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

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

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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

OA D

Source: Magoosh

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BTGmoderatorDC 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
For these kinds of questions, I typically follow the approach that Fabio used (assign a nice value to the distance).
But we can also solve the question algebraically:

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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BTGmoderatorDC 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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scott@targettestprep.com



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BTGmoderatorDC 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
Source: Magoosh
$$\left. \matrix{
{V_{A \to B}} =\,\, 5\,\,{{{\rm{km}}} \over {\rm{h}}} \hfill \cr
{V_{B \to A}} =\,\, ?\,\,{{{\rm{km}}} \over {\rm{h}}}\,\, \hfill \cr} \right\}\,\,\,{\rm{for}}\,\,{V_{{\rm{average}}}} = 6\,\,{{{\rm{km}}} \over {\rm{h}}}$$

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

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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English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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BTGmoderatorDC 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
Let the distance between A and B = the LCM of the two speeds = 5*6 = 30 kilometers, implying that the distance for the entire round trip = 60 kilometers.
At a rate of 5 kph, the time to travel the 30 kilometers from A to B = d/r = 30/5 = 6 hours.
At a rate of 6 kph, the time to travel the entire 60-kilometer round trip = d/r = 60/6 = 10 hours.
Time to travel from B to A = (time for the entire round trip) - (time to travel from A to B) = 10-6 = 4 hours.
Since it takes 4 hours to travel the 30 kilometers from B to A, the rate from B to A = d/t = 30/4 = 7.5 kph.

The correct answer is D.

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