Last visit was: 18 Nov 2025, 14:15 It is currently 18 Nov 2025, 14:15
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
 1   2   3   4   
User avatar
emmak
User avatar
Joined: 09 Feb 2013
Last visit: 08 Jul 2014
Posts: 103
Own Kudos:
4,727
 [437]
Given Kudos: 17
Posts: 103
Kudos: 4,727
 [437]
On a partly cloudy day, Derek decides to walk back from work 11 Feb 2013, 08:44
45
Kudos
Add Kudos
391
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

33% (03:11) correct 67% (03:04) wrong based on 1817 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 479
Own Kudos:
3,340
 [115]
Given Kudos: 141
Posts: 479
Kudos: 3,340
 [115]
On a partly cloudy day, Derek decides to walk back from work 12 Feb 2013, 01:03
69
Kudos
Add Kudos
42
Bookmarks
Bookmark this Post
We know that the average speed is 2.8 mph. Thus,

as \(s<2.8<s+1\)

or \(s>1.8\) and \(s<2.8\). As s is an integer, the only value can be s=2. Thus,

Let the total length be d. Let the path where the sun shone on him be d1 = kd (0<k<1)

\(\frac{d}{2.8}\)=\(\frac{d1}{2}\)+\(\frac{d2}{3}\) =\(\frac{kd}{2}+\frac{(1-k)d}{3}\)

or \(\frac{10}{28}\)=\(\frac{k}{2}+\frac{(1-k)}{3}\)

By simple substitution, we can find that k = 1/7.
User avatar
SVaidyaraman
User avatar
Joined: 17 Dec 2012
Last visit: 11 Jul 2025
Posts: 576
Own Kudos:
1,795
 [69]
Given Kudos: 20
Location: India
Expert
Expert reply
Posts: 576
Kudos: 1,795
 [69]
On a partly cloudy day, Derek decides to walk back from work 02 Aug 2013, 18:26
39
Kudos
Add Kudos
27
Bookmarks
Bookmark this Post
WholeLottaLove


I do get it a bit more. These concepts can be quite tricky!!! Thanks!!!
Show more

1.Let the distance traveled by Derek when it is sunny be x and when it is cloudy be y. We need to find x / x+y
2. The average speed is given as 2.8 . The speed at which Derek travels when it is sunny and when it is cloudy have to be 2 and 3 resp as others have reasoned.
3. Average speed = total distance traveled/ total time taken , which is, x+y / (x/s + y/s+1) = 2.8 where s=2
4. Solving (3) we get x/y = 1/6 i.e., x / x+y = 1/7

Only a few approaches are there to be followed. Once you see which one to apply for a problem, it gets really easy.
User avatar
sambam
User avatar
Joined: 18 Oct 2011
Last visit: 09 Jan 2014
Posts: 58
Own Kudos:
416
 [28]
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Posts: 58
Kudos: 416
 [28]
On a partly cloudy day, Derek decides to walk back from work 11 Feb 2013, 12:50
16
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,265
Own Kudos:
76,980
 [23]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,265
Kudos: 76,980
 [23]
On a partly cloudy day, Derek decides to walk back from work Updated on: 08 Aug 2023, 04:32
Originally posted by KarishmaB on 06 Mar 2013, 22:04.
Last edited by KarishmaB on 08 Aug 2023, 04:32, edited 1 time in total.
20
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
swarman
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 :)
Show more

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.

Check this video: https://www.youtube.com/watch?v=_GOAU7moZ2Q
and this post: https://anaprep.com/arithmetic-weighted-averages/
General Discussion
User avatar
VeritasPrepRon
User avatar
Veritas Prep GMAT Instructor
Joined: 11 Dec 2012
Last visit: 16 Nov 2025
Posts: 306
Own Kudos:
695
 [8]
Given Kudos: 66
Expert
Expert reply
Posts: 306
Kudos: 695
 [8]
On a partly cloudy day, Derek decides to walk back from work Updated on: 29 Jul 2013, 22:14
Originally posted by VeritasPrepRon on 11 Feb 2013, 17:59.
Last edited by VeritasPrepRon on 29 Jul 2013, 22:14, edited 1 time in total.
3
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
sambam
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E
Show more

The key to this question is indeed that s must be 2, and therefore s+1 must be 3. It is impossible to average out a speed of 2.8 with any other two consecutive integers. The algebraic solution outlined above using D=RT gives the correct answer very quickly, but this can also be solved by using the concept and backsolving.

