document.write( "Question 881687: a man reaches his office from his home 40 minutes too late.if he walks at 3 km an hour and 30 minutes too soon.if he walks 4 km an hour.how far is his office frm his home? \n" ); document.write( "
Algebra.Com's Answer #532407 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! a man reaches his office from his home 40 minutes too late if he walks at 3 km an hour. \n" ); document.write( " And 30 minutes too soon, if he walks 4 km an hour. \n" ); document.write( "how far is his office from his home? \n" ); document.write( ": \n" ); document.write( "Let d = distance from his home to his office \n" ); document.write( "the difference between being 40 min late and 30 min to soon, is 70 min \n" ); document.write( "70 min \n" ); document.write( ": \n" ); document.write( "Write time equation; time = dist/speed \n" ); document.write( "Late time - early time = 70 min \n" ); document.write( " \n" ); document.write( "multiply equation by 24 \n" ); document.write( "24* \n" ); document.write( "Cancel the denominators and you have \n" ); document.write( "8d - 6d = 4(7) \n" ); document.write( "2d = 28 \n" ); document.write( "d = 28/2 \n" ); document.write( "d = 14 km from home to work \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "See if that checks out, find the time at each speed \n" ); document.write( "14/3 = 4.67 hrs \n" ); document.write( "14/4 = 3.5 hrs \n" ); document.write( "--------------- \n" ); document.write( "time dif: 1.17 hrs which is about 70 min \n" ); document.write( " \n" ); document.write( " |