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

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.

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.