Three boxes, B, C and A, contained 228 balls. Luis added some balls into Box B and the number of balls in Box B tripled. He took out half of the number of balls from Box C and removed 24 balls from Box A. As a result, the ratio of the number of balls in Box B, Box C and Box A became 6 : 4 : 7. What was the ratio of the number of balls in Box A to the total number of balls in Box B and Box C at first? Give the answer in its lowest term.
|
Box B |
Box C |
Box A |
Total |
Before |
1x2 = 2 u |
2x4 = 8 u |
7 u + 24 |
228 |
Change |
+ 2x2 = + 4 u |
- 1x4 = - 4 u |
- 24 |
|
After |
3x2 = 6 u |
1x4 = 4 u |
|
|
Comparing the 3 boxes |
6 u
|
4 u |
7 u |
|
The number of balls in Box B in the end is the same. Make the number of balls in Box B in the end the same. LCM of 3 and 6 is 6.
The number of balls in Box C in the end is the same. Make the number of balls in Box C in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 2 u + 8 u + 7 u + 24
= 17 u + 24
17 u + 24 = 228
17 u = 228 - 24
17 u = 204
1 u = 204 ÷ 17 = 12
Number of balls in Box A at first
= 7 u + 24
= 7 x 12 + 24
= 84 + 24
= 108
Number of balls in Box B and Box C at first
= 228 - 108
= 120
Box A : Box B and Box C
108 : 120
(÷12)9 : 10
Answer(s): 9 : 10