document.write( "Question 296126: Samuel left the house at noon heading to David's store, 60 miles away. David left his store at 2 pm, heading to Samuel's house and traveling at twice Samuel's speed. The two men met at 3 p.m. How fast was David traveling? \n" ); document.write( "
Algebra.Com's Answer #213476 by MathTherapy(10555)![]() ![]() You can put this solution on YOUR website! Samuel left the house at noon heading to David's store, 60 miles away. David left his store at 2 pm, heading to Samuel's house and traveling at twice Samuel's speed. The two men met at 3 p.m. How fast was David traveling?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since the journey from David's store to Samuel's house is 60 miles, then the sum of the distances traveled by the 2 will total 60, or, \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Samuel's distance covered + David's distance covered = 60 miles\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let Samuel's speed be S, then David's speed will be 2S, since David is traveling at twice Samuel's speed\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then Samuel's distance traveled = S(3), since it took him 3 hours to cover the distance to the meeting point, and\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "David's distance covered = 2S(1), since it took David 1 hour to cover the distance to the meeting point\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, we'll have: 3S + 2S = 60\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "5S = 60\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "S = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now, since S, or Samuel's speed is 12 mph, then David's speed = |