Three boxes, C, A and B, contained 219 beads. Riordan added some beads into Box C and the number of beads in Box C tripled. He took out half of the number of beads from Box A and removed 29 beads from Box B. As a result, the ratio of the number of beads in Box C, Box A and Box B became 12 : 4 : 7. What was the ratio of the number of beads in Box B to the total number of beads 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 + 29 |
219 |
Change |
+ 2x4 = + 8 u |
- 1x4 = - 4 u |
- 29 |
|
After |
3x4 = 12 u |
1x4 = 4 u |
|
|
Comparing the 3 boxes |
12 u
|
4 u |
7 u |
|
The number of beads in Box C in the end is the same. Make the number of beads in Box C in the end the same. LCM of 3 and 12 is 12.
The number of beads in Box A in the end is the same. Make the number of beads in Box A in the end the same. LCM of 1 and 4 is 4.
Total number of beads at first
= 4 u + 8 u + 7 u + 29
= 19 u + 29
19 u + 29 = 219
19 u = 219 - 29
19 u = 190
1 u = 190 ÷ 19 = 10
Number of beads in Box B at first
= 7 u + 29
= 7 x 10 + 29
= 70 + 29
= 99
Number of beads in Box C and Box A at first
= 219 - 99
= 120
Box B : Box C and Box A
99 : 120
(÷3)33 : 40
Answer(s): 33 : 40