[lbo-talk] math problem...

Michael Pollak mpollak at panix.com
Wed Feb 22 22:38:00 PST 2006


On Wed, 22 Feb, Joanna wrote:


> Sorry, it's driving me crazy. Can anyone help with below. The equations I
> came up with don't seem to be right.
>
> Boat can travel 5 times speed of current.
> It can travel 375 miles downriver in 1 hour more than it takes to travel
> 225 miles upriver.
> What is speed of boat in still water?
>
> r = speed of current
> t = time
>
> 375 = 6r ( t + 1)
> 275 = 4r (t)

Your equations look fine to me. The boat's top speed is 5r. Going upriver it goes 6r (its top speed plus the current) and downriver 4r (its top speed minus the current)

The time down river is t hours, the time up river is t+1 hours.

So far so good. The only problem is you wrote "225" in the word problem part and "275" in the equation part. I'm going to assume the second is the typo. But the method works either way.

I think the trick you might have missed is to use "rt" as a variable and get it to drop out. I.e., if

4rt = 225, then

rt = 225/4

and you can simply substitute that into your first equation to solve for r

6r(t+1) = 375 6rt + 6r = 375 6(225/4) + 6r = 375 337.5 + 6r = 375 6r = 37.5 r = 6.25

Then substitute that solution into the second equation:

4rt = 225 4(6.25)(t) = 225 25(t) = 225 t = 9

Then check that these numbers work in both original equations:

4rt = 4(6.25)(9) = 25(9) = 225 CHECK 6r(t+1) = 6(6.25)(10) = 37.5(10) = 375 CHECK

Speed of the boat in still water = 5r = 5(6.25) = 31.25

Michael



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