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Bicyclist's average speed

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Bicyclist's average speed

by bhumika.k.shah » Fri Jan 29, 2010 4:08 am
A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?


1) 4 4/5
2)5
3)5 1/5
4)5 2/5
5)5 3/5
Last edited by bhumika.k.shah on Sun Jan 31, 2010 4:19 am, edited 1 time in total.
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by sars72 » Fri Jan 29, 2010 4:52 am
i thought i knew how to solve these problems, but my answer is not among the answer choices :( . Please point out where i am going wrong.


Total Distance Traveled going up = 4*2 = 8 = Same distance traveled coming down

Time taken going up = 2 hrs
Time taken coming down = distance/speed= 8/6 = 4/3 hrs

Average Speed = Total Distance/ Total Time = 16/(4/3 + 2) = 16/(10/3) = 16*3/10 = 48/10 = 24/5

24/5 is not among the answer choices

I tried the same using the harmonic mean formula
harmonic mean
average speed = 2xy/(x+y), where x and y are the two speeds

--> average speed = (2*4*6)/(4+6) = 48/10 = 24/5

Please help!!
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by ajith » Fri Jan 29, 2010 12:43 pm
bhumika.k.shah wrote:A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?


A.4*4/5
B.5
C.5*1/5
D.5*2/5
E.5*3/5

I am too bad at this topic! :-(
Total distance between A&B = 8miles (4*2)

Time taken for the return trip = 8/6 = 4/3 hours

now average speed = total distance/total time = 16/(2+4/3) = 48/10 = 4&4/5 {answer choices have a mistake]
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by bhumika.k.shah » Sun Jan 31, 2010 4:19 am
Edited.Got it wrong the second time also :(
ajith wrote:
bhumika.k.shah wrote:A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?


A.4*4/5
B.5
C.5*1/5
D.5*2/5
E.5*3/5

I am too bad at this topic! :-(
Total distance between A&B = 8miles (4*2)

Time taken for the return trip = 8/6 = 4/3 hours

now average speed = total distance/total time = 16/(2+4/3) = 48/10 = 4&4/5 {answer choices have a mistake]
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by bhumika.k.shah » Sun Jan 31, 2010 4:26 am
Hey
i know i m not really good at this. but i did the same thing what u did and got my answer as 24/5
Which is option A :D

P.s.: whats this harmonic mean formula and when can u use it ???
sars72 wrote:i thought i knew how to solve these problems, but my answer is not among the answer choices :( . Please point out where i am going wrong.


Total Distance Traveled going up = 4*2 = 8 = Same distance traveled coming down

Time taken going up = 2 hrs
Time taken coming down = distance/speed= 8/6 = 4/3 hrs

Average Speed = Total Distance/ Total Time = 16/(4/3 + 2) = 16/(10/3) = 16*3/10 = 48/10 = 24/5

24/5 is not among the answer choices

I tried the same using the harmonic mean formula
harmonic mean
average speed = 2xy/(x+y), where x and y are the two speeds

--> average speed = (2*4*6)/(4+6) = 48/10 = 24/5

Please help!!
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by bhumika.k.shah » Sun Jan 31, 2010 4:27 am
got it.how simple can it be.

Any link where speed,time & distance sums are consolidated into one file???

That would be of great help.!
bhumika.k.shah wrote:Edited.Got it wrong the second time also :(
ajith wrote:
bhumika.k.shah wrote:A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?


A.4*4/5
B.5
C.5*1/5
D.5*2/5
E.5*3/5

I am too bad at this topic! :-(
Total distance between A&B = 8miles (4*2)

Time taken for the return trip = 8/6 = 4/3 hours

now average speed = total distance/total time = 16/(2+4/3) = 48/10 = 4&4/5 {answer choices have a mistake]
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by harsh.champ » Mon Feb 01, 2010 2:12 am
Basically,in speed-time-distance ques. 2 are constants and 1 is the variable.....the important part lies in identifying them.....
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by harsh.champ » Mon Feb 01, 2010 2:14 am
I think the 1st answer 4 4/5 is correct.......................4 4/5 can be alternatively written as 48/10 or 24/5(the answer which you had got -bhumika)
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by harsh.champ » Mon Feb 01, 2010 2:17 am
I have seen many people have developed fear regarding the SDT(Speed Distance Time) questions.......that does hinder your thought process......just believe in yourself ,identify the constants and the variables and it will seem easy to you.....its people's psychology not their ability that restricts them from solving the problems....
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by bhumika.k.shah » Mon Feb 01, 2010 3:18 am
thanks :-)
harsh.champ wrote:I have seen many people have developed fear regarding the SDT(Speed Distance Time) questions.......that does hinder your thought process......just believe in yourself ,identify the constants and the variables and it will seem easy to you.....its people's psychology not their ability that restricts them from solving the problems....
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by harsh.champ » Tue Feb 02, 2010 1:10 am
bhumika.k.shah wrote:thanks :-)
harsh.champ wrote:I have seen many people have developed fear regarding the SDT(Speed Distance Time) questions.......that does hinder your thought process......just believe in yourself ,identify the constants and the variables and it will seem easy to you.....its people's psychology not their ability that restricts them from solving the problems....
_________________

anytime...actually i also said this from my personal experience.............sometimes after i retrospect i find that it was my mind only that was creating the ceiling....not my ability..
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