Three cartons, A, B and C, contained 333 balls. Reggie added some balls into Carton A and the number of balls in Carton A tripled. He took out half of the number of balls from Carton B and removed 78 balls from Carton C. As a result, the ratio of the number of balls in Carton A, Carton B and Carton C became 12 : 4 : 5. What was the ratio of the number of balls in Carton C to the total number of balls in Carton A and Carton B at first? Give the answer in its lowest term.
| |
Carton A |
Carton B |
Carton C |
Total |
| Before |
1x4 =
4 u |
2x4 =
8 u |
5 u + 78 |
333 |
| Change |
+ 2x4 =
+ 8 u |
- 1x4 =
- 4 u |
- 78 |
|
| After |
3x4 =
12 u |
1x4 =
4 u |
|
|
Comparing the
3 cartons |
12 u
|
4 u |
5 u |
|
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 3 and 12 is 12.
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 1 and 4 is 4.
Total number of balls at first
= 4 u + 8 u + 5 u + 78
= 17 u + 78
17 u + 78 = 333
17 u = 333 - 78
17 u = 255
1 u = 255 ÷ 17 = 15
Number of balls in Carton C at first
= 5 u + 78
= 5 x 15 + 78
= 75 + 78
= 153
Number of balls in Carton A and Carton B at first
= 333 - 153
= 180
Carton C : Carton A and Carton B
153 : 180
(÷9)17 : 20
Answer(s): 17 : 20
