u
|
2 u |
9 u |
|
The number of balls in Carton B in the end is the same. Make the number of balls in Carton B in the end the same. LCM of 3 and 6 is 6.
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 1 and 2 is 2.
Total number of balls at first
= 2 u + 4 u + 9 u + 37
= 15 u + 37
15 u + 37 = 232
15 u = 232 - 37
15 u = 195
1 u = 195 ÷ 15 = 13
Number of balls in Carton A at first
= 9 u + 37
= 9 x 13 + 37
= 117 + 37
= 154
Number of balls in Carton B and Carton C at first
= 232 - 154
= 78
Carton A : Carton B and Carton C
154 : 78
(÷2)77 : 39
Answer(s): 77 : 39
