Average Speed on GMAT

Responding to a pm:
Solving this using weighted averages.

When talking about speeds, what are the weights? The weight given to each speed is the time for which that speed was maintained.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Avg Speed = (Speed1*Time1 + Speed2*Time2)/(Time 1 + Time2)
Avg Speed = (Distance1 + Distance2)/(Time 1 + Time2)
Avg Speed = Total Distance/Total Time

So the weighted avg also ultimately boils down to this simple formula only.

Distance traveled cannot be the weight.

Avg Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2) makes no physical sense.

Since this question gives you the fraction of distance, you will need to convert it to time to use weighted avgs. Easier would be to use Total Distance/Total Time

Say total distance is D

Time taken 1 = Distance1/Speed1 \(= \frac{(x% of D)}{40} = \frac{xD}{4000}\)
Time taken 2 = Distance2/Speed2 \(= \frac{[(1 - x%) of D]}{60} = \frac{(100-x)D}{6000}\)

Avg Speed = Total Distance/Total Time \(= \frac{D}{\frac{xD}{4000} + \frac{(100-x)D}{6000}} = \frac{12000}{200+x}\)

Answer (E)

Another way to deal with this question would be to plug in values in the options.
When you put x = 100, you should get the avg speed as 40 (entire distance traveled at 40 mph)
When you put x = 0, you should get the avg speed as 60 (entire distance traveled at 60 mph)

A. (180-x)/2 -------- Ruled out
B. (x+60)/4 -------- Ruled out
C. (300-x)/5
D. 600/(115-x) -------- Ruled out
E. 12,000/(x+200)

C. (300-x)/5 --------- When you put x = 50 here, you get avg speed = 50. This is not correct. Avg speed will be 50 only when time taken in both cases is equal. Here distance traveled in both cases will be equal. When you put x = 50, you should get avg speed = 48 (When distance traveled is the same, avg speed = 2ab/(a+b) = 2*40*60/(40+60) = 48)

Answer must be (E) by elimination.