First of all, the classic trap of 1/4 of the time of the total time above is misleading. In fact, it implies that Derek spent 1/5 of his time at the slower speed and 4/5 at the higher speed. This is because 0.8 is analogous to 4/5, and can be demonstrated by (1/5 * 2) + (4/5 * 3) = 2/5 + 12/5 = 14/5 or 2.8. Once we know that Derek spent 1/5 of his time walking at his sunny-weather rate, and we know that the other 4/5 of the time he was walking faster, we can deduce that he covered less than 1/5 of the distance at the sunny-weather rate. If we understand this concept, we are down to two answer choices, D or E.

We can now backsolve by converting time into distance. Using D (1/6 of distance), we can assume 6 miles of distance, 1 of which at 2 mph and the other 5 at 3 mph. This leaves us with 0.5 hours of timeat speed s and 1.67 hours at speed s+1. These need to be in the ratio of 1:4 (or 1/5 to 4/5), and therefore don't work. Once this doesn't work, we know the answer is E. (Note: Backsolving for E gives 7 miles, 1 at 2 mph and 6 at 3 mph, yielding totals of 0.5 hours and 2 hours, exactly what we're looking for.

Undoubtedly the algebraic solution is faster, however, the concept alone leaves this at a 50/50 choice between D and E. Afterward, solving using one answer choice will confirm which one of the two must be correct.
User avatar
Rock750
User avatar
Joined: 25 Oct 2012
Last visit: 20 Sep 2016
Posts: 185
Own Kudos:
1,419
 [1]
Given Kudos: 85
Status:Final Lap
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE:Project Management (Retail Banking)
Posts: 185
Kudos: 1,419
 [1]
On a partly cloudy day, Derek decides to walk back from work 12 Feb 2013, 02:16
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
vinaymimani
We know that the average speed is 2.8 mph. Thus,

as \(s<2.8<s+1\)

or \(s>1.8\) and \(s<2.8\). As s is an integer, the only value can be s=2. Thus,

Let the total length be d. Let the path where the sun shone on him be d1 = kd (0<k<1)

\(\frac{d}{2.8}\)=\(\frac{d1}{2}\)+\(\frac{d2}{3}\) =\(\frac{kd}{2}+\frac{(1-k)d}{3}\)

or \(\frac{10}{28}\)=\(\frac{k}{2}+\frac{(1-k)}{3}\)

By simple substitution, we can find that k = 1/7.
Show more

Hi vinaymimani
How did u get that : as \(s<2.8<s+1\)

or \(s>1.8\) and \(s<2.8\).

Regards
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 479
Own Kudos:
3,340
 [5]
Given Kudos: 141
Posts: 479
Kudos: 3,340
 [5]
On a partly cloudy day, Derek decides to walk back from work Updated on: 19 Mar 2013, 04:13
Originally posted by mau5 on 12 Feb 2013, 02:35.
Last edited by mau5 on 19 Mar 2013, 04:13, edited 1 time in total.
5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rock750
vinaymimani
We know that the average speed is 2.8 mph. Thus,

as \(s<2.8<s+1\)

or \(s>1.8\) and \(s<2.8\). As s is an integer, the only value can be s=2. Thus,

Let the total length be d. Let the path where the sun shone on him be d1 = kd (0<k<1)

\(\frac{d}{2.8}\)=\(\frac{d1}{2}\)+\(\frac{d2}{3}\) =\(\frac{kd}{2}+\frac{(1-k)d}{3}\)

or \(\frac{10}{28}\)=\(\frac{k}{2}+\frac{(1-k)}{3}\)

By simple substitution, we can find that k = 1/7.

Hi vinaymimani
How did u get that : as \(s<2.8<s+1\)

or \(s>1.8\) and \(s<2.8\).

Regards
Show more

Hi Rock750

I will give a general proof for the above question :

From the given picture, let n1 and n2(n1,n2 are not equal to 0) be the speed for the length d1 and d2 respectively.

Also,

d=d1+d2 , d1 = kd, where 0<k<1

Now, average speed for this length is

\(\frac{d}{[d1/n1+d2/n2]}\) =\(\frac{d}{[kd/n1+d(1-k)/n2]}\) = let's call this value as AVG


Now, Considering n1>n2;lets assume that this average speed will be always between n1 & n2 or n2<AVG<n1



\(n2<\frac{n1*n2}{[n2*k+n1*(1-k)]}\) ..... I

and

\(\frac{n1*n2}{[n2*k+n1*(1-k)]}<n1\).........II

Thus, from I, our assumption will be right iff
\([n2*k+n1*(1-k)]<n1\)

or (n1-n2)k>0.

