Three boxes, C, A and B, contained 360 balls. Wesley added some balls into Box C and the number of balls in Box C tripled. He took out half of the number of balls from Box A and removed 37 balls from Box B. As a result, the ratio of the number of balls in Box C, Box A and Box B became 12 : 4 : 7. What was the ratio of the number of balls in Box B to the total number of balls in Box C and Box A at first? Give the answer in its lowest term.
|
Box C |
Box A |
Box B |
Total |
Before |
1x4 = 4 u |
2x4 = 8 u |
7 u + 37 |
360 |
Change |
+ 2x4 = + 8 u |
- 1x4 = - 4 u |
- 37 |
|
After |
3x4 = 12 u |
1x4 = 4 u |
|
|
Comparing the 3 boxes |
12 u
|
4 u |
7 u |
|
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 3 and 12 is 12.
The number of balls in Box A in the end is the same. Make the number of balls in Box A in the end the same. LCM of 1 and 4 is 4.
Total number of balls at first
= 4 u + 8 u + 7 u + 37
= 19 u + 37
19 u + 37 = 360
19 u = 360 - 37
19 u = 323
1 u = 323 ÷ 19 = 17
Number of balls in Box B at first
= 7 u + 37
= 7 x 17 + 37
= 119 + 37
= 156
Number of balls in Box C and Box A at first
= 360 - 156
= 204
Box B : Box C and Box A
156 : 204
(÷12)13 : 17
Answer(s): 13 : 17