A. stmt I aline is sufficeint.
Am not sure why you are posting the options in wrong order. Please note that the order of options in GMAT DS questions are fixed.
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I am not posting in that order It was in software, I know its bad but I think we all are aware of the options i just don't wana to type the question that's why i added the attachment.this_time_i_will wrote:A. stmt I aline is sufficeint.
Am not sure why you are posting the options in wrong order. Please note that the order of options in GMAT DS questions are fixed.
hope you don't mind.
Man can you please explain because i think both are required....
i think it wud be helpful if u jsut take the screenshot of question only.Wud save the members from any confusion 
For the question, please note that to one way of finding the avergae speed is to caluclate the Harmonic mean, provided distance traveled in both the case is same.
so if speed from town to village = x and speed from village to town = y, then
avergae speed = Harmonic Mean = 2xy/(x+y)
For the question, please note that to one way of finding the avergae speed is to caluclate the Harmonic mean, provided distance traveled in both the case is same.
so if speed from town to village = x and speed from village to town = y, then
avergae speed = Harmonic Mean = 2xy/(x+y)
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thanks for the formula.this_time_i_will wrote:i think it wud be helpful if u jsut take the screenshot of question only.Wud save the members from any confusion
For the question, please note that to one way of finding the avergae speed is to caluclate the Harmonic mean, provided distance traveled in both the case is same.
so if speed from town to village = x and speed from village to town = y, then
avergae speed = Harmonic Mean = 2xy/(x+y)
But is there any other to find the average speed i am very bad at remembering formula........
but still if there is any other way that you know then please post it.
what i thought was (x+y)/2 But i think its wrong in this case......
you may try to solve this way
av speed=total distance/ total time
let it be distance D. and total distance will be D+D=2D (forward and back)
time forward=D/60, ( distance/rate)
time back=D/x whre x is speed back
total time=D/60+D/x=D*((x+60)/60x)
and now divide total distance=2D on total time D((x+60)/60x), cancel D and left with 2*60/(x+60) you have one unknown x that is his speed back
anf the 1 st gives that x=40 suff
2 st gives you distance=2*60 but as you see you don`t need it
av speed=total distance/ total time
let it be distance D. and total distance will be D+D=2D (forward and back)
time forward=D/60, ( distance/rate)
time back=D/x whre x is speed back
total time=D/60+D/x=D*((x+60)/60x)
and now divide total distance=2D on total time D((x+60)/60x), cancel D and left with 2*60/(x+60) you have one unknown x that is his speed back
anf the 1 st gives that x=40 suff
2 st gives you distance=2*60 but as you see you don`t need it
Very good work man i understand fully,clock60 wrote:you may try to solve this way
av speed=total distance/ total time
let it be distance D. and total distance will be D+D=2D (forward and back)
time forward=D/60, ( distance/rate)
time back=D/x whre x is speed back
total time=D/60+D/x=D*((x+60)/60x)
and now divide total distance=2D on total time D((x+60)/60x), cancel D and left with 2*60/(x+60) you have one unknown x that is his speed back
anf the 1 st gives that x=40 suff
2 st gives you distance=2*60 but as you see you don`t need it
But just missed one X in above equation. I know this not make any difference but for future reference and surely in good faith.
