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Noelle walks from point A to point B at an average speed of 5 kilomete

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Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 19 Feb 2015, 08:43
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Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.

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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 20 Feb 2015, 03:08
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1
We know that, Average speed = Total distance / Total time

So, 6 = 2d/(d/5 + d/v) ; where v = average speed from pt B to pt A.
Solving, we get, 2v = 15
or v = 7.5km/hr
Hence and is (D)
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 22 Feb 2015, 11:49
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Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION
Attachment:
averagespeedwalker_text.PNG
averagespeedwalker_text.PNG [ 23.69 KiB | Viewed 7493 times ]

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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 06 Jun 2015, 12:40
Let's assume that total distance from A to B is 5 km. Then t=(5/5)=1 hour. Hence we need to solve an equation 6=(10)/(1+x) where x is time from B to A . Solving we get x=(2/3). Hence speed from B to A should be 5/(2/3)=7.5.
P.S i remember this method from one of Bunuel's solutions. Really help! Thanks=))
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 06 Jun 2015, 19:20
2
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.


Since both ways (to and fro) the distance to be travelled is same, therefore

Average Speed = 2ab / (a+b)

where a and b are the speeds at which the equal distances have been travelled

i.e. 6 = Average Speed = 2*5*b/ (5+b)

i.e. 6 * (5+b) = 10 b

i.e. 30 + 6b = 10b

i.e. 4b = 30

i.e. b = 7.5

Answer: Option
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Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 11 Jun 2015, 04:52
1
I think I have a similar solution, but I understand mine better so I am posting it:

1) Adding what we know to the chart:
....................R..........T...........D
A to B...........5.........................
B to A......................................
Both.............6.........................

2) Since the distance both ways is the same we pick a number. I picked 30 for distance one way. The chart becomes:
....................R..........T..........D
A to B...........5......................30
B to A...................................30
Both.............6......................60

3) We will now calculate the Time for the Rates and Distances we have. Then we add the results to the chart:
A to B: 5T = 30 --> T = 6
Both: 6T = 60 --> T = 10.

....................R..........T...........D
A to B...........5..........6...........30
B to A......................4...........30
Both.............6.........10..........60

We now see that we can subtrackt 6 from 10, leading to T=4, for B to A.

We calculate the missing R using this last information:
R = 30/4
R = 7.5 ANS D.
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 15 Jun 2015, 09:33
HI

Average Speed = 2xy/(x+y) where x and y are speeds . Given the average speed from A to B is 5 kmph and Average speed for to and fro journey is 6 kmph , so we have x = 5 and Average Speed is 6.
Substituting the above formula we get y = 7.5
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 15 Nov 2015, 05:29
Harmonic mean approach:

2/(1/5+1/x)=6

1/5+1/x=1/3

1/x=1/3-1/5=2/15=1/7.5, so x=7.5

D
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 02 Jun 2016, 02:22
Average speed for round trip=2xy/x+y

6=2*5*x/5+x

or x=7.5 km/hr

x-speed while moving from A to B and y-speed while moving from B to A. It would take less than 30 sec to solve
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 02 Jun 2016, 08:13
1
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.


Let's suppose that speed while returning was xkm/h

Since the distance is same, we can apply the formula of avg speed

Avg speed= 2S1S2/S1+S2

6= 2*5*x/5+x
x= 7.5
D is the answer
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 10 Jun 2017, 07:14
1
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


We can let the distance from point A to point B = d, so the entire trip is 2d.

Thus, the time to go from point A to point B is d/5. If we let r = Noelle’s walking rate from point B to point A, then her time is d/r. We can plug all of our values into the average rate formula:

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

6 = 2d/(d/5 + d/r)

6(d/5 + d/r) = 2d

6d/5 + 6d/r = 2d

6/5 + 6/r = 2

6/r = 4/5

30 = 4r

7.5 = r

Answer: D
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Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 10 Jun 2017, 09:33
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.


let x=B→A kph
let 2 k=total distance
2/(x+5/5x)=6 kph
x=7.5 kph
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 05 Aug 2018, 09:21
Let distance be 10 from A to B

so, 10/5 +10/x = 20/6

therefore x = 7.5
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 24 Oct 2018, 03:16
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75

\(\left. \matrix{
{V_{A \to B}} = \,\,5\,\,{{{\rm{km}}} \over {\rm{h}}} \hfill \cr
{V_{B \to A}} = \,\,?\,\,{{{\rm{km}}} \over {\rm{h}}}\,\, \hfill \cr} \right\}\,\,\,{\rm{for}}\,\,{V_{{\rm{average}}}} = 6\,\,{{{\rm{km}}} \over {\rm{h}}}\)

\({\rm{Take}}\,\,{\rm{dist}}\left( {A,B} \right) = 30\,\,{\rm{km}}\)

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

\(\left. \matrix{
A \to B\,\,\,:\,\,\,{\rm{30}}\,\,{\rm{km}}\left( {{{1\,\,{\rm{h}}} \over {5\,\,{\rm{km}}}}} \right) = 6\,\,{\rm{h}} \hfill \cr
A \to B \to A\,\,\,:\,\,\,2 \cdot {\rm{30}}\,\,{\rm{km}}\left( {{{1\,\,{\rm{h}}} \over {6\,\,{\rm{km}}}}} \right) = 10\,\,{\rm{h}}\,\,\,\, \hfill \cr} \right\}\,\,\,\, \Rightarrow \,\,\,\,B \to A = 10 - 6 = 4\,\,{\rm{h}}\)

\(? = {{30} \over 4} = {{28 + 2} \over 4} = 7{1 \over 2}\,\,\,\,\,\,\left[ {{{{\rm{km}}} \over {\rm{h}}}} \right]\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Noelle walks from point A to point B at an average speed of 5 kilomete  [#permalink]

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New post 24 Oct 2018, 09:51
Top Contributor
Bunuel wrote:
Noelle walks from point A to point B at an average speed of 5 kilometers per hour. At what speed, in kilometers per hour, must Noelle walk from point B to point A so that her average speed for the entire trip is 6 kilometers per hour?

A. 6.75
B. 7
C. 7.25
D. 7.5
E. 7.75


Kudos for a correct solution.


Let d = the distance between Points A and B.
Let x = the Noelle's speed from Point B to Point A.

We want the average speed for the entire trip to be 6 kilometers per hour.
So, we want: (TOTAL distance)/(TOTAL travel time) = 6 kmh
In other words: (TOTAL distance)/(travel time from A to B + travel time from B to A) = 6 kmh

Travel time = distance/speed
So, we get: (d + d)/(d/5 + d/x) = 6 kmh
Simplify to get: (2d)/(d/5 + d/x) = 6
Rewrite fractions with same denominator: (2d)/(dx/5x + 5d/5x) = 6
Combine fractions: (2d)/[(dx + 5d)/5x] = 6
Simplify: (2d)(5x)/(dx + 5d) = 6
Simplify: 10dx/[d(x + 5)] = 6
Divide top and bottom by d to get: 10x/(x + 5) = 6
Multiply both sides by (x+5) to get: 10x = 6(x + 5)
Expand: 10x = 6x + 30
So: 4x = 30
x = 30/4 = 7.5

Answer: D

Cheers,
Brent
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