Find all School-related info fast with the new School-Specific MBA Forum

It is currently 23 Sep 2014, 06:31

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
14 KUDOS received
Manager
Manager
avatar
Joined: 09 Feb 2013
Posts: 121
Followers: 1

Kudos [?]: 238 [14] , given: 17

On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 11 Feb 2013, 07:44
14
This post received
KUDOS
12
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (04:28) correct 64% (02:47) wrong based on 506 sessions
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
[Reveal] Spoiler: OA

_________________

Kudos will encourage many others, like me.
Good Questions also deserve few KUDOS.

2 KUDOS received
Manager
Manager
avatar
Joined: 18 Oct 2011
Posts: 92
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Followers: 2

Kudos [?]: 24 [2] , given: 0

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 11 Feb 2013, 11:50
2
This post received
KUDOS
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
Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 11 Dec 2012
Posts: 313
Followers: 59

Kudos [?]: 188 [1] , given: 66

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 11 Feb 2013, 16:59
1
This post received
KUDOS
Expert's post
sambam wrote:
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


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.
_________________

Ron Awad
Veritas Prep | GMAT Instructor
Save $100 on Veritas Prep GMAT Courses and Admissions Consulting Services
Veritas Prep Reviews


Last edited by VeritasPrepRon on 29 Jul 2013, 21:14, edited 1 time in total.
Expert Post
16 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 627
Followers: 41

Kudos [?]: 579 [16] , given: 135

Premium Member
Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 12 Feb 2013, 00:03
16
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Senior Manager
Senior Manager
User avatar
Status: Final Lap
Joined: 25 Oct 2012
Posts: 293
Concentration: General Management, Entrepreneurship
Schools: Oxford
GPA: 3.54
WE: Project Management (Retail Banking)
Followers: 2

Kudos [?]: 102 [0], given: 85

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 12 Feb 2013, 01:16
vinaymimani wrote:
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
_________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Expert Post
3 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 627
Followers: 41

Kudos [?]: 579 [3] , given: 135

Premium Member
Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 12 Feb 2013, 01:35
3
This post received
KUDOS
Expert's post
Rock750 wrote:
vinaymimani wrote:
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


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 6552 times ]


_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions


Last edited by mau5 on 19 Mar 2013, 03:13, edited 1 time in total.
Intern
Intern
avatar
Joined: 17 Jan 2013
Posts: 37
Followers: 1

Kudos [?]: 3 [0], given: 67

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 06 Mar 2013, 02:17
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 :)
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 22981
Followers: 3517

Kudos [?]: 26731 [0], given: 2714

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 06 Mar 2013, 02:26
Expert's post
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 :)


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

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
9 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4791
Location: Pune, India
Followers: 1123

Kudos [?]: 5102 [9] , given: 164

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 06 Mar 2013, 21:04
9
This post received
KUDOS
Expert's post
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Intern
Intern
avatar
Joined: 17 Jan 2013
Posts: 37
Followers: 1

Kudos [?]: 3 [0], given: 67

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 07 Mar 2013, 09:03
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! :)
1 KUDOS received
Intern
Intern
avatar
Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Followers: 1

Kudos [?]: 11 [1] , given: 19

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 08 Mar 2013, 11:48
1
This post received
KUDOS
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
Intern
Intern
avatar
Joined: 30 Jan 2012
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 6

GMAT ToolKit User
Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 20 May 2013, 11:10
Bunuel 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 :)


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



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
_________________

No looking back !

1 KUDOS received
Intern
Intern
avatar
Joined: 19 Jul 2012
Posts: 22
Location: United States
Concentration: Operations, Entrepreneurship
Schools: INSEAD '14
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 9 [1] , given: 8

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 29 Jul 2013, 12:10
1
This post received
KUDOS
1
This post was
BOOKMARKED
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 :)
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 51 [0], given: 134

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 02 Aug 2013, 09:05
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!!
Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 11 Dec 2012
Posts: 313
Followers: 59

Kudos [?]: 188 [2] , given: 66

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 02 Aug 2013, 11:36
2
This post received
KUDOS
Expert's post
WholeLottaLove wrote:

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!!


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
_________________

Ron Awad
Veritas Prep | GMAT Instructor
Save $100 on Veritas Prep GMAT Courses and Admissions Consulting Services
Veritas Prep Reviews

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 51 [0], given: 134

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 02 Aug 2013, 11:53
VeritasPrepRon wrote:
WholeLottaLove wrote:

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!!


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:

Silly question, but are we assuming that he walks for one hour here?

\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}

Where did the 10/28 come from?


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


I do get it a bit more. These concepts can be quite tricky!!! Thanks!!!
2 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 17 Dec 2012
Posts: 395
Location: India
Followers: 13

Kudos [?]: 182 [2] , given: 10

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 02 Aug 2013, 17:26
2
This post received
KUDOS
WholeLottaLove wrote:

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


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.
_________________

Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravna.com/courses.php

Classroom Courses in Chennai
Online and Correspondence Courses

1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 51 [1] , given: 134

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 03 Aug 2013, 09:29
1
This post received
KUDOS
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
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 51 [0], given: 134

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 15 Aug 2013, 13:25
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
Senior Manager
Senior Manager
avatar
Joined: 10 Jul 2013
Posts: 343
Followers: 3

Kudos [?]: 114 [0], given: 102

Re: On a partly cloudy day, Derek decides to walk back from work [#permalink] New post 15 Aug 2013, 13:51
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



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......
_________________

Asif vai.....

Re: On a partly cloudy day, Derek decides to walk back from work   [#permalink] 15 Aug 2013, 13:51
    Similar topics Author Replies Last post
Similar
Topics:
3 Although the vaccine, made partly from the brains of other s blueseas 9 08 Jul 2013, 07:43
3 Back to school Math days!! Altruistic 1 28 Oct 2012, 02:16
The hiker walked for two days. on the second day the hiker blover 4 09 Sep 2008, 04:15
A hiker walked for two days. On the second day the hiker alimad 4 02 May 2008, 06:01
A hiker walks for two days. On the second day the hiker étudiant 2 07 Oct 2005, 05:13
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 28 posts ] 



cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.