SOLUTION: A computer technician travels 220 on a service call. He averaged 40 miles per hour for part of the trip and averaged 50 mph for the rest of the trip. If the total trip took five h

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A computer technician travels 220 on a service call. He averaged 40 miles per hour for part of the trip and averaged 50 mph for the rest of the trip. If the total trip took five h      Log On


   

Question 1197605: A computer technician travels 220 on a service call. He averaged 40 miles per hour for
part of the trip and averaged 50 mph for the rest of the trip. If the total trip took five hours
how long did he travel at a rate of 40 miles per hour?
Found 3 solutions by greenestamps, ewatrrr, josgarithmetic:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


First a standard algebraic method for solving the problem....

x = hours at 40mph
5-x = hours at 50mph

40(x) = distance traveled at 40mph
50(5-x) = distance traveled at 50mph

The total distance was 220 miles:

40x+50(5-x) = 220
40x+250-50x = 220
30 = 10x
x = 3

ANSWER: He traveled for 3 hours at 40mph (and 2 hours at 50mph)

CHECK: 3(40)+2(50) = 120+100 = 220

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

(1) The average speed was 220/5 = 44mph
(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)
(3) that means 2/5 of the total time was spent at the higher rate

ANSWER: 2/5 of 5 hours, or 2 hours, at 50mph; the other 3 hours at 40mph

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi  
Total distance traveled = 220mi in a total of 5hr
D = rt
50t + 40(5-t) = 220mi
 10t = 20
t = 2hr, time traveled at 50mph
  5hr - 2hr = 3hr traveling at 40mph
 checking:
 100mi + 120mi = 220mi  checks

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Review this analysis, understand it, and you should know what to do with it.
                SPEED     TIME      DISTANCE

FIRSTPART       40         x        40x

LASTPART        50        5-x       50(5-x)

TOTAL                      5        220