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

Question

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

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

Bunuel · Accepted Answer

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

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

Since the average speed is 2.8 mph and  the two speeds are consecutive integers , they must be 2 mph and 3 mph.

Now, his speed = distance/time = 2.8 = 14/5, so suppose Derek covered distance = 14 miles and walked for time = 5 hours.

If he had walked all 5 hours at 3 mph, he would have covered 15 miles. He actually covered 1 mile less.

Every hour spent walking at 2 mph instead of 3 mph reduces the distance by 1 mile, so he must have walked at 2 mph for exactly 1 hour.

Thus, while the sun was shining, he covered 2 miles out of 14:

2/14 = 1/7.

Answer: E