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

Question

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

A. 7 mph
B. 8 mph
C. 9 mph
D. 10 mph
E. 11 mph

JeffTargetTestPrep · Accepted Answer

We can let Jill’s initial rate going up the hill be x and the rate going down be 2x. We can also let the distance each way = d. Thus, the time going up was d/x and the time coming down was d/2x. Let’s plug all of this into the average rate formula:

average rate = (distance 1 + distance 2)/(time 1 + time 2)

6 = 2d/(d/x + d/2x)

3 = d/(2d/2x + d/2x)

3 = d/(3d/2x)

3 = 2x/3

2x = 9

So, her rate going down the hill is 2x = 9 mph.

Answer: C

pushpitkc · Answer

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)

gracie · Answer

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

stonecold · Answer

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.

AkshdeepS · Answer

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

Avg. speed =  \frac{2xy}{(x + y)}

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)

TargetMBA007 · Answer

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

KarishmaB

KarishmaB · Answer

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/

Usernamevisible · Answer

Let uphill speed = v mph
Then downhill speed = 2v mph

Average speed for equal distances = (2ab) / (a + b)

So,
6 = (2 × v × 2v) / (v + 2v)
6 = (4v2) / (3v)
6 = 4v / 3

v = 18 / 4 = 4.5

Downhill speed = 2v = 9 mph

Answer: C. 9 mph

Mars1303 · Answer

6 = 2d/(d/x + d/2x)

3 = d/(2d/2x + d/2x)

how is this process correct? Could you please explain the rules you used.

Saurav1408 · Answer

We could also do this question using weighted average. Directly using the options and taking the half speed for the uphill speed. The ratio of time for uphill:downhill::2:1.
So it would be easier and quicker to analyze which option would give us the ratio of time as 2:1. What do you think on this?

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