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Meredith jogged to the top of a steep hill at an average pace of 6 mil

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Meredith jogged to the top of a steep hill at an average pace of 6 mil [#permalink]

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Meredith jogged to the top of a steep hill at an average pace of 6 miles per hour. She took the same trail back down. To her relief, the descent was much faster; her average speed rose to 14 miles per hour. If the entire run took Meredith exactly one hour to complete and she did not make any stops, how many miles, approximately, is the trail one way?

A 2
B 3
C 4
D 5
E 6
[Reveal] Spoiler: OA

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Re: Meredith jogged to the top of a steep hill at an average pace of 6 mil [#permalink]

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New post 18 Sep 2017, 02:46
SajjadAhmad wrote:
Meredith jogged to the top of a steep hill at an average pace of 6 miles per hour. She took the same trail back down. To her relief, the descent was much faster; her average speed rose to 14 miles per hour. If the entire run took Meredith exactly one hour to complete and she did not make any stops, how many miles, approximately, is the trail one way?

A 2
B 3
C 4
D 5
E 6



Hi...

some flaw in the way the Q is written..

If the Q is taken the way it is written..
Quote:
her average speed rose to 14 miles per hour

average speed for entire distance is 14 miles per hour
time taken for entire route is 1hr
so distance = 14*1=14 miles..
one way = 14/2=7miles

but I am sure the Q means..
Quote:
her average speed rose to 14 miles per hour on return

here let the distance be x..
so time upwards = x/6 and downwards = x/14..
total time =\(\frac{x}{4}+\frac{x}{16}= \frac{20x}{84}\)
but total time given is 1 hr..
so \(\frac{20x}{84}=1.......x=\frac{84}{20}=4.2miles\)
C
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Re: Meredith jogged to the top of a steep hill at an average pace of 6 mil [#permalink]

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New post 18 Sep 2017, 11:38
While going up hill : speed = 6 miles/hour
Let time be = \(t_1\)
total distance =d= \(6t_1\)

While going down hill : speed = 14 miles/hour
Let time be = \(t_2\)
total distance =d= \(14t_2\)

Total time = t = 1 hour = \(t_1+ t_2\)

As distance up hill and down hill is same => \(6t_1 = 14t_2\)
=> \(t_1 = \frac{14}{6}t_2\)

as \(t_1+ t_2 = 1\) => \(\frac{14}{6} * t_2+ t_2 = 1\) => \(t_2 = \frac{3}{10}\)

distance one way = d = \(14t_2\) = \(14 * \frac{3}{10}\) = 4.2

Answer: C

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Re: Meredith jogged to the top of a steep hill at an average pace of 6 mil [#permalink]

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New post 22 Sep 2017, 07:29
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SajjadAhmad wrote:
Meredith jogged to the top of a steep hill at an average pace of 6 miles per hour. She took the same trail back down. To her relief, the descent was much faster; her average speed rose to 14 miles per hour. If the entire run took Meredith exactly one hour to complete and she did not make any stops, how many miles, approximately, is the trail one way?

A 2
B 3
C 4
D 5
E 6


We use the basic formula for distance: distance = rate x time, which can be re-expressed as time = distance/rate. We can let each one-way distance = d, so her time uphill = d/6 and her time downhill = d/14. We sum the uphill time and the downhill time to get 1 hour; thus:

d/6 + d/14 = 1

Multiplying by 84, we have:

14d + 6d = 84

20d = 84

d = 4.2, which is about 4.

Answer: C
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Re: Meredith jogged to the top of a steep hill at an average pace of 6 mil   [#permalink] 22 Sep 2017, 07:29
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