document.write( "Question 1115037: Jessie paddled her canoe 20 miles upstream, then paddled back. If the speed of the current was
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document.write( "3 mph and the total trip took 7 hours, what was Jessie’s speed? \n" );
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Algebra.Com's Answer #729991 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "If x is her speed in still water, then her speed upstream is x-3, and her speed downstream is x+3. \n" ); document.write( "The time for the upstream trip is 20/(x+3); the time for the downstream trip is 20/(x-3). \n" ); document.write( "Since the total time is 7 hours... \n" ); document.write( " \n" ); document.write( "Multiply both sides of the equation by the common denominator, (x+3)(x-3); you will get a quadratic equation which has one positive root and one negative. Obviously the positive root is the answer to the question. \n" ); document.write( "I leave the details of the algebra to you. \n" ); document.write( "Here is a way to solve the problem without the algebra by using logical reasoning. \n" ); document.write( "Note that the total time for the trip upstream and back is 7 hours. Not 7.2 hours; not 7 hours, 3 minutes, and 43 seconds. EXACTLY 7 hours. \n" ); document.write( "Because the total trip time is a whole number of hours, the times for each leg of the trip are almost certain to be whole numbers. \n" ); document.write( "Now look at the basic equation for solving the problem. 20 divided by x+3 should be a whole number; and 20 divided by x-3 should be a whole number. \n" ); document.write( "So you need to find two numbers whose difference is 6 (the difference between x+3 and x-3) that are both divisors of 20. \n" ); document.write( "The divisors of 20 are 1, 2, 4, 5, 10, and 20. The only two that have a difference of 6 are 4 and 10. \n" ); document.write( "So her upstream speed is 4mph and her downstream speed is 10mph; 20/4 + 20/10 = 5+2 = 7 hours. \n" ); document.write( "And since the speed of the current is 3mph, her speed in still water is 7mph. \n" ); document.write( " |