document.write( "Question 515751: Victor and Rich live 100 miles apart. They agree to meet at the bridge between their two houses. Victor runs at 20 mph but Rich just walks at 8 mph. If each leaves home at 11 a.m, when do they meet? \n" ); document.write( "
Algebra.Com's Answer #344148 by oberobic(2304)![]() ![]() ![]() You can put this solution on YOUR website! d = distance \n" ); document.write( "r = rate \n" ); document.write( "t = time \n" ); document.write( ". \n" ); document.write( "d = r*t is the basic distance equation \n" ); document.write( ". \n" ); document.write( "You will use this to solve the problem. \n" ); document.write( ". \n" ); document.write( "d = 100, which is known. \n" ); document.write( ". \n" ); document.write( "The two guys travel at different rates, but the rates are known: 20 mph and 8 mph \n" ); document.write( ". \n" ); document.write( "If they leave at the same time, then their elapsed times will be the same. \n" ); document.write( ". \n" ); document.write( "20 mph * t hr + 8 mph * t hr = 100 miles \n" ); document.write( ". \n" ); document.write( "t(20+8) = 100 \n" ); document.write( ". \n" ); document.write( "t = 100/28 \n" ); document.write( ". \n" ); document.write( "t= 25/7 = 3 4/7 hr \n" ); document.write( ". \n" ); document.write( "They will meet in 3 4/7 hr. \n" ); document.write( ". \n" ); document.write( "4/7 of an hour = 4/7 * 60 min/hr = about 34.3 minutes \n" ); document.write( ". \n" ); document.write( "They left at 11 a.m. \n" ); document.write( "so \n" ); document.write( "11 a.m. + 3 hr 34.3 min = about 2:34 p.m. \n" ); document.write( ". \n" ); document.write( "We can check the distances traveled by each to be sure the answer is correct. \n" ); document.write( ". \n" ); document.write( "25/7 hr * 20 mph = 500/7 miles \n" ); document.write( "25/7 hr * 8 mph = 200/7 miles \n" ); document.write( "500/7 + 200/7 = 700/7 = 100 miles \n" ); document.write( "Correct. \n" ); document.write( ". \n" ); document.write( "Done. \n" ); document.write( " |