Problem Solving

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

The OA is B.

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

Let the distance between A and B = 120 miles.
Since the 120-mile distance between A and B is twice the distance between B and C, the distance between B and C = 60 miles, implying that the total distance = 120+60 = 180 miles.

Since the average speed for the 120-mile leg between A and B = 60 mph, the time to travel from A to B = d/r = 120/60 = 2 hours.
Since the average speed for the entire 180-mile trip = 36 mph, the time to travel the entire trip = d/r = 180/36 = 5 hours.

Time to travel from B to C = (time for the entire trip) - (time to travel from A to B) = 5-2 = 3 hours.
Since it takes 3 hours to travel the 60-mile leg between B and C, the driving speed from B to C = d/t = 60/3 = 20 mph.

The correct answer is B.

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.