When you travel to work going 50 mph, you arrive too early when you leave at a specified time. When you travel 40 mph, you arrive too late, leaving at that same time. In both cases, you arrive exactly as early as you arrive late (i.e., time early = time late).

How fast should you go to get to work on time, leaving at the same time?

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x/50 – y = x/40 + y – 1

x is the speed that gets you there on time, y is the difference how late/early you arrive when going 40 & 50mph

If x = 45, the values for y are not equal. Only at 44.444… does this equation solve with y having the same value on both sides (0.1111)

So, travelling at 50 gets you there 0.111 units of time late, and travelling 40 gets you there 0.111 units of time early, travelling at 44.44… (.4 repeated technically) gets you there right on time.

My equation may not be correct, but the logic is sound. It may be best to start by representing the problem by two equations:

x/50 = 1-y

x/40 = 1+y

Solve the second for y, would be , then substitute that for y in the first equation

x/50 = 1-(x/40 -1)

This reduces to x = 400/9, which is 44.44444444 repeated.

Crabman, you are today’s winner!

44.44 mph. For the sake of easy mathematics, let’s say you need to go 200 miles to get to work. At 50 mph it will take you 4 hours; at 40 mph it will take you 5 hours. The essence of the problem is that you need to find the average speed of traveling 400 miles in 9 hours, which is 44.444….mph.