Three containers, B, C and A, contained 165 balls. Fred added some balls into Container B and the number of balls in Container B tripled. He took out half of the number of balls from Container C and removed 22 balls from Container A. As a result, the ratio of the number of balls in Container B, Container C and Container A became 12 : 3 : 1. What was the ratio of the number of balls in Container A to the total number of balls in Container B and Container C at first? Give the answer in its lowest term.
| |
Container B |
Container C |
Container A |
Total |
| Before |
1x4 =
4 u |
2x3 =
6 u |
1 u + 22 |
165 |
| Change |
+ 2x4 =
+ 8 u |
- 1x3 =
- 3 u |
- 22 |
|
| After |
3x4 =
12 u |
1x3 =
3 u |
|
|
Comparing the
3 containers |
12 u
|
3 u |
1 u |
|
The number of balls in Container B in the end is the same. Make the number of balls in Container B in the end the same. LCM of 3 and 12 is 12.
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 1 and 3 is 3.
Total number of balls at first
= 4 u + 6 u + 1 u + 22
= 11 u + 22
11 u + 22 = 165
11 u = 165 - 22
11 u = 143
1 u = 143 ÷ 11 = 13
Number of balls in Container A at first
= 1 u + 22
= 1 x 13 + 22
= 13 + 22
= 35
Number of balls in Container B and Container C at first
= 165 - 35
= 130
Container A : Container B and Container C
35 : 130
(÷5)7 : 26
Answer(s): 7 : 26
