Three containers, A, B and C, contained 158 marbles. Perry added some marbles into Container A and the number of marbles in Container A tripled. He took out half of the number of marbles from Container B and removed 68 marbles from Container C. As a result, the ratio of the number of marbles in Container A, Container B and Container C became 6 : 2 : 3. What was the ratio of the number of marbles in Container C to the total number of marbles in Container A and Container B at first? Give the answer in its lowest term.
|
Container A |
Container B |
Container C |
Total |
Before |
1x2 = 2 u |
2x2 = 4 u |
3 u + 68 |
158 |
Change |
+ 2x2 = + 4 u |
- 1x2 = - 2 u |
- 68 |
|
After |
3x2 = 6 u |
1x2 = 2 u |
|
|
Comparing the 3 containers |
6 u
|
2 u |
3 u |
|
The number of marbles in Container A in the end is the same. Make the number of marbles in Container A in the end the same. LCM of 3 and 6 is 6.
The number of marbles in Container B in the end is the same. Make the number of marbles in Container B in the end the same. LCM of 1 and 2 is 2.
Total number of marbles at first
= 2 u + 4 u + 3 u + 68
= 9 u + 68
9 u + 68 = 158
9 u = 158 - 68
9 u = 90
1 u = 90 ÷ 9 = 10
Number of marbles in Container C at first
= 3 u + 68
= 3 x 10 + 68
= 30 + 68
= 98
Number of marbles in Container A and Container B at first
= 158 - 98
= 60
Container C : Container A and Container B
98 : 60
(÷2)49 : 30
Answer(s): 49 : 30