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.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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