Three cartons, C, A and B, contained 206 balls. Tom 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 66 balls from Carton B. As a result, the ratio of the number of balls in Carton C, Carton A and Carton B became 9 : 2 : 3. 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 |
1x3 =
3 u |
2x2 =
4 u |
3 u + 66 |
206 |
| Change |
+ 2x3 =
+ 6 u |
- 1x2 =
- 2 u |
- 66 |
|
| After |
3x3 =
9 u |
1x2 =
2 u |
|
|
Comparing the
3 cartons |
9 u
|
2 u |
3 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 9 is 9.
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 2 is 2.
Total number of balls at first
= 3 u + 4 u + 3 u + 66
= 10 u + 66
10 u + 66 = 206
10 u = 206 - 66
10 u = 140
1 u = 140 ÷ 10 = 14
Number of balls in Carton B at first
= 3 u + 66
= 3 x 14 + 66
= 42 + 66
= 108
Number of balls in Carton C and Carton A at first
= 206 - 108
= 98
Carton B : Carton C and Carton A
108 : 98
(÷2)54 : 49
Answer(s): 54 : 49
