document.write( "Question 346579: . A man left A at 6:00 a.m. expecting to reach B at 9:00 a.m. But after walking one hour, he was delayed for half an hour and so he had to increase his rate by 1 mile per hour to reach B at 9:00 a.m. Find his speed before the delay and the distance between A and B.\r
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Algebra.Com's Answer #247878 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A man left A at 6:00 a.m. expecting to reach B at 9:00 a.m. \n" ); document.write( " But after walking one hour, he was delayed for half an hour and so he had to \n" ); document.write( " increase his rate by 1 mile per hour to reach B at 9:00 a.m. \n" ); document.write( " Find his speed before the delay and the distance between A and B. \n" ); document.write( ": \n" ); document.write( "Let d = distance from A to B \n" ); document.write( "Given that he left at 6, and expected to be there at 9, therefore: \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "He walked for 1 hr, then stopped for .5 hrs, therefore \n" ); document.write( "he continued at a speed of ( \n" ); document.write( ": \n" ); document.write( "dist = speed * time \n" ); document.write( "total dist = slow speed dist + faster speed dist \n" ); document.write( "d = \n" ); document.write( "d = \n" ); document.write( "d = \n" ); document.write( "multiply both sides by 3 \n" ); document.write( "3d = 2.5d + 3(1.5) \n" ); document.write( "3d - 2.5d = 4.5 \n" ); document.write( ".5d = 4.5 \n" ); document.write( "d = \n" ); document.write( "d = 9 miles from A to B \n" ); document.write( ": \n" ); document.write( "Find the original speed: \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Confirm this by finding the distance at each speed \n" ); document.write( "1 * 3 = 3 miles \n" ); document.write( "1.5 * 4 = 6 mile \n" ); document.write( "---------------- \n" ); document.write( "total dist: 9 mi \n" ); document.write( " |