Question 1197605
<br>
First a standard algebraic method for solving the problem....<br>
x = hours at 40mph
5-x = hours at 50mph<br>
40(x) = distance traveled at 40mph
50(5-x) = distance traveled at 50mph<br>
The total distance was 220 miles:<br>
40x+50(5-x) = 220
40x+250-50x = 220
30 = 10x
x = 3<br>
ANSWER: He traveled for 3 hours at 40mph (and 2 hours at 50mph)<br>
CHECK: 3(40)+2(50) = 120+100 = 220<br>
And next an alternative method that I like to use on problems like this; look at it and see if it "works" for you.<br>
(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<br>
ANSWER: 2/5 of 5 hours, or 2 hours, at 50mph; the other 3 hours at 40mph<br>