There were some kiwis and starfruits in Container R and Container S. In Container R, the ratio of the kiwis to the number of starfruits was 4 : 1. In Container S, the ratio of the number of kiwis to the number of starfruits was 3 : 2. There were 2 times as many fruits in Container R as in Container S. After another 28 starfruits were put into Container S, the ratio of the number of kiwis to the number of starfruits in Container S became 1 : 2. How many fruits were there in Container S in the end?
| Container R |
Container S |
2x5 =
10 u |
1x5 =
5 u |
| Kiwis |
Starfruits |
Kiwis |
Starfruits |
| 4x2 |
1x2 |
3 |
2 |
| 8 u |
2 u |
3 u |
2 u |
The total number of fruits in Container R is repeated. Make the total number of fruits in Container R the same. LCM of 2 and 5 is 10.
The total number of fruits in Container S at first is repeated. Make the total number of fruits in Container S the same. LCM of 1 and 5 is 5.
| |
Container R |
Container S |
| |
Kiwis |
Starfruits |
Kiwis |
Starfruits |
| Before |
8 u |
2 u |
3 u |
2 u |
| Change |
|
|
|
+ 28 |
After
|
8 u
|
2 u
|
1x3 =
3 u |
2x3 =
6 u |
Number of kiwis in Container S remains unchanged. Make the number of kiwis in Container S the same. LCM of 3 and 1 is 3.
Number of starfruits put into Container S
= 6 u - 2 u
= 4 u
4 u = 28
1 u = 28 ÷ 4 = 7
Number of fruits in Container S in the end
= 3 u + 6 u
= 9 u
= 9 x 7
= 63
Answer(s): 63
