document.write( "Question 1197605: A computer technician travels 220 on a service call. He averaged 40 miles per hour for
\n" ); document.write( "part of the trip and averaged 50 mph for the rest of the trip. If the total trip took five hours
\n" ); document.write( "how long did he travel at a rate of 40 miles per hour?
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Algebra.Com's Answer #830933 by greenestamps(13203)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "First a standard algebraic method for solving the problem....

\n" ); document.write( "x = hours at 40mph
\n" ); document.write( "5-x = hours at 50mph

\n" ); document.write( "40(x) = distance traveled at 40mph
\n" ); document.write( "50(5-x) = distance traveled at 50mph

\n" ); document.write( "The total distance was 220 miles:

\n" ); document.write( "40x+50(5-x) = 220
\n" ); document.write( "40x+250-50x = 220
\n" ); document.write( "30 = 10x
\n" ); document.write( "x = 3

\n" ); document.write( "ANSWER: He traveled for 3 hours at 40mph (and 2 hours at 50mph)

\n" ); document.write( "CHECK: 3(40)+2(50) = 120+100 = 220

\n" ); document.write( "And next an alternative method that I like to use on problems like this; look at it and see if it \"works\" for you.

\n" ); document.write( "(1) The average speed was 220/5 = 44mph
\n" ); document.write( "(2) the average of 44mph is 4/10 = 2/5 of the way from 40mph to 50mph (use a number line to see that, if it helps)
\n" ); document.write( "(3) that means 2/5 of the total time was spent at the higher rate

\n" ); document.write( "ANSWER: 2/5 of 5 hours, or 2 hours, at 50mph; the other 3 hours at 40mph

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