As we had already assumed that, this stands true.

Similarly, from II, we have our assumption to be true iff

\((n2-n1)*(1-k)<0\); which is again true.

Thus for any positive value of n1,n2 the inequality n2<AVG<n1 will always holds.

In the given sum, n2 = s, n1 = s+1 and AVG = 2.8 mph. Thus, 2.8 will always lie between s and (s+1). The question mentions that "s" is an integer for this very purpose.

I hope it was clear enough.
Attachments

3.jpg
3.jpg [ 17.97 KiB | Viewed 76404 times ]

avatar
swarman
avatar
Joined: 17 Jan 2013
Last visit: 15 Feb 2019
Posts: 41
Own Kudos:
92
Given Kudos: 109
Location: India
Schools: Cambridge"20 (A) LBS '21 (D) Ross '21 (S)
Schools: Cambridge"20 (A) LBS '21 (D) Ross '21 (S)
Posts: 41
Kudos: 92
On a partly cloudy day, Derek decides to walk back from work 06 Mar 2013, 03:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 :)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
 [3]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,062
 [3]
On a partly cloudy day, Derek decides to walk back from work 06 Mar 2013, 03:26
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
swarman
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 :)
Show more

Where, in which post, is the average speed calculated as the average of the two speeds?
avatar
swarman
avatar
Joined: 17 Jan 2013
Last visit: 15 Feb 2019
Posts: 41
Own Kudos:
92
 [2]
Given Kudos: 109
Location: India
Schools: Cambridge"20 (A) LBS '21 (D) Ross '21 (S)
Schools: Cambridge"20 (A) LBS '21 (D) Ross '21 (S)
Posts: 41
Kudos: 92
 [2]
On a partly cloudy day, Derek decides to walk back from work 07 Mar 2013, 10:03
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Hi Bunuel
Actually i thought the way we got s as 2 was:
(s+s+1)/2= 2.8 and by solving it we get value of s as 2 (closest integer).. but now its clear thanks to Karishma!

thank you both! :)
avatar
SpotlessMind
avatar
Joined: 22 Dec 2012
Last visit: 27 Nov 2013
Posts: 11
Own Kudos:
40
 [5]
Given Kudos: 19
GMAT 1: 720 Q49 V39
GMAT 1: 720 Q49 V39
Posts: 11
Kudos: 40
 [5]
On a partly cloudy day, Derek decides to walk back from work 08 Mar 2013, 12:48
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Hi,

Nice question!!!

I solved it like this..

let t1 be sunny time and t2 be cloudy time
then we are asked to find t1/ (t1+t2) (say X)

and avg speed = 2.8 = (st1+ st2+ t2)/(t1+t2)
so 2.8 = s + t2/(t1+t2)
2.8 = S + 1 - X
So our Req Qty X = S - 1.8
and since S is an integer the only value it can take is 2 (If its 3, the fraction will become absurd and if its 0, it becomes -ve!)
so
X = 0.2 or 1/5
avatar
spaceman29spiff
avatar
Joined: 30 Jan 2012
Last visit: 26 Jul 2013
Posts: 3
Own Kudos:
3
 [2]
Given Kudos: 6
Posts: 3
Kudos: 3
 [2]
On a partly cloudy day, Derek decides to walk back from work 20 May 2013, 12:10
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
swarman
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 :)

Where, in which post, is the average speed calculated as the average of the two speeds?
Show more


Hi Bunuel,
I took the following classical equations route to solve

Sunny:
Speed = s
Distance = x
hence, time = x/s

Cloudy:
Speed = s+1
dist = y
hence, time = y/(s+1)

Average:
Speed = 2.8
Total dist = x+y
total time = (x/s) + y/(s+1)

solving,
x+y = 2.8 * [(x/s) + (y/s+1)]

i got the final equation as;

s*(s-1.8) / 2.8 = x/(x+y)

How to solve it from here?

thanks,
Rohan
User avatar
chechaxo
User avatar
Joined: 19 Jul 2012
Last visit: 15 Oct 2013
Posts: 16
Own Kudos:
40
 [12]
Given Kudos: 8
Location: United States
Concentration: Operations, Entrepreneurship
Schools: INSEAD '14
WE:Consulting (Manufacturing)
Schools: INSEAD '14
Posts: 16
Kudos: 40
 [12]
On a partly cloudy day, Derek decides to walk back from work 29 Jul 2013, 13:10
7
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
As we know the average speed is total distance/total time.
A quicker method to solve the problem would be through the allegations technique..

