document.write( "Question 907728: Elizabeth Suco and her friend Harold Tanner are riding their bicycles in a fundraiser for charity. Harold leaves the starting point at 7:00 AM and rides at 15 miles per hour. Elizabeth leaves the starting point at 7:30 AM and rides at 18 miles per hour.
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document.write( "How long after Elizabeth leaves the starting point will she catch up?
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document.write( "How far from the starting point will they be? \n" );
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Algebra.Com's Answer #550520 by mananth(16946)![]() ![]() You can put this solution on YOUR website! the distance traveled by both is same when one catches up with the other. \n" ); document.write( "Let them meet at distance x \n" ); document.write( "Suco speed= 15 mph , starts at 00:00 am \n" ); document.write( "Tanner speed= 18 mph starts at 07:30 am \n" ); document.write( " \n" ); document.write( "Time difference between the starting of the two= 0.50 hours \n" ); document.write( " \n" ); document.write( "Time taken by Suco = X / 15 \n" ); document.write( "Time taken by Tanner = X / 18 \n" ); document.write( " \n" ); document.write( "X / 15 - X / 18 = 0.5 \n" ); document.write( " \n" ); document.write( "LCD - 90 \n" ); document.write( "Multiply equation by 90 90 \n" ); document.write( " \n" ); document.write( "6 x - 5 x = 45 \n" ); document.write( "1 x = 45 \n" ); document.write( "/ 1 \n" ); document.write( "x= 45 miles \n" ); document.write( "Tanner speed= 18 mph \n" ); document.write( "Distance = 45 miles \n" ); document.write( "Time taken = 45 / 18 \n" ); document.write( "Time taken to catch up = 2.5 hour \n" ); document.write( " \n" ); document.write( " |