speed for first leg --- x mph
speed for 2nd leg = x+10 mph
time for first leg = 120/x
time for 2nd leg = 150/(x+10)
120/x - 150/(x+10) = 6/60 = 1/10
multiply each term by 10x(x+10)
10(x+10)(120) - 10x(150) = x(x+10)
1200x + 12000 - 1500x = x^2 + 10x
x^2 + 310x + 12000 = 0
I don't think it factors, use the quadratic formula to solve, reject the negative value of x
A salesman drives from Ajax to
Barrington, a distance of 120 mi, at a steady speed. He then
increases his speed by 10 mi/h to drive the 150 mi from Bar-
rington to Collins. If the second leg of his trip took 6 min
more time than the first leg, how fast was he driving between
Ajax and Barrington?
1 answer
1 answer