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

Question

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 the 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.

B.

C.

D.

E.

Show Answer

Official Answer

E

A.  40 mph
 B.  35 mph
 C.  30 mph
 D.  22.5 mph
 E.  17.5 mph

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EgmatQuantExpert · Accepted Answer

Solution

Given:  
 • 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 the half of the remaining journey at an average speed of 35 mph.
• The overall average speed of the whole journey is 20 mph

To find:  
 • Average speed of the bus in the remaining part of the journey

Approach and Working:  
We know, average speed of the journey is the ratio of total distance covered and the total time taken to cover that distance.

Total distance = 120 miles
Average speed of the whole journey = 20 mph
 • Total journey time =  \frac{120}{20}  = 6 hours

Now, as per the individual journey times, 
 • Time taken in first part of journey =  \frac{50}{20}  = 2.5 hours
• Time taken in second part of journey =  \frac{0.5(120-50)}{35}  = 1 hour
• Halt time = 0.5 hours
• Hence, time taken in the last part of journey = 6 – (2.5 + 1 + 0.5) = 2 hours
• Distance covered in the last part of journey = 120 –   = 35 miles

Therefore, the average speed in the last part of journey =  \frac{35}{2}  = 17.5 mph

Hence, the correct answer is option E.

Answer: E

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bangu · Answer

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 the 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.

B.

C.

D.

E.

Show Answer

Official Answer

E

Wanted overall average speed of journey = 20 mph
Total distance = 120 miles
Time required to cover for total distance = 120/20 = 6 Hours

First 50 mile  20  mph => Time taken for trip = 50/20 = 2.5 Hours
Halt of 30 minutes or 0.5 Hour.

Remaining distance = 120 - 50 = 70 miles.
Half of remaining journey = 70/2 = 35 miles at 35 mph => Time taken for trip = 35/35 = 1 Hour

Time Elapsed so far = 2.5 + 0.5 + 1 = 4 Hours
Time that should be spent to achieve 20 mph = Time required to cover for total distance - Time Elapsed so far => 6 - 4 = 2

Remaining distance 70 - 35 = 35 miles
Average Speed = 35/2 = 17.5 mph

Answer E

JourneyToTheTop · Answer

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.

OC2910 · Answer

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

fskilnik · Answer

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.

nazii · Answer

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

LucienH · Answer

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

nazii · Answer

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??

bumpbot · Answer

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

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