logic problem?

Question

45 watermelons.
a boy has to take them 15 miles to a fair
he eats one watermellon per mile
he can leave as many as he wants at each mile
he can only carry 15 at a time
how many get to the fair?

the highest i got was 8

he can leave some at each mile

you have to explain why

you can go back for watermellons

mt watermelons

Answer 1

The boy takes 11 watermelons to the fair.

He starts with 15 melons.Eats 1 every mile.He has 14 at 1st mile 13 at 2nd,12 at 3rd,11 at 4th,10 at 5th,9 at 6th,8 at 7th.

He drops all 8 at the 7th mile and walks back to his starting point,where he has another 30 melons.

He carriess them 15 at a time,twice in the same fashion as his first trip.

Now at the 7th mile he has 24 melons.(8 X 3)

He carries 15 from the 7th mile.He has 14 at 8th mile,13 at 9th,12 at 10,11 at 11.He drops the 11 melons and comes back to the 7th mile.
He carries the remaining 9 melons.He has 8 melons at 8th mile,7 at 9th,6 at 10th,5 at 11th.

At the 11th mile he now has 16 melons(11+ 5)

He carries 15 melons.He has 14 melons at 12th mile,13 at 13th,12 at 14th,and 11 at the 15 th mile,where the fair is located.

He need not go back for the 1 melon at the 11 th mile.He will eat that on his way home.

Answer 2

He get 8 to the fair. Since the problem doesn't state that he can't go back for more watermelons here is how he does it.

At mile 1 he carries 15 melons and leaves 13, eating 1 on the way there and bringing 1 back to eat on the way. He carries 15 of the 30 he left at the start point and leaves 13 more at mile 1 as before and eats the one he brings back on the way. He then carries the remaining 15 to mile one eating one on the way leaving 14. There are now 40 melons at mile 1.

at Mile 2 he carries 15 melons from mile 1 leaving 13 and eating 1 back to mile 1. then carries 15 to mile 2 leaving 13 and eating one back to mile 1. Carries the remaining 10 to mile 2 and have 4 when he get there leaving 35 melons at mile 2.

at mile 3 he carries 15 leaving 13 eating 1 back to mile 2, carries 15 more to mile 3 leaving 13, eating 1 on the way back for the remaining 5. Upon his return to mile 3 he has 4 left making 30 melons at mile 3.

at mile 4 he carries 15 melons leaving 13 and eating one back for the remaining 15 and coming back with 14. Leaving a total of 27 melons.

at mile 5 he carries 15 and leaves 13 eating one on the way back for the remaining 12 and arriving with only 11. He now has 24 melons left.

at mile 6 he carries 15 and leave 13 eating one on the way back for the remaining 9 and arriving with 8. with 21 melons left he goes to mile 7.

at mile 7 he carries 15 leaves 13 eats 1 on the way for the remaining 6 and arrives with 5. He has 18 melons left.

at mile 8 he carries 15 melons leaves 13 eats 1 to to get the last 3 arrives with 2 leaving him a total of 15 watermelons to carry the rest of the 7 miles to the fair eating 1 every mile and arriving at the fair with 8 watermelons and in serious need of medical attention.

Answer 3

Eating one per mile can mean either he eats one for every mile he actually walks or that he eats one for each mile he progresses towards the market.

Assuming that he eats one for each mile he progresses. He makes three trips to carry 45 watermelons the first mile and stops to eat one. He then carries the remaining 44 to the second mile in three trips. With this assumption, he can only eat 15 so he'd have 30 remaining at the end.

Assuming he eats one for every mile traveled: He carries 15 to mile 1, eats 1 and leaves 14. He returns to the start where there are 30, eats one, takes 15 and leaves 14. He carries these 15 to mile 1, eats one and leaves 28 at mile 1. He returns to the start, eats one and takes the remaining 13 to mile 1. When he gets there, he eats one and adds the remaining 12 to the 28 sitting at mile 1.

There are now 40 melons, he has eaten 5 and traveled one mile. So long as he has to make 3 trips, he will consume 5 watermelons. When he only has to make 2 trips, he will only consume 3 per mile progressed.

So, by the end of mile three he will be down to 30 watermelons. From this point for several miles he will only consume 3 melons per mile. Five miles farther down, at mile 8 he will be down to 15 melons. He will travel 7 more miles, consuming one watermelon as he completes each mile. This will reduce his load by 7, leaving him with 8 watermelons.

45 = 5 melons * 3 miles + 3 melons * 5 miles + 1 melon * 7 miles + 8 melons remaining.

Answer 4

He carries 15 the first mile, eating one, so 14 make it to the first mile. Then he takes one with him as he walks back to the start. Now there are 13 at the first mile. He takes another 15 with him to the first mile, eating one, so another 14 make it to the first mile (total = 27). He takes one back with him to the beginning again (total at the first mile = 26). He takes the last 15 to the first mile, eating one along the way. Total at the first mile = 40.

Using the same method, the total at the second mile = 35, the total at the third mile would be 30.

Now the rules change. He takes 15 to the fourth mile, eating one along the way. This leaves 14 at the fourth mile. He takes one for the walk back, leaving 13 at the fourth mile. Then he takes the 15 to the fourth mile, eating one along the way. Now there are 27 at the fourth mile.

Likewise, there would be 24 at the fifth mile, 21 at the sixth, 18 at the seventh, and 15 at the eighth mile.

From here, he eats one per mile, so the grand total that he would have at the fair would be 8 watermelons.

Answer 5

Say he ferries 15 water melons 1 mile, eating one. Then he makes to more trips so he has 42 watermelons at the 14 mile mark. So now say he moves them all 1 mile closer, again losing three., so now he is down to 39 with 12 miles to go. He keeps ferrying his watermelons until he has 30 at the 10 mile mark. At wich point he takes 15, carries them to the fair, eating one a mile, so 5 get to the fair. Then he goes back and gets the remaing 15, and 5 of those make it to the fair.

So all in all, up to ten make it to the fair.

Answer 6

Since at the 15th mile he will already reach the fair he won't eat the last one. He makes three rounds to carry all 45 to the fair, thus only 3 get to the fair.(but honestly I am not sure abotu the answer, because the information "he can leave as many as he wants at each mile" is a bit confusing)

Answer 7

Well he can leave 14 after the first mile( as he has ate 1), walk back and get the next 15, take them, leave them after a mile, walk back and get the next 15.....so altoger he eats 15 and takes 30...you never stated how far he had to walk....you just said every mile he eats a watermelon....

If that's not right then none....it's impossible i he has to eat one every time he travels a mile.

Answer 8

he can only carry 15 watermelons, so 15 to the fair, and 1 watermelon per mile, 15-15=0 no watermelons got to the fair.

Answer 9

If he travels 15 miles, can only carry 15 watermelons, and eats one every mile, then none make it to the fair, because he ate them all.

Answer 10

At face value, the answer is zero. Its implied that he eats a watermelon (sp!!) DURING each mile walk and, hence, completes one at the end of each mile length. Of course, if he rests & eats each mile, he'll have one remaining at the end of each 15 mile trek - hence the answer would be 3.
Thanks.