SOLUTION: Charles made a business trip of 200.5 miles. He averaged 53mph for the first part of the trip and 59mph for the second part. If the trip took 3.5 hours, how long did he t

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Charles made a business trip of 200.5 miles. He averaged 53mph for the first part of the trip and 59mph for the second part. If the trip took 3.5 hours, how long did he t      Log On

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Question 1203986: Charles made a business trip of 200.5
miles. He averaged 53mph
for the first part of the trip and 59mph
for the second part. If the trip took 3.5
hours, how long did he travel at each rate?
Found 2 solutions by MathLover1, josgarithmetic:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!



Let +t+= time traveled at +53mph
then
+%283.5-t%29 = time traveled at +59mph
Total time is given as +3.5h, therefore

Write distance equation;
+d+=+s%2At
+53t+%2B+59%283.5-t%29+=200.5
+206.5+-+6+t+=+200.5
+206.5+-200.5+=+6+t+
+6+=+6+t
+t=1
so,
+t=1=>Charles traveled +1h+at +53mph
+%283.5-t%29=3.5-1=2.5=>Charles traveled +2.5h at +59mph


Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
You can arrange the described information this way:
                            (hiour)      (miles)
               SPEED        TIME        DISTANCE

part one        53

part two        59

TOTAL                        3.5         200.5


A variable can be used and the other information can be made like here:
                            (hiour)      (miles)
               SPEED        TIME        DISTANCE

part one        53         3.5-x        53(3.5-x)

part two        59            x           59x

TOTAL                        3.5         200.5

Use that and continue....