jordanshl wrote:
I kind of agree with the dissenters that it cannot be A.
If his average round trip was 80mph then it could have taken 30mph and 130mph meaning that the first leg was less than 30mph
It could also be 60 and 100 meaning he did it faster than 40
in sufficient
What you missed to note is that avg of 30 mph and 130 mph is 80 mph when he travels at the two speeds for the same TIME.
If someone travels at two speeds for the same amount of time, say an hr at one speed and an hr at another speed, then the average speed is (Speed1 + Speed 2)/2
Here, the case is different. The two speeds are for the same distance. He travels at Speed1 from A to B and at Speed2 from B to A (distance in the two cases are same, time taken would be different). So here, the avg is not (Speed1 + Speed 2)/2.
If D is the distance from A to B,
Avg Speed = 2D/(D/Speed1 + D/Speed2) = Total distance/Total time
Avg Speed = 2*Speed1*Speed2/(Speed1 + Speed2) (this is the avg when one travels for the same distance)
Now, what happens if one of the speeds is half the avg speed?
Say avg speed was 80 mph. Say distance from A to B is 80 miles. Since avg speed for to and fro journey is 80 mph, we need 2 hrs to cover the journey. Now, what happens when the speed while going from A to B is 40 mph? He takes 2 hrs to go from A to B i.e. the entire time allotted for the return trip has gotten used up on the first leg of the journey itself. Of course, it doesn't matter what your speed is in the second leg, you will take more than 2 hrs for the entire journey and hence your avg speed will be less than 80 mph.
Hence, in any one leg, your speed cannot be less than half the average.
If you want to see it algebraically,
From above, A = 2*S1*S2/(S1 + S2)
If S1 = A/2,
A = 2*(A/2)*S2/(A/2 + S2)
S2 = A/2 + S2
There is no value of S2 which will satisfy this equation.
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