Three cartons, B, C and A, contained 142 balls. Liam 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 22 balls from Carton A. As a result, the ratio of the number of balls in Carton B, Carton C and Carton A became 9 : 2 : 3. 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 |
1x3 = 3 u |
2x2 = 4 u |
3 u + 22 |
142 |
Change |
+ 2x3 = + 6 u |
- 1x2 = - 2 u |
- 22 |
|
After |
3x3 = 9 u |
1x2 = 2 u |
|
|
Comparing the 3 cartons |
9 u
|
2 u |
3 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 9 is 9.
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
= 3 u + 4 u + 3 u + 22
= 10 u + 22
10 u + 22 = 142
10 u = 142 - 22
10 u = 120
1 u = 120 ÷ 10 = 12
Number of balls in Carton A at first
= 3 u + 22
= 3 x 12 + 22
= 36 + 22
= 58
Number of balls in Carton B and Carton C at first
= 142 - 58
= 84
Carton A : Carton B and Carton C
58 : 84
(÷2)29 : 42
Answer(s): 29 : 42