All we need to remember that Speed is inversely proportional to time.

Cloudy Sunny
A|------------------------|-----------------------|B
time <-------y------------->|<----------x--------->
speed <-----s+1 ---------->|<---------s---------->
distance <---y(s+1)--------->|<------xs------------> Distance=Speed*Time[/color]

Question asks = xs/(xs+y(s+1) ??

From Allegation : (s+1 - 2.8)/(2.8-s)=x/y
I see three unknowns! I realize that I have to carefully look for the value of one of the unknowns...
Ahh..Average Speed! S+1 < 2.8< S (Problem mentions that S is an integer) Hence Average speed has to lie between 3 and 2. ie 3<2.8<2 (notice relationship between S and S+1)

Substitute S=2
x/y=0.2/.08=1/4 (Ratio of time)
Hence
Distance covered when Cloudy = y(s+1) = 4*(1+2) = 12 miles
Distance covered when Sunny = xs = 1*2 = 2 miles

Required answer = 2/(2+12) = 1/7

Option E :)
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
626
 [1]
Given Kudos: 134
Posts: 305
Kudos: 626
 [1]
On a partly cloudy day, Derek decides to walk back from work 02 Aug 2013, 10:05
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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?

s (and s+1) are both integers, therefore, the two speeds are consecutive integers. Because we know the average speed is 2.8 and the two speeds are consecutive integers, the only possible solution is the slower speed = 2 and the faster speed = 3.
rate=distance/time
Cloudy rate: s/t
2/t
2 miles/1 hour
Sunny rate: (s+1)/t
3/t
3 miles / 1 hour

Average speed = Distance/Time

Average speed = 2.8 miles/1 hour

So we know the average speed and the rate at which he walked for both sunny and cloudy times. In a way, I view this as a weighted average problem: "how many units of 2 miles/hour + 3miles/hour = 2.8 miles/hour total"

Past that, I am completely lost on the algebra and logic. An explanation of both would be mighty helpful!!

Thanks!!
User avatar
VeritasPrepRon
User avatar
Veritas Prep GMAT Instructor
Joined: 11 Dec 2012
Last visit: 16 Nov 2025
Posts: 306
Own Kudos:
695
 [3]
Given Kudos: 66
Expert
Expert reply
Posts: 306
Kudos: 695
 [3]
On a partly cloudy day, Derek decides to walk back from work 02 Aug 2013, 12:36
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
WholeLottaLove


So we know the average speed and the rate at which he walked for both sunny and cloudy times. In a way, I view this as a weighted average problem: "how many units of 2 miles/hour + 3miles/hour = 2.8 miles/hour total"

Past that, I am completely lost on the algebra and logic. An explanation of both would be mighty helpful!!

Thanks!!
Show more

Hi WholeLottaLove,

Mau5 started the algebra solution above. Let me continue it while being consistent with the variables set up. The logic has been covered before in this post many times (including by me), hopefully at least one of them will make sense either in conjunction with the algebra or on its own:

Algebraically, we know Derek spent some portion of his time walking at 2 MPH (k) and the rest of the time (1-k) at 3 MPH. Set the total time to be d and set up d1 for the slow walk and d2 for the fast walk, all we have to do is solve for variable k:

\(\frac{d}{2.8}\)=\(\frac{d1}{2}\)+\(\frac{d2}{3}\) =\(\frac{kd}{2}+\frac{(1-k)d}{3}\)

This can be rewritten as:

\(\frac{10}{28}\) = \(\frac{k}{2}+\frac{(1-k)}{3}\)

Putting the right side on a common denominator to (eventually) isolate K:

\(\frac{10}{28}\) = \(\frac{3k}{6}+\frac{(2-2k)}{6}\)

Multiplying both sides by 6:

\(\frac{60}{28}\) = \(3k+(2-2k)\)


Simplifying:

\(\frac{15}{7}\) = \(k + 2\)

Subtracting 2 from both sides:

\(\frac{1}{7}\) = \(k\)

I hope this makes sense to you. The picking numbers or backsolving approaches might be faster in this case, so feel free to use whichever method you want. As the old saying goes, though, algebra always tells the truth.

