There were some dragonfruits in 3 baskets, F, G and H. 25% of the number of dragonfruits in Basket F was equal to 10% of the number of dragonfruits in Basket G. The number of dragonfruits in Basket H was 70% of the total number of dragonfruits. After Oliver removed 20% of the dragonfruits in Basket H, there were 968 more dragonfruits in Basket H than in Basket G. In the end, how many dragonfruits should be transferred from Basket G to Basket H so that the number of dragonfruits in Basket F would be the same as Basket G?
Basket F |
Basket G |
Basket H |
2x3 |
5x3 |
|
3x7 |
7x7 |
6 u |
15 u |
49 u |
|
Basket F |
Basket G |
Basket H |
Before |
6 u |
15 u |
49 u |
Change |
|
|
- 9.8 u |
After |
6 u |
15 u |
39.2 u |
25% =
25100 =
14 10% =
10100 =
110 14 Basket F =
110 Basket G
Basket F : Basket G
4 : 10
2 : 5
70% =
70100 =
710The total number of dragonfruits in Basket F and Basket G is the combined repeated identity. Make the total number of dragonfruits in Box F and Basket G the same. LCM of 7 and 3 = 21
Number of dragonfruits removed from Basket H
=
20100 x 49 u
= 9.8 u
Number of dragonfruits left in Basket H
= 49 u - 9.8 u
= 39.2 u
Number of more dragonfruits in Basket H than Basket G
= 39.2 u - 15 u
= 24.2 u
24.2 u = 968
1 u = 968 ÷ 24.2 = 40
|
Basket F |
Basket G |
Basket H |
Before |
6 u |
15 u |
39.2 u |
Change |
|
- 9 u |
+ 9 u |
After |
6 u |
6 u |
48.2 u |
Number of dragonfruits to be transferred from Basket G to Basket H so that the number of dragonfruits in Basket F would be the same as Basket G.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360
Answer(s): 360