There were some pears and kiwis in Container E and Container F. In Container E, the ratio of the pears to the number of kiwis was 4 : 1. In Container F, the ratio of the number of pears to the number of kiwis was 3 : 2. There were 3 times as many fruits in Container E as in Container F. After another 20 kiwis were put into Container F, the ratio of the number of pears to the number of kiwis in Container F became 1 : 2. How many fruits were there in Container F in the end?
| Container E |
Container F |
3x5 =
15 u |
1x5 =
5 u |
| Pears |
Kiwis |
Pears |
Kiwis |
| 4x3 |
1x3 |
3 |
2 |
| 12 u |
3 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 3 and 5 is 15.
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 |
| |
Pears |
Kiwis |
Pears |
Kiwis |
| Before |
12 u |
3 u |
3 u |
2 u |
| Change |
|
|
|
+ 20 |
After
|
12 u
|
3 u
|
1x3 =
3 u |
2x3 =
6 u |
Number of pears in Container F remains unchanged. Make the number of pears in Container F the same. LCM of 3 and 1 is 3.
Number of kiwis put into Container F
= 6 u - 2 u
= 4 u
4 u = 20
1 u = 20 ÷ 4 = 5
Number of fruits in Container F in the end
= 3 u + 6 u
= 9 u
= 9 x 5
= 45
Answer(s): 45
