Jill went up a hill at an unknown constant speed. She then immediately tumbled down the hill along the same route but twice as fast. If Jill’s average speed for the round trip was 6 mph, what was her average speed tumbling down the hill?

A. 7 mph
B. 8 mph
C. 9 mph
D. 10 mph
E. 11 mph
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Average speed = (2*a*b)/(a+b) when the distance travelled is equal
Let the speed while travelling up the hill be x
Hence, speed while tumbling down the hill is 2x
Given average speed = 6
6 = (2*x*2x)/(x+2x) = 4x/3
Solving for x, x = 4.5. Speed tumbling down =2x = 9(Option C)
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As always, we introduce the Speed-Distance-Time relationship:

Distance = Speed x Time

Now, we should note some elements of the question:

1. The distance upwards and downwards is the same

We can translate this as:Distance Upwards = Distance Downwads -> \(S_u * T_u = S_d * T_d\)

2. The speed downwards is 2 times the speed upwards.

We can modify the previous equation as \(2*S_d * T_u = S_d * T_d\)

Now, doing easy algebra we have the following relation: \(2* T_u = T_d\)


We are being told about the average speed we need to remember that the average speed is the total distance betwwen the total time: \(D_{total}/T_{total}\).

Total distance: \(D_u+D_d=2*D\)
Total Time: \(T_u+T_d=T_u+2*T_u=3T_u\)

\(S_{average}=2*D/(3*T_u)=6\)

We are being asked for speed downwards:

\(D/T_u=9\)

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)
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Can anyone please explain why this particular approach is wrong?
Avg speed= (Speed1+Speed2) /2; since the distance is the same.
Avg speed= (x+2x)/2=6
3x=12
x=4
Tumbling down speed=2x=8