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# Lexy walks 5 miles from point A to point B in one hour, then

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Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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21 Jul 2013, 14:16
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Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s average speed. How many minutes does Ben spend on his round trip?

A. 40
B. 80
C. 120
D. 160
E. 180
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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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21 Jul 2013, 14:24
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kingflo wrote:
Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s average speed. How many minutes does Ben spend on his round trip?

A. 40
B. 80
C. 120
D. 160
E. 180

$$average \ speed=\frac{total \ distance}{total \ time}=\frac{5+5}{1+\frac{5}{15}}=\frac{15}{2}$$ miles per hour.

Ben's average speed = $$\frac{15}{4}$$ miles per hour (half of Lexy's).

Ben's time = $$\frac{distance}{speed}=\frac{10}{(\frac{15}{4})}=\frac{8}{3}$$ hours, or 8/3*60=160 minutes.

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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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02 Aug 2013, 12:16
Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s average speed. How many minutes does Ben spend on his round trip?

Average speed = total distance/total time

We know the total distance is 5+5 (going from A to B then from B to A)

Time = distance/speed
Time = 5/15
Time = 1/3 hour

Average speed = (10)/(1hr + 1/3hr)
Average speed (lexi) = 10 / 1.33
Therefore, if Ben's average speed is 1/2 of lexi's then his speed = 10/2.66.

This means it took him 2.66 hours to cover the same 10 miles Lexi did. 2.66 hours = roughly 160 minutes.

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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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29 Oct 2013, 20:20
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Lexy travels 10 miles (5+5) in 80 Mins (60 + 60/3)
So, speed of Lexy = 10/80 = 1/8

Speed of Ben is half of speed of Lexy = 1/16

So, time taken by Ben = 10 / (1/16) = 160 Minutes

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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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29 Oct 2013, 22:29
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kingflo wrote:
Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s average speed. How many minutes does Ben spend on his round trip?

A. 40
B. 80
C. 120
D. 160
E. 180

Time taken by Lexy: 1 hour + 5/15 = 1 + 1/3 hours i.e 1 hour and 20 minutes

If Ben travels at half the speed then the time taken will be doubled: 2 hours and 40 minutes which is 120 + 40 = 160 minutes

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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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06 Aug 2015, 07:44
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Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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09 Aug 2015, 08:32
Harmonic average for average speed: 2/(1/5+1/15)=15/2

the half speed: 15/2:2=15/4

10/(15/4)=8/3h*60=160min.

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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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09 Aug 2015, 09:02
Time taken by Lexy to walk from point B to A => 60*5/15 = 20 mins.

Average speed of Lexy = 5+5/(60+20) = 10/80 = 7.5 mph.

Ben's average speed = 7.5/2 = 3.75 mph.

Time taken by Ben = 10*60/3.75 = 160 mins. Ans (D).
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Re: Lexy walks 5 miles from point A to point B in one hour, then [#permalink]

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03 Dec 2016, 01:59
Total time taken by Ben will be twice the total time taken by Lexy
ie from 2 * ( 1+15/5) = 2* ( 4/3) = 8/3 hours = 8*60/3 minutes = 160 minutes
Re: Lexy walks 5 miles from point A to point B in one hour, then   [#permalink] 03 Dec 2016, 01:59
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