Three cartons, B, C and A, contained 375 balls. Seth 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 90 balls from Carton A. As a result, the ratio of the number of balls in Carton B, Carton C and Carton A became 6 : 4 : 5. 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 |
2x4 =
8 u |
5 u + 90 |
375 |
| Change |
+ 2x2 =
+ 4 u |
- 1x4 =
- 4 u |
- 90 |
|
| After |
3x2 =
6 u |
1x4 =
4 u |
|
|
Comparing the
3 cartons |
6 u
|
4 u |
5 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 4 is 4.
Total number of balls at first
= 2 u + 8 u + 5 u + 90
= 15 u + 90
15 u + 90 = 375
15 u = 375 - 90
15 u = 285
1 u = 285 ÷ 15 = 19
Number of balls in Carton A at first
= 5 u + 90
= 5 x 19 + 90
= 95 + 90
= 185
Number of balls in Carton B and Carton C at first
= 375 - 185
= 190
Carton A : Carton B and Carton C
185 : 190
(÷5)37 : 38
Answer(s): 37 : 38
