A family of four was stranded on an island, two adults and two children. They then discivered a boat, which can only hold 200 pounds(the most) at a time. Each parent weighs 150 pounds and each child 50 pounds. How mant trips will it take to get the who family to the nearby city?
Kid + parent, drop parent off, kid goes back, gets other kid, drops off, comes back for last parent. 3 trips.
if kid + parent go first and drop off parent. thats one trip. Then you have to go back for the other kid 2 trips. Then drop that kid to the city three trips. At this piont everyone is not at the city so it cant be 3 trips.
I thought a trip would be there and back.
When you go on vacation, you don't call it two trips, it's just one trip. There, then back.
So I misunderstood.
You're not counting the trips back.
Parent + child go to city = 1 trip
Parent left behind and child returns for 2nd parent = 1 trip
2nd parent and first child go to city = 1 trip
child left behind and a parent returns for final child = 1 trip
parent and 2nd child go to city = 1 trip
5 trips total, assuming the child is old enough or capable enough to make that 2nd trip back for parent #2...
One parent could go to the nearby city and get larger boat with better capacity, or even return with helicopter?
If you can only use the boat, then it could be in done in one trip with a little ingenuity, both children in boat and parent swimming or slightly holding on to the outside.
Since there is no determination of distance to the nearby city and trip is not defined either. I would go with one trip.
I have actually participated in such a venture only it involved a lot of camping equipment and trespassing on the property of a dead president.
Dontcha just hate it when someone starts a thread like this and leaves us all hanging?! Oh, well....my hand is going down...
I agree, richwf. Let's just consider ourselves winners and move on!
i guess the answer is five then, thats what i said when i was asked the question, thanks for answering this question guys
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