Posted by Dani on .
I need to know how to set up this problem:
A salesman drives from Ajax to Barrington, a distance of 120 miles, at a steady speed. He then increases his speed by 10 mph to drive the 150 miles from Barrington to Collins. If the second leg of his trip took 6 minutes more time than the first leg, how fast was he driving between Ajax and Barrington?

Math (Word Problems) 
Reiny,
let his first steady speed by x mph
so the time for the first leg of the trip was 120/x hours
on the second leg of the trip his speed was x+10 mph
so the time taken for that was 150/(x+10)
but this took 6 min or 1/10 hour longer
so 150/(x+10)  120/x = 1/10
multiply each term by x(x+10) to get a quadratic equation.
one of the roots should be extraneous, let me know what you got 
Math (Word Problems) 
drwls,
Let the two speeds be V1 and V2. The times required to travel the two legs of the journey are
120/V1 and 150/V2 = 150/(V1+10)
The second leg took 0.1 hours (6 minutes) more, so
150/(V1+10) = (120/V1) + 0.1
That equation can be solved for V1.
150/(V1 + 10) = (120 + 0.1V1)/V1
150 V1 = 120 V1 + 1200 + V1 + 0.1V1^2
0.1V1^2 29 V1 + 1200 = 0
V1 = [29 +/19]/0.2 = 50 or 240 mph
The higher of the two roots is unrealistic. Go with 50 mph