document.write( "Question 58492: On a 130 mile trip, a car traveled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 mph? \n" ); document.write( "
Algebra.Com's Answer #40073 by ankor@dixie-net.com(22740)\"\" \"About 
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On a 130 mile trip, a car traveled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took 2.5 hours
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\n" ); document.write( "Let t = no. of hrs at 55 mph
\n" ); document.write( "(2.5-t) = no of hrs at 40 mph
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\n" ); document.write( "Dist = speed * time
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\n" ); document.write( "Dist at 55 mph = 55t
\n" ); document.write( "Dist at 40 mph = 40(2.5-t)
\n" ); document.write( "These two distances = 130 mph
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\n" ); document.write( "55t + 40(2.5-t) = 130
\n" ); document.write( "55t + 100 - 40t = 130
\n" ); document.write( "55t - 40t = 130 - 100
\n" ); document.write( "15t = 30
\n" ); document.write( "t = 30/2
\n" ); document.write( "t = 2 hrs at 55 mph
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\n" ); document.write( "Then 2.5 - 2 = .5 hrs at 40 mph
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\n" ); document.write( "Dist at 40 mph = .5*40 = 20 miles
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\n" ); document.write( "Check using distance:
\n" ); document.write( "55(2) + .5(40) =
\n" ); document.write( "110 + 20 = 130 miles \n" ); document.write( "
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