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We have:

\(\frac{Total distance}{total times} = 5\) ,

also we have the time for the first trip (delivering mails) is\(1\) hour and let the time for the return trip be \(x\). Therefore, we have

\(\frac{6}{1+x} = 5 => 6=5x + 5 => 5x=1 => x= \frac{1}{5}\)the time taken for the return trip, Therefore,

\(\frac{distance for the return trip}{take taken for the return trip} = speed\)

\(\frac{3}{1/5} = 3*5=15\)

Option E.
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Imo E
2d/[(d/3)+(d/x)]=5
Solving above equation we have
15


Sent from my ONE E1003 using GMAT Club Forum mobile app
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Since the average speed of the entire trip is given to be 5
and the distance traveled while going towards delivery and back is same,

Average speed = \(\frac{{2*speed(delivery)*speed(back home)}}{{speed(delivery)+speed(back home)}}\)

Let the speed(back home) be x

\(5 =\frac{{2*3*x}}{{(3+x)}}\) (because Joey travels 3 miles in 1 hour)
15+5x = 6x
x = 15(Option E)
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Hi,

Can anyone please explain to me how everyone got 2*3? I am bit confused regarding 2 as I can't seem to make out where it came from. Please help!

Thank You!
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csaluja
Hi,

Can anyone please explain to me how everyone got 2*3? I am bit confused regarding 2 as I can't seem to make out where it came from. Please help!

Thank You!

Hi,

Distance one way is 3 miles so both ways it will be 2*3..
And this is how 2*3 comes. Hope it helps
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csaluja
Hi,

Can anyone please explain to me how everyone got 2*3? I am bit confused regarding 2 as I can't seem to make out where it came from. Please help!

Thank You!

Hi,

Distance one way is 3 miles so both ways it will be 2*3..
And this is how 2*3 comes. Hope it helps


I see, thanks alot! Kudos given!
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Bunuel
It takes Joey the postman 1 hours to run a 3 mile long route every day. He delivers packages and then returns to the post office along the same path. If the average speed of the round trip is 5 mile/hour, what is the speed with which Joey returns?

A. 11
B. 12
C. 13
D. 14
E. 15
\(5 = \frac{2*3*s}{(3+s)}\)

\(15 +5s = 6s\)

So, \(x = 15\), Answer must be (E)
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