Steven needs to be at a meeting in x minutes. His house is x miles away from his meeting. At which speed will he arrive late? (x is the same number in both instances)

At 50mph with x miles distance, time = x/50 hours. Since he arrives late for x minutes = x/60 hours, we have x/50 > x/60 (which is true since 1/50 > 1/60). For x+1 mph, time = x/(x+1) which for x > 0 is less than 1 hour, so he's on time. At 60mph, time = x/60 hours exactly, which is exactly x minutes. At x/2 mph, he's much slower (twice the 60mph time).