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

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Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 24 Apr 2017, 04:15
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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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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 27 Apr 2017, 17:13
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Bunuel wrote:
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


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
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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 24 Apr 2017, 07:01
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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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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 27 Apr 2017, 19:00
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Bunuel wrote:
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


let 2d=total distance
3t=total time
2d/3t=6 mph average speed
d/t=6/(2/3)=9 mph average downhill speed
C
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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 27 Apr 2017, 19:06
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:

\(D/T_u=9\)
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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 28 Apr 2017, 03:19
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.

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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 15 Mar 2019, 09:49
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Bunuel wrote:
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


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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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 15 Mar 2019, 10:23
Poor Jill. This is mine

Rate(x)....Distance(D)....Time (=D/R)
X.............D..................D/x
2X...........D..................D/2x

T:..............2D................3D/2X

Solve for X = 4.5. Multiply by 2.
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Re: Jill went up a hill at an unknown constant speed. She then immediately  [#permalink]

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New post 22 Nov 2019, 07:11
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

Bunuel wrote:
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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Re: Jill went up a hill at an unknown constant speed. She then immediately   [#permalink] 22 Nov 2019, 07:11
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