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

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Manager
Joined: 14 Oct 2012
Posts: 166
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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23 Apr 2017, 16:09
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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23 Apr 2017, 19:53
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
Manager
Joined: 31 Dec 2016
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On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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05 Jul 2017, 17:27
D=distance,
R= Rate
T= Time

D1=Distance (1)
D2= Distance (2)
T1 = Time (1)
T2 = Time (2)
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Manager
Joined: 09 Jun 2015
Posts: 94
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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24 Dec 2017, 21:37
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.
CEO
Joined: 11 Sep 2015
Posts: 3127
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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03 Apr 2018, 09:27
Top Contributor
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

Cheers,
Brent
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Intern
Joined: 04 Apr 2017
Posts: 19
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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07 May 2018, 14:42
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?
CEO
Joined: 11 Sep 2015
Posts: 3127
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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07 May 2018, 15:49
1
Top Contributor
hisho wrote:
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?

Your solution assumes that average speed is based on the proportions of the distances traveled at each speed.
So, for example, if we traveled half the distance at 40 mph and traveled the other half the distance at 60 mph, then the average speed would be 50 mph.
However, this is not how average speeds work.

For more on this, watch the following video:

Cheers,
Brent
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Location: Pune, India
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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07 May 2018, 19:38
1
1
hisho wrote:
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?

Average Speed = Total Distance / Total Time

Average Speed = (Distance1 + Distance2) / (Time1 + Time2)

Average Speed = (Speed1*Time1 + Speed2*Time2) / (Time1 + Time2)

What you have done is

Average Speed = (Speed1*Distance1 + Speed2*Distance2) / (Distance1 + Distance2)
which is illogical.
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Joined: 21 Jan 2017
Posts: 33
Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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28 Jun 2018, 03:32
VeritasPrepKarishma wrote:
swarman wrote:
What i m not getting is that why are we considering average speed as arithmetic mean of the speeds?
average speed is calculated by total distance divided by total time taken, isnt it??

kindly help

Average Speed lies in between the two speeds. It may not be in the center since the time taken at the two speeds might be different but it does lie somewhere in between them. You cannot drive at two speeds: 50 mph and 60 mph and still expect to average 70 mph. Your average will lie somewhere between 50 and 60.

Similarly, if the average speed is 2.8 and the two speeds are consecutive integers, the speeds must be 2 and 3. You cannot have the speeds as (1 and 2) or (3 and 4) since they cannot average out to be 2.8.

Thanks a lot Karishma . I like the way you explain things
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Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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02 Jul 2018, 08:42
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 the total distance = d and x = the distance he covered when it was sunny. Thus, the distance he covered when it was cloudy is d - x. We need to determine the value of x/d.

Furthermore, we see that s must be 2. That is because if s = 1, then s + 1 = 2. We can see that if both speeds are 2 mph or less, then the average speed can never be 2.8 mph. A similar situation exists if s = 3 (or more), and s + 1 = 4 (or more). We can see that if both speeds are 3 mph or greater, then the average speed again can never be 2.8 mph. Therefore, s must be 2 since it given that it’s an integer. Therefore, his “sunny” speed is 2 mph, and his “cloudy” speed is 3 mph.

Since (total distance)/(total time) = average speed, we have:

d/[x/2 + (d - x)/3] = 2.8

d/2.8 = x/2 + (d - x)/3

Multiplying the equation by 84, we have:

30d = 42x + 28(d - x)

30d = 42x + 28d - 28x

2d = 14x

2/14 = x/d

1/7 = x/d

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Re: On a partly cloudy day, Derek decides to walk back from work &nbs [#permalink] 02 Jul 2018, 08:42

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