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

Well if you have 20mph, then 35 mph, then x mph.
You want the average to be 20 again, then x has to be under 20. Hence E.
I think this is a poor quality question, you should update the answers to be more misleading.

total time according to 20mph and 120 mi distance= 120/20=6hrs
let speed in final leg be x
now
time of leg1(50 mi at the speed of 20mi/hr)+wait time +time of leg 2 (35 mi at speed of 35mi/hr)+ time of remaining leg=6 hrs

(50/20)+0.5+(35/35)+((120-85)/x)=6 hrs
4+(35/x)=6
x=17.5mi/hr

Excellent opportunity for UNITS CONTROL, one of the most powerful tools of our method!

\(? = x\,\,{\rm{mph}}\,\,\,\left( {{\rm{final}}\,\,{\rm{miles}}} \right)\)

\(120\,\,{\rm{miles}}\,\,\left( {{{1\,\,{\rm{h}}} \over {20\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,6\,{\rm{h}}\,\,\,\, \to \,\,\,\,{\rm{trip}}\,\,{\rm{total}}\,\,{\rm{time}}\,\,\)


\(\left. \matrix{\\
50\,\,{\rm{miles}}\,\,\left( {{{1\,\,{\rm{h}}} \over {20\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,2.5\,{\rm{h}}\,\,\,\, \to \,\,\,\,\,{\rm{first}}\,\,{\rm{distance}}\,\,{\rm{time}}\,\,\,\, \hfill \cr \\
{1 \over 2}{\rm{h}}\,\,\,\, \to \,\,\,\,\,{\rm{halt}}\,\,{\rm{time}} \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,6 - \left( {2.5 + 0.5} \right) = 3{\mathop{\rm h}\nolimits} \,\,\,\,{\rm{last}}\,\,{\rm{distance}}\,\,\left( {120 - 50 = 70\,\,{\rm{miles}}} \right)\,\,{\rm{time}}\)


\({{70} \over 2}\,\,{\rm{miles}}\,\,\left( {{{1\,\,{\rm{h}}} \over {35\,\,{\rm{miles}}}}} \right)\,\,\,\, = \,\,\,\,1\,{\rm{h}}\,\,\,\, \to \,\,\,\,\,{\rm{last}}\,\,\underline {{\rm{half}}} \,\,{\rm{distance}}\,\,{\rm{time}}\)

\({\rm{Last}}\,\,{\rm{2}}\,\,{\rm{h}}\,\,{\rm{for}}\,\,35\,\,{\rm{miles}}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = x = {{30 + 5} \over 2}\,\,{\rm{ = }}\,\,{\rm{17}}{\rm{.5}}\,\,{\rm{mph}}\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hello, would you pls explain do we need to consider stop time (halt time= the time the bus was not on the move?) for calculating the average speed?? fROm the first part of the question we understand the first part of trip taken in 2.5 hour, in order to average speed be 20 for a 120-mile trip, the total time taken for trip should be 6. so if we deduct 6-2.5= 3.5 is the time taken for second part of the trip. .... Pls explain the halt time part and its role in question . Thanks in advance

By Skimming through the answer choices, you directly know it has to be E

The speed of the first part of the trip is 20mph
The second is 35mph
The only way for the overall average speed to be 20mph is if the third speed is lower than 20mph

Hello my friend
would you pls explain do we need to consider stop time (halt time= the time the bus was not on the move?) for calculating the average speed??