Hope this helps!
-Ron
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
626
 [10]
Given Kudos: 134
Posts: 305
Kudos: 626
 [10]
On a partly cloudy day, Derek decides to walk back from work 03 Aug 2013, 10:29
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
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?

Looking for the fraction of his total distance that was covered when the sun was shining on him. We're going to have to assign variables to sun distance and cloud distance so we can ultimately plug into the formula for average speed. We will also need to get the time taken for each leg of the journey which means we also need to know the speed traveled for each leg.
x=sun
y=clouds
The fraction of time spent walking when it is sunny = x/(x+y) (x divided by the total distance)

We have established that the speed for the first part = 2 and the speed for the second part =3

We are given average speed so it is fair to assume that we will apply it to a formula: Average speed = total distance/total time taken
Time taken:
sunny part of the journey: x/s (distance/speed)
cloudy part of the journey: x/s+1

2.8 = (x+y) / (x/2 + y/3)
2.8x/2 + 2.8y/3 = (x+y)
8.4x/6 + 5.6y/6 = x+y
8.4x+5.6y/6 = x+y
8.4x+5.6y = 6x+6y
2.4x = .4y
6x=y
x=1/6y
x/y=1/6

Answer: E. 1/7
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
626
Given Kudos: 134
Posts: 305
Kudos: 626
On a partly cloudy day, Derek decides to walk back from work 15 Aug 2013, 14:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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?

There are two portions to Derek's walk: the sunny portion (S) and the cloudy portion (C).

speed (S): s/1hour
speed (C): (s+1)/1hour

what fraction of the total distance did he cover while the sun was shining on him?

We need to find total distance, then solve for the distance he covered while the sun was shining.

Average speed = total distance/total time
2.8 = d / (t)
The problem is, we don't know how long it took Derek to walk the distance.

(s is an integer)

This gives us a needed clue (and makes the problem damn tricky!) If his average, round trip speed is 2.8 miles per hour and S is an integer value, it severely limits the possible values of S (especially when considering that his round trip speed was (s+1)

If s=1 then on the first leg of the trip he walked at 1 mile/hour and on the return he walked at (1+1) = 2 miles/hour. That wouldn't be fast enough to attain an average of 2.8 miles/hour round trip.

If s=3 then on the first leg of the trip he walked at 3 miles/hour and on the return he walked at (3+1) = 4 miles/hour. That would mean his average would be greater than 2.8 miles/hour.

If s=2 then on the first leg of the trip he walked at 2 miles/hour and on the return he walked at (2+1) = 3 miles/hour. This would allow for an average speed of 2.8 miles/hour.

speed (S): 2/1hour
speed (C): (3)/1hour

Average speed = total distance/total time taken
time = distance/speed

2.8 = d/[ (d/2) + (d/3) ]
2.8 = d/[ (3d/6) + (2d/6) ]
2.8 = d/[5d/6]
2.8 = d/1 * [6/5d]
2.8 = 6d/5d
2.8 = d

He walked a total of 2.8 miles. If his average speed was 2.8 miles then the entire walk took 1 hour.

I know I already solved this problem, but I tried solving it a different way and got stock above. Can someone explain to me why I am wrong or where I went wrong in solving this problem? Thanks! :-D
avatar
Asifpirlo
avatar
Joined: 10 Jul 2013
Last visit: 26 Jan 2014
Posts: 222
Own Kudos:
1,150
Given Kudos: 102
Posts: 222
Kudos: 1,150
On a partly cloudy day, Derek decides to walk back from work 15 Aug 2013, 14:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
emmak
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
Show more


2.8 = (S1+S2)/(t1+t2)
or, 2-8t1+2.8t2 = 2t1+3t2
or, t1 =t2/4

We need, S1/(S1+S2) = 2t1/(2t1+3t2) = (2t2/4) / (7t2/2) = t2/2 * 2/7t2 = 1/7

special thanks to KARISHMA for the subtle analysis of the matter......
User avatar
VeritasPrepRon
User avatar
Veritas Prep GMAT Instructor
Joined: 11 Dec 2012
Last visit: 16 Nov 2025
Posts: 306
Own Kudos:
695
Given Kudos: 66
Expert
Expert reply
Posts: 306
Kudos: 695
On a partly cloudy day, Derek decides to walk back from work 21 Aug 2013, 14:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hopefully at this point one of the solutions listed above will make sense to you and you can get to the right answer on this question. If there are still any doubts, I formally went through three different solutions on my weekly blog:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/08 ... -the-gmat/

Hope this helps!
-Ron
 1   2   3   4   
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts