BTGmoderatorDC wrote:The Lewiston Road Race is how many kilometers long?

(1) It takes 2 more hours to complete the race at an average speed of 40 kilometers per hour than at an average speed of 50 kilometers per hour.

(2) At an average speed of 20 kilometers per hour, it took Brett 10 hours to complete half the race.

Statement 1:
Rate and time have a RECIPROCAL relationship.
A rate ratio of (40kph) : (50kph) = 4:5 implies a time ratio of 5:4 = 10:8.
The time ratio in blue implies that -- for the time at 40kph to be 2 hours more than the time at 50kph -- the time at 40kph must be 10 hours, while the time at 50kph must be 8 hours.
Since it takes 10 hours to complete the race at a speed of 40kph, the race = rt = 40*10 = 400 kilometers.
SUFFICIENT.

Statement 2:
When Brett travels at a speed of 20kph for 10 hours, the distance = rt = 20*10 = 200 kilometers.
Since these 200 kilometers constitute half the race, the entire race = 400 kilometers.
SUFFICIENT.

The correct answer is D.

BTGmoderatorDC wrote:The Lewiston Road Race is how many kilometers long?

(1) It takes 2 more hours to complete the race at an average speed of 40 kilometers per hour than at an average speed of 50 kilometers per hour.

(2) At an average speed of 20 kilometers per hour, it took Brett 10 hours to complete half the race.
Source: Princeton Review

$$? = D\,\,\,\,\,\left[ {{\rm{km}}} \right]$$

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

$$\left( 1 \right)\,\,\,\left( {\,\,\left[ {{\rm{km}}} \right] \cdot {{\left[ {\rm{h}} \right]} \over {\left[ {{\rm{km}}} \right]}}\,\, = \,\,\left[ {\rm{h}} \right]\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,D \cdot {1 \over {40}} - D \cdot {1 \over {50}} = 2\,\,\,\,\left[ {\rm{h}} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\,\,\,\,\,\,\,\,\,\left( {{1^{{\rm{st}}}}\,\,{\rm{degree}}\,\,{\rm{equation}}\,\,{\rm{in}}\,\,D} \right)$$

$$\left( 2 \right)\,\,\,?\,\, = \,\,2 \cdot 10\,{\rm{h}}\,\, \cdot \,\,{{20\,\,{\rm{km}}} \over {1\,\,{\rm{h}}}}\,\, = 400\,\,\,\,\left[ {\,{\rm{km}}\,} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

From 1: for average speed of 50 kilometers per hour time =t
for average speed of 40 kilometers per hour time =t+2
50*t=40*(t+2).
From here we will get t and multiply with 50 to get distance.
No need to solve till last.

Sufficient
From 2: 10*20=d/2
hence sufficient.

So option D.