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Natasha climbs up a hill, and descends along the same way she went up.

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Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 09:13
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Natasha climbs up a hill, and descends along the same way she went up. It takes her 3 hours to reach the top and 2 hours to come back down. If her average speed along the whole journey is 3 kilometers per hour, what was her average speed (in kilometers per hour) while climbing to the top?

A. 1.5
B. 2.5
C. 3.75
D. 5
E. 7.5

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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 09:55
2
Lets assume distance to top as X, so the total distance travelled by Natasha = 2x

Total Time taken = 3+ 2 = 5 Hrs

Avg speed = Total Dist/Total time taken = 2x/5

Avg speed of complete journey is given as = 3 Hrs

2x/5 = 3

X = 7.5 miles

Avg speed while climbing = Distance/Time = 7.5/3 = 2.5

Option B
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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 12:21
Climbing to the mountain top

V1 = x/3 kmph ---- (1)

Coming down the mountain

V2 = x/2 kmph ----(2)

V = V1 + V2 = 5x/6

Total average speed = [5x/6] /2
= 5x/12 kmph

But this average speed is given as 3 kmph (in Q stem)

=> 5x/12 = 3 kmph
=> x = 36/5 kmph

Therefore, average speed while climbing at the mountain top = x/3 from (1)
= 36/15
= 2.4 Option B
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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 13:39
Bunuel wrote:
Natasha climbs up a hill, and descends along the same way she went up. It takes her 3 hours to reach the top and 2 hours to come back down. If her average speed along the whole journey is 3 kilometers per hour, what was her average speed (in kilometers per hour) while climbing to the top?

A. 1.5
B. 2.5
C. 3.75
D. 5
E. 7.5


3kph*5 total hours=15k total distance
(15/2)/3=2.5kph average uphill speed
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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 16:58
Since the distance travelled both up and down is the same.
We can use the formula
Avg speed = (2 * speed while going up * speed while coming down)/sum of their speeds

Let x be the distance traveled up/down. Total distance = 2x
Substituting values
3 = \((2 * x/3 * x/2)/5x/6\)
\(2*x^2*6*6/30x\)= 3
x = 7.5

Speed upwards -> x/3 = 2.5(Option B)
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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 19 Oct 2016, 23:33
answer: B

D=3*T = 3*5 = 15 (total distance)

one way distance= 15/2

15/2 = 3*speed required


speed required = 5/2 = 2.5 8-)
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Re: Natasha climbs up a hill, and descends along the same way she went up.  [#permalink]

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New post 02 Dec 2019, 19:16
Bunuel wrote:
Natasha climbs up a hill, and descends along the same way she went up. It takes her 3 hours to reach the top and 2 hours to come back down. If her average speed along the whole journey is 3 kilometers per hour, what was her average speed (in kilometers per hour) while climbing to the top?

A. 1.5
B. 2.5
C. 3.75
D. 5
E. 7.5


Since average = total distance/total time, we have:

3 = (d + d)/(3 + 2)

3 = 2d/5

15 = 2d

7.5 = d

Thus, the average speed climbing up is 7.5/3 = 2.5 kmph.

Answer: B
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Re: Natasha climbs up a hill, and descends along the same way she went up.   [#permalink] 02 Dec 2019, 19:16
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