Three containers, C, A and B, contained 240 balls. Fred added some balls into Container C and the number of balls in Container C tripled. He took out half of the number of balls from Container A and removed 64 balls from Container B. As a result, the ratio of the number of balls in Container C, Container A and Container B became 12 : 2 : 3. What was the ratio of the number of balls in Container B to the total number of balls in Container C and Container A at first? Give the answer in its lowest term.
| |
Container C |
Container A |
Container B |
Total |
| Before |
1x4 =
4 u |
2x2 =
4 u |
3 u + 64 |
240 |
| Change |
+ 2x4 =
+ 8 u |
- 1x2 =
- 2 u |
- 64 |
|
| After |
3x4 =
12 u |
1x2 =
2 u |
|
|
Comparing the
3 containers |
12 u
|
2 u |
3 u |
|
The number of balls in Container C in the end is the same. Make the number of balls in Container C in the end the same. LCM of 3 and 12 is 12.
The number of balls in Container A in the end is the same. Make the number of balls in Container A in the end the same. LCM of 1 and 2 is 2.
Total number of balls at first
= 4 u + 4 u + 3 u + 64
= 11 u + 64
11 u + 64 = 240
11 u = 240 - 64
11 u = 176
1 u = 176 ÷ 11 = 16
Number of balls in Container B at first
= 3 u + 64
= 3 x 16 + 64
= 48 + 64
= 112
Number of balls in Container C and Container A at first
= 240 - 112
= 128
Container B : Container C and Container A
112 : 128
(÷16)7 : 8
Answer(s): 7 : 8
