Robin drove from Townville to Villageton. Upon arriving in

[swerve]

Robin drove from Townville to Villageton. Upon arriving in Villageton, she immediately returned to Townville. What was Robin's average speed for the entire trip?

1) While driving from Townville to Villageton, Robin's average speed was 40 miles per hour. While driving from Villageton to Townville, Robin's average speed was 60 miles per hour.
2) The distance from Townville to Villageton is 120 miles.

The OA is A

Source: Magoosh

[Ian Stewart]

The distance each way is the same. If we call it 'd', then since time = distance/speed, Robin's time for the first part of the trip was d/40 and for the return trip was d/60. So her total time was d/40 + d/60 = 5d/120 = d/24. Her total distance was 2d, so her average speed was (total distance)/(total time) = 2d / (5d/120) = 240d/5d = 48 mph. So Statement 1 is sufficient. Statement 2 is clearly not sufficient so the answer is A.

[deloitte247]

$$Average\ speed=\ \frac{Dis\tan ce}{time\ taken}$$
What was Robin's Average speed for the entire drop

Statement 1
While driving from Townville to Villageton
Robin's average speed was 40 miles per hour while driving from villageton to Townvile Robin's average speed was 60 miles per hour.

Let the distance = D; between hours = 2D
Time = d/40 ( From Townsville to villageton )
= d/60 ( From Villageton to Townsville)
$$\frac{d}{40}+\frac{d}{60}=\frac{d}{24}$$
$$average\ speed\ =\frac{2D}{\frac{d}{24}}=2D\cdot\frac{24}{D}=\frac{48D}{D}=48$$
Statement 1 is SUFFICIENT.

statement 2
Distance from Townsville to Villageton is 120 miles, this is not enough to calculate speed because, distance is missing
Statement 2 is INSUFFICIENT.

$$Answer\ is\ Option\ A$$