document.write( "Question 356612: Fred walks 3 miles an hour. John walks 2 miles an hour. If they live 4 miles apart, and leave their house at the same time, when will they meet? \n" ); document.write( "
Algebra.Com's Answer #254548 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Fred walks 3 miles an hour. John walks 2 miles an hour. If they live 4 miles apart, and leave their house at the same time, when will they meet? \n" ); document.write( ".. \n" ); document.write( "fred = 3mph \n" ); document.write( "john = 2mph \n" ); document.write( "they walk towards each other \n" ); document.write( "so their effective speed = 3+2 = 5mph \n" ); document.write( ".. \n" ); document.write( "distance = 4 miles \n" ); document.write( "time = distance / speed \n" ); document.write( "4/5 hours \n" ); document.write( "=48 minutes \n" ); document.write( "... \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |