The math tea traveled to a math competition. They averaged 40mph there and 55mph back. Whats was the average speed for the entire trip?

Question

He said it would be wrong if you were to add them up and divide by 2.

Answer 1

I tried posting a comment but it never got posted. In doing the problem I ended up with the simple form of the harmonic mean. 2/([1/x]+[1/y])=2/([x+y]/xy)=2xy/(x+y) where x and y are 40 and 55

2*40*55/(40+55)=4400/95

d=40T₁

d=55T₂

d/40=T₁

d/55=T₂

2d=r(T₁+T₂)

2d=r([d/40]+d/55])

2d=r([55d+40d]/40*55)

2d=r(95d/2200)

(2200/95d)2d=r(95d/2200)(2200/95d)

(2*2200)/95=r

4400/95=r

46.3 mph=r

Answer 2

Yes, average speed needs to be TOTAL miles divided by TOTAL hours, neither of which we appear to know at this point.  However, the miles must be the same there and back, so we can work with that.  Let's start by calling that distance x.

 

Distance = rate * time, or D=rt.  So t = D/r.  "Flipping" the rates, we get:

 

(1 hr/40 mi)*(x miles) = time 1, or t₁

and

(1 hr/55 mi)*(x miles) = time 2, or t₂

 

It looks messy, but I think we have what we need.  Again, average speed is total miles divided by total hours:

 

• Total miles = 2x (there and back)
• Total time = t₁+t₂ = x/40+x/55
• Average speed = 2x/(x/40+x/55)
• (The tricky part will be doing this math.  We'll need a common denominator first.  Note that the unknown x's will all cancel)
• S_avg=2/(11/440+8/440)
• S_avg=2/(19/440)=2*440/19 = 880/19 mph , or roughly 46.3 mph  
• This makes sense, as we would expect the average to be closer to the 40 than the 55, since more time was spent driving at this lower speed.