Three boxes, C, A and B, contained 316 marbles. Ken added some marbles into Box C and the number of marbles in Box C tripled. He took out half of the number of marbles from Box A and removed 76 marbles from Box B. As a result, the ratio of the number of marbles in Box C, Box A and Box B became 12 : 3 : 5. What was the ratio of the number of marbles in Box B to the total number of marbles 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 |
2x3 =
6 u |
5 u + 76 |
316 |
| Change |
+ 2x4 =
+ 8 u |
- 1x3 =
- 3 u |
- 76 |
|
| After |
3x4 =
12 u |
1x3 =
3 u |
|
|
Comparing the
3 boxes |
12 u
|
3 u |
5 u |
|
The number of marbles in Box C in the end is the same. Make the number of marbles in Box C in the end the same. LCM of 3 and 12 is 12.
The number of marbles in Box A in the end is the same. Make the number of marbles in Box A in the end the same. LCM of 1 and 3 is 3.
Total number of marbles at first
= 4 u + 6 u + 5 u + 76
= 15 u + 76
15 u + 76 = 316
15 u = 316 - 76
15 u = 240
1 u = 240 ÷ 15 = 16
Number of marbles in Box B at first
= 5 u + 76
= 5 x 16 + 76
= 80 + 76
= 156
Number of marbles in Box C and Box A at first
= 316 - 156
= 160
Box B : Box C and Box A
156 : 160
(÷4)39 : 40
Answer(s): 39 : 40
