Three boxes, B, C and A, contained 194 balls. Ian 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 77 balls from Box A. As a result, the ratio of the number of balls in Box B, Box C and Box A became 6 : 2 : 3. 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 |
2x2 = 4 u |
3 u + 77 |
194 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
- 77 |
|
After |
3x2 = 6 u |
1x2 = 2 u |
|
|
Comparing the 3 boxes |
6 u
|
2 u |
3 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 2 is 2.
Total number of balls at first
= 2 u + 4 u + 3 u + 77
= 9 u + 77
9 u + 77 = 194
9 u = 194 - 77
9 u = 117
1 u = 117 ÷ 9 = 13
Number of balls in Box A at first
= 3 u + 77
= 3 x 13 + 77
= 39 + 77
= 116
Number of balls in Box B and Box C at first
= 194 - 116
= 78
Box A : Box B and Box C
116 : 78
(÷2)58 : 39
Answer(s): 58 : 39