document.write( "Question 1171616: Paddlewheel Pat has a motorboat. He can travel a distance of 10 miles boating into the current in the same time it takes him to travel 15 miles with the current. If the current has a speed of 3 mph, what is the speed of Pat’s boat in still water? \n" ); document.write( "
Algebra.Com's Answer #796570 by greenestamps(13209)\"\" \"About 
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\n" ); document.write( "Here is a highly unusual way for solving this kind of problem that I personally find easy to use.

\n" ); document.write( "Take a look at it and see if it \"works\" for you. If not, there are of course many other ways to solve the problem.

\n" ); document.write( "The times are the same, and the distances are in the ratio 10:15 = 2:3. That means the speeds are in the ratio 2:3.

\n" ); document.write( "If the speed of the boat in still water is x, then the upstream speed is x-3 and the downstream speed is x+3.

\n" ); document.write( "Solve for x by writing the ratio 2:3 as an equivalent ratio in which the difference between the two numbers is the difference between the upstream and downstream speeds:

\n" ); document.write( "\"2%2F3+=+%28x-3%29%2F%28x%2B3%29\"

\n" ); document.write( "On the left, the difference between numerator and denominator is 1; on the right is is 6. To get the desired equivalent ratio, multiply numerator and denominator on the left by 6.

\n" ); document.write( "\"2%2F3+=+12%2F18+=+%28x-3%29%2F%28x%2B3%29\"

\n" ); document.write( "So x-3 is 12 and x+3 is 18; that means x is 15.

\n" ); document.write( "ANSWER: The speed of the boat in still water is 15 mph.

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