There were some starfruits and kiwis in Box E and Box F. In Box E, the ratio of the starfruits to the number of kiwis was 6 : 1. In Box F, the ratio of the number of starfruits to the number of kiwis was 4 : 3. There were 2 times as many fruits in Box E as in Box F. After another 50 kiwis were put into Box F, the ratio of the number of starfruits to the number of kiwis in Box F became 1 : 2. How many fruits were there in Box E?
| Box E |
Box F |
2x7 =
14 u |
1x7 =
7 u |
| Starfruits |
Kiwis |
Starfruits |
Kiwis |
| 6x2 |
1x2 |
4 |
3 |
| 12 u |
2 u |
4 u |
3 u |
The total number of fruits in Box E is repeated. Make the total number of fruits in Box E the same. LCM of 2 and 7 is 14.
The total number of fruits in Box F at first is repeated. Make the total number of fruits in Box F the same. LCM of 1 and 7 is 7.
| |
Box E |
Box F |
| |
Starfruits |
Kiwis |
Starfruits |
Kiwis |
| Before |
12 u |
2 u |
4 u |
3 u |
| Change |
|
|
|
+ 50 |
After
|
12 u
|
2 u
|
1x4 =
4 u |
2x4 =
8 u |
Number of starfruits in Box F remains unchanged. Make the number of starfruits in Box F the same. LCM of 4 and 1 is 4.
Number of kiwis put into Box F
= 8 u - 3 u
= 5 u
5 u = 50
1 u = 50 ÷ 5 = 10
Number of fruits in Box E
= 12 u + 2 u
= 14 u
= 14 x 10
= 140
Answer(s): 140
