document.write( "Question 1185319: Wilma drove at an average speed of 45 mi/h from her home in City A to visit her sister in City B. She stayed in City B 10 hours, and on the trip back averaged 50 mi/h. She returned home 48 hours after leaving. How many miles is City A from City B? \n" ); document.write( "
Algebra.Com's Answer #816139 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "First a standard algebraic setup for solving the problem....

\n" ); document.write( "d = distance between A and B

\n" ); document.write( "time going from A to B: d/45
\n" ); document.write( "time going from B to A: d/50

\n" ); document.write( "Total driving time: 48-10=38

\n" ); document.write( "\"d%2F45%2Bd%2F50=38\"

\n" ); document.write( "Solve using basic algebra; probably start by multiplying both sides by a common denominator to clear fractions.

\n" ); document.write( "I leave it to you to finish solving the problem by that method.

\n" ); document.write( "Here is a very different way of solving the problem....

\n" ); document.write( "The distances both directions are the same; the ratio of speeds is 50:45=10:9. That means the ratio of times spent at the two speeds is 9:10.

\n" ); document.write( "So let the time at 50mph be 9x and the time at 45mph be 10x.

\n" ); document.write( "The total time is 38 hours:

\n" ); document.write( "\"9x%2B10x=38\"
\n" ); document.write( "\"19x=38\"
\n" ); document.write( "\"x=2\"

\n" ); document.write( "The time at 50mph is 9x=18 hours; that means the distance between A and B is 50(18)=900 miles.

\n" ); document.write( "Note also the time at 45mph is 10x=20 hours, which means the distance between A and B is 45(20)=900 miles -- which of course agrees with the answer we already found.

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