There were some avocados in 3 boxes, E, F and G. 30% of the number of avocados in Box E was equal to 60% of the number of avocados in Box F. The number of avocados in Box G was 60% of the total number of avocados in Box E and F. After Oscar removed 20% of the avocados in Box G, there were 88 more avocados in Box F than in Box G. In the end, how many avocados should be transferred from Box F to Box G so that the avocados in Box G would be the same as Box E?

Guest Apr 25, 2022

#1**+1 **

There were some avocados in 3 boxes, E, F and G. 30% of the number of avocados in Box E was equal to 60% of the number of avocados in Box F. The number of avocados in Box G was 60% of the total number of avocados in Box E and F. After Oscar removed 20% of the avocados in Box G, there were 88 more avocados in Box F than in Box G. In the end, how many avocados should be transferred from Box F to Box G so that the avocados in Box G would be the same as Box E?

I do not think that this scenario is possible.

30% of the number of avocados in Box E was equal to 60% of the number of avocados in Box F

so

0.3E=0.6F

E=2F

So in the beginning there are 2 times more in box E than in box F (And each one has a muliple of 10)

G=0.6(E+F)

G=0.6(3F)

G=1.8F

Now 20% are removed from box G

so box G will then have 0.8*1.8F = 1.44F

So now

box1 (originally called box E) | Box2 (originally called box F) | box3 (originally called box G) |

2F | F | 1.44F |

It is clear that there are now mow avocados in the last box than in the middle one.

So there cannot be 88 more in the middle one, that is a contradiction.

Melody May 5, 2022