BTGmoderatorLU wrote:Source: Economist GMAT
Tom drives from town A to town B, driving at a constant speed of 60 miles per hour. From town, B Tom immediately continues to town C. The distance between A and B is twice the distance between B and C. If the average speed of the whole journey was 36 mph, then what is Tom's speed driving from B to C in miles per hour?
A. 12
B. 20
C. 24
D. 30
E. 36
$$?\,\, = \,\,\,{V_{B \to C}}\,\,\,\,\,\,\left[ {\,{\rm{mph}}\,} \right]$$
Let´s use
UNITS CONTROL, one of the most powerful tools of our method!
$${\rm{Let}}\,\,B \to C = 180\,\,{\rm{miles}}\,\,\,\,\,\left[ {\,180 = LCM\left( {60,36} \right)\,} \right]$$
$$\left. \matrix{
\left. {\matrix{
{A \to B} \cr
{2 \cdot 180\,\,{\rm{miles}}\,\,} \cr
} } \right\}\,\,\,\,\,\,\,2 \cdot 180\,\,{\rm{miles}}\,\,\, \cdot \,\,\,\left( {{{1\,\,{\rm{h}}} \over {60\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,6\,\,\,{\rm{h}}\,\,\, \hfill \cr
\left. {\matrix{
{B \to C} \cr
{180\,\,{\rm{miles}}\,\,} \cr
} } \right\}\,\,\,\,\,\,\,180\,\,{\rm{miles}}\,\,\, \cdot \,\,\,\left( {{{1\,\,{\rm{h}}} \over {{V_{B \to C}}\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,{{180} \over {{V_{B \to C}}}}\,\,\,{\rm{h}}\,\,\,\,\, \hfill \cr
\left. {\matrix{
{A \to C} \cr
{3 \cdot 180\,\,{\rm{miles}}\,\,} \cr
} } \right\}\,\,\,\,\,\,\,3 \cdot 180\,\,{\rm{miles}}\,\,\, \cdot \,\,\,\left( {{{1\,\,{\rm{h}}} \over {36\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,15\,\,\,{\rm{h}} \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,6 + {{180} \over {{V_{B \to C}}}} = 15\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{180} \over {{V_{B \to C}}}} = 9\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{V_{B \to C}} = 20\,\,\,\,\,\,\,\left[ {\rm{h}} \right]$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
