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On a partly cloudy day, Derek decides to walk back from work

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Joined: 14 Oct 2012
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]

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New post 23 Apr 2017, 17:09
My 2 cents:
My solution based on the veritas article: https://www.veritasprep.com/blog/2013/0 ... -the-gmat/
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]

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New post 23 Apr 2017, 20:53
emmak wrote:
On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

A. 1/4
B. 4/5
C. 1/5
D. 1/6
E. 1/7


let s=sunny distance
c=cloudy distance
total distance=s+c miles
total time=s/2+c/3=(3s+2c)/6 hours
(s+c)/[(3s+2c)/6]=2.8 mph average speed
s/c=1/6
s/(s+c)=1/7
E
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On a partly cloudy day, Derek decides to walk back from work [#permalink]

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New post 05 Jul 2017, 18:27
D=distance,
R= Rate
T= Time

D1=Distance (1)
D2= Distance (2)
T1 = Time (1)
T2 = Time (2)
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]

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New post 24 Dec 2017, 22:37
emmak wrote:
On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

A. 1/4
B. 4/5
C. 1/5
D. 1/6
E. 1/7

Since the average speed is 2.8, s has to be 2. The speed ratio is 2:3 and the time ratio is 1:4; therefore, the required ratio is 2/(2+12) = 1/7.
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]

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New post 03 Apr 2018, 10:27
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emmak wrote:
On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

A. 1/4
B. 4/5
C. 1/5
D. 1/6
E. 1/7


Let's assign a nice value to the total distance traveled.

If Derek's average speed is 2.8 mph, then let's say that he traveled a total of 28 miles.
At an average rate of 2.8 mph, a 28 mile trip will take 10 hours.

Since Derek's average speed is BETWEEN 2 mph and 3 mph, we can conclude that Derek walked 2 mph when it was sunny, and he walked 3 mph when it was cloudy.

Let's t = number of hours walking while sunny
So, 10 - t = number of hours walking while cloudy

We'll begin with a word equation: (distance traveled while sunny) + (distance traveled while cloudy) = 28
Since distance = (speed)(time), we can now write:
(2)(t) + (3)(10 - t) = 28
Expand: 2t + 30 - 3t = 28
Solve: t = 2
In other words, Derek walked for 2 hours while sunny.

At a walking speed of 2 mph, Derek walked for 4 miles while sunny.
So, Derek walked 4/28 of the total distance while the sun was shining on him.
4/28
= 1/7

Answer: E

Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com

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Re: On a partly cloudy day, Derek decides to walk back from work   [#permalink] 03 Apr 2018, 10:27

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