Oh no! Buzzy's powerboat is caught in the current and is being sucked towards the waterfall. The boat can run at 16mph but the current is pulling him towards the falls at 9mph. Buzzy's motor uses 5 gallons of gas an hour , he has 16 gallons left and the first safe anding is 21 miles back up river. Will he make it ? regards
Originally posted by HS Thomas: Oh I forgot to write the question "How?" regards
Wanted to be the first to answer the question...thatz why Here is the solution. He can sail at a speed of 7 mph (16 - 9). At this speed, it will take him 3 hours to reach the safe landing ares (21/7). So, to travel for 3 hours, he needs 15 gallons (3X5) and he has an extra gallon to spare. Thus he can make it and donate the remaining gallon to me [ October 24, 2003: Message edited by: Mani Ram ]
i must be missing something. if he heads straight upstream, at 16 mph, and the current is pulling him back at 9mph, his effective speed is 7mph (relative to the shore). he has 16 gallons of fuel, being burned at 5 gallons/hour. this give him 3 hours and 12 minutes to travel upstream. that should allow him to go 22.4 miles, plenty far enough to make the landing... what am i missing?
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Joined: May 15, 2002
what am i missing?
I don't think you are missing anything. The safe landing is 21 miles upstream as indicated in the question. You just calculated Buzzy has enough fuel to land 22.4 miles back up river. both of you regards
That's it. Do you think there should be more ? Perhaps, if you are going against the current in a power boat I am not sure if the drift (or drag) is a simple substraction. But that was the answer given.
I’ve looked at a lot of different solutions, and in my humble opinion Aspose is the way to go. Here’s the link: http://aspose.com