There were some starfruits and pears in Container E and Container F. In Container E, the ratio of the starfruits to the number of pears was 4 : 1. In Container F, the ratio of the number of starfruits to the number of pears was 3 : 2. There were 2 times as many fruits in Container E as in Container F. After another 56 pears were put into Container F, the ratio of the number of starfruits to the number of pears in Container F became 1 : 3. How many fruits were there in Container F in the end?
| Container E |
Container F |
2x5 =
10 u |
1x5 =
5 u |
| Starfruits |
Pears |
Starfruits |
Pears |
| 4x2 |
1x2 |
3 |
2 |
| 8 u |
2 u |
3 u |
2 u |
The total number of fruits in Container E is repeated. Make the total number of fruits in Container E the same. LCM of 2 and 5 is 10.
The total number of fruits in Container F at first is repeated. Make the total number of fruits in Container F the same. LCM of 1 and 5 is 5.
| |
Container E |
Container F |
| |
Starfruits |
Pears |
Starfruits |
Pears |
| Before |
8 u |
2 u |
3 u |
2 u |
| Change |
|
|
|
+ 56 |
After
|
8 u
|
2 u
|
1x3 =
3 u |
3x3 =
9 u |
Number of starfruits in Container F remains unchanged. Make the number of starfruits in Container F the same. LCM of 3 and 1 is 3.
Number of pears put into Container F
= 9 u - 2 u
= 7 u
7 u = 56
1 u = 56 ÷ 7 = 8
Number of fruits in Container F in the end
= 3 u + 9 u
= 12 u
= 12 x 8
= 96
Answer(s): 96
