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Old 05-14-2008   #1
the girl
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Default can anyone help with this problem? i'm sooo frustrated with this!!

A boy has 45 watermelons in the desert. He needs to get them across to
the Oasis fair, 15 miles away. He can only carry 15 watermelons at a
time, and he eats one watermelon every mile he walks, including walking
back to where he started from. He can also leave watermelons at any
mile he has walked, but no fractions of a mile. How many watermelons
can he possibly take to the fair?
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Old 05-15-2008   #2
chris
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well the boy could possibly take maybe no watermelons at all considering tha fact that he eats one every mile he walks and even if he puts them down at every mile he still wouldnt be able to take any cause he would eat them all
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Old 05-15-2008   #3
chris
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i change my answer he can take 3 watermelons to the fair thats if he drops the watermelons at every mile
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Old 09-10-2008   #4
Math Tyrant
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Lightbulb Whoa!!!

He'd have no watermelons.
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Old 09-16-2008   #5
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Ok...he'd have 5 watermelons when he got to the fair.

He takes his first 15 watermelons and walks 1 mile. When he gets to his first mile, he has 14 left. He drops 13 and takes one back with him to eat on his return trip. He does this for two more trips for his other 30 watermelons. Now at his first mile, he has 39 watermelons.

He takes 13 watermelons and walks another mile (so now he is at mile 2). He drops 11 and takes one for the walk back. He does this two more times. At his second mile, he has 33 watermelons.

He takes 11 watermelons and walks another mile (so now he is at mile 3). He drops 9 and takes one for the return trip. He does this two more times. At his third mile, he has 27 watermelons.

He takes 9 watermelons and walks another mile (he is now at mile 4), drops 7 an takes one for the return trip. After two more trips like this, he has 21 watermelons at mile 4.

He takes 7 watermelons and walks one more mile. He drops 5 and takes one for the return trip. After two more trips, he now has 15 watermelons at mile number 5.

Since he has 15 watermelons now, he can walk the last 10 miles, eat one every mile and arrive at the fair with 5 uneaten, but probably very warm and possibly rotten watermelons! AND he has walked a total of 100 miles!!!
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Old 09-17-2008   #6
ladeewzl
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Default I change my answer!!!!

Ok...so I told this math problem to my husband and he came up with a different (and correct) answer!!

The boy will arrive at the fair with 8 watermelons.

He still travels by the mile and he will make three trips - the first and second trips he eats two watermelons each, but for the third trip (since he doesn't have a return trip to the starting point) he only eats 1 watermelon. So by mile number 1, he has eaten 5 watermelons.

The same thing happens for the trips between miles 2 and 3. So by mile number 3 he has eaten a total of 15 watermelons. Now, he only has to make two trips for each mile (since he only has 30 watermelons left).

Following the same logic as for the first three miles, at miles 4, 5, 6, 7, and 8, he will eat another 15 watermelons (2 for the first trip to the mile and 1 for the last for a total of 3 for each mile). So for those 5 miles, he has eaten another 15 watermelons.

Now he only has 15 watermelons left, 7 miles to go, which leaves him with 8 watermelons at the fair.
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Old 09-19-2008   #7
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Default Macman2008

r u doing CTRL+C/CTRL+V? Becuse someone else posted the exact same thing on this forum.
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Old 09-19-2008   #8
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Default Copy/paste?

No...I did NOT copy paste my answer...I distinctly remember typing it. Where else is it posted? I only see my post/response?????
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Old 04-12-2013   #9
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He eats 15 watermelons and he carries 15 watermelons to Oasis
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My life is always lonely and sad It needs friends(maybe even boyfriends)to fill it.
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