Jill went up a hill at an unknown constant speed. She then immediately

let 2d=total distance
3t=total time
2d/3t=6 mph average speed
d/t=6/(2/3)=9 mph average downhill speed
C

Great Question.
Here is what i did on this one =>


Using Distance = Speed * Time

Distance Speed Time

D X D/X


D 2X D/2X

Total Distance => 2D
Total Time => D/X + D/2X => 3D/2X

Average speed => (2D)/(3D/2X) => 4X/3 = 6
Hence X=18/4 =4.5

We are asked about the value of 2X => 2*4.5=9mph.

Smash that C.

Formula for average speed when distance of round trip is equal :

Avg. speed = \(\frac{2xy}{(x + y)}\) [x and y are two rates of up and down respectively]

Let Jill's speed while going UP = x

Jill's speed while coming down = 2x

Now, \(6 = 2 * x * \frac{2x}{x} + \frac{2x}{1}\)

6 = \(\frac{4x}{3}\)

x = 9/2

Therefore downward speed = 2*9/2 = 9

(C)

Wonder if this can also be solved using ratios or weighted averages?

KarishmaB

When averaging speed, say s and 2s here, the weights have to be 'time taken', not the distance. Hence we cannot find the average speed with distance as weights.

When distance travelled is the same, the Average Speed is given by \(\frac{2ab}{a+b}\) where a and b are the two speeds.
So here \(6 = \frac{2s*2s}{s+2s}\) giving 2s = 9

Alternatively, Avg Speed = Total Distance/Total Time works in every situation since it pretty much defines average speed.
So assuming one way distance as d, we can say
\(6 = \frac{(d + d)}{(d/s + d/2s)}\) which is actually what gives us the formula 2ab/a+b.
Here too, obviously, 2s = 9 (d in numerator will get cancelled with d in the denominator)

How to find the weights in a weighted average problem is discussed here: https://anaprep.com/arithmetic-weights- ... d-average/

Is there a way to use the numbers of the answer choices?