Problema Solution
A salesman drives from Ajax to Barrington, a distance of 129 mi, at a steady speed. He then increases his speed by 16 mi/h to drive the 183 mi from Barrington to Collins. If the second leg of his trip took 8 min more time than the first leg, how fast was he driving between Ajax and Barrington?
Answer provided by our tutors
First trip: 129 miles in x hours at a speed of v:
129/x = v
Second trip: 183 miles in 8 minutes more at a speed increased by 16:
183/(x+8/60)=v+16
Since v = 129/x from the first equation, we have:
183/(x+8/60)=129/x + 16
183 = (x+8/60) * 129/x + 16
...which leads to this quadratic equation:
778x - 240x^2 - 258 = 0
Solving for x, we get: { 43/15, 3/8 } = {2.87, 0.375}
It is highly unlikely that the salesman was "cruising" so fast that he arrived in .375 hours, so we choose 2.87 hours as the correct time from Ajax to Barrington.
His speed would have been: 129/2.87 = ~45mph