A bus completed first 50 miles of a 120-mile trip at an

Question

e-GMAT A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed half of the remaining journey at an average speed of 35 mph. At what average speed it should complete the remaining journey so that the overall average speed of the whole journey becomes 20 mph? A. 40 mph B. 35 mph C. 30 mph D. 22.5 mph E. 17.5 mph The OA E

Rich.C@EMPOWERgmat.com · Answer

Hi All,

We're told that a bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed half of the remaining journey at an average speed of 35 mph. We're asked for the average speed it should complete the REMAINING part of the journey so that the overall average speed of the WHOLE journey is 20 mph. This question is awkwardly-worded, but the intent is that we're supposed to include the 30-minute 'stop time' in the overall calculation.

To start, we can break this bus ride down into 'pieces':

-For the first piece, the bus traveled 50 miles at an average speed of 20 miles/hour.
D = (R)(T)
50 miles = (20 mph)(T)
50/20 = T
2.5 hours = T
Thus, this first piece of the trip took 2.5 hours

-Then we stop for 30 minutes, which = 0.5 hours

-Next, travel HALF of the resining distance at 35 miles/hour.
Remaining distance = 120 - 50 = 70 miles; half of that distance = 35 miles
D = (R)(T)
35 miles = (35 mph)(T)
35/35 = T
1 hour = T
Thus, this piece of the trip took 1 hour

At this point, the bus has traveled 85 miles and taken 4 total hours. The current average speed is 85/4 = 21.25 mph. To reduce this average speed to 20 mph, we will need to travel the final 35 miles at a speed that is LESS than 20 mph. There's only one answer that fits...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

fskilnik@GMATH · Answer

AAPL wrote:e-GMAT A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed half of the remaining journey at an average speed of 35 mph. At what average speed it should complete the remaining journey so that the overall average speed of the whole journey becomes 20 mph? A. 40 mph B. 35 mph C. 30 mph D. 22.5 mph E. 17.5 mph Excellent opportunity for UNITS CONTROL, one of the most powerful tools of our method! $$? = x\,\,{m{mph}}\,\,\,\left( {{m{final}}\,\,{m{miles}}} ight)$$ $$120\,\,{m{miles}}\,\,\left( {{{1\,\,{m{h}}} \over {20\,\,{m{miles}}}}} ight)\,\,\,\, = \,\,\,\,6\,{m{h}}\,\,\,\, 	o \,\,\,\,{m{trip}}\,\,{m{total}}\,\,{m{time}}\,\,$$ $$\left. \matrix{ 50\,\,{m{miles}}\,\,\left( {{{1\,\,{m{h}}} \over {20\,\,{m{miles}}}}} ight)\,\,\,\, = \,\,\,\,2.5\,{m{h}}\,\,\,\, 	o \,\,\,\,\,{m{first}}\,\,{m{distance}}\,\,{m{time}}\,\,\,\, \hfill \cr {1 \over 2}{m{h}}\,\,\,\, 	o \,\,\,\,\,{m{halt}}\,\,{m{time}} \hfill \cr} ight\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,6 - \left( {2.5 + 0.5} ight) = 3{\mathop{m h}
olimits} \,\,\,\,{m{last}}\,\,{m{distance}}\,\,\left( {120 - 50 = 70\,\,{m{miles}}} ight)\,\,{m{time}}$$ $${{70} \over 2}\,\,{m{miles}}\,\,\left( {{{1\,\,{m{h}}} \over {35\,\,{m{miles}}}}} ight)\,\,\,\, = \,\,\,\,1\,{m{h}}\,\,\,\, 	o \,\,\,\,\,{m{last}}\,\,\underline {{m{half}}} \,\,{m{distance}}\,\,{m{time}}$$ $${m{Last}}\,\,{m{2}}\,\,{m{h}}\,\,{m{for}}\,\,35\,\,{m{miles}}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = x = {{30 + 5} \over 2}\,\,{m{ = }}\,\,{m{17}}{m{.5}}\,\,{m{mph}}$$ We follow the notations and rationale taught in the GMATH method. Regards, Fabio.

Scott@TargetTestPrep · Answer

AAPL wrote:e-GMAT A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed half of the remaining journey at an average speed of 35 mph. At what average speed it should complete the remaining journey so that the overall average speed of the whole journey becomes 20 mph? A. 40 mph B. 35 mph C. 30 mph D. 22.5 mph E. 17.5 mph The OA E We see that the bus completed the first 50 miles in 50/20 = 2.5 hours. It also completed the next (120 - 50)/2 = 70/2 = 35 miles in 35/35 = 1 hour. Let t = the time it took to complete the last 35 miles. We can create the following equation (keep in mind that the bus also took a 30-minute (or 0.5 hour) break): 120/(2.5 + 0.5 + 1 + x) = 20 120/(4 + x) = 20 120 = 20(4 + x) 120 = 80 + 20x 40 = 20x 2 = x Since it takes the bus 2 hours to drive the last 35 miles, its speed for the last 35 miles is 35/2 = 17.5 mph. Answer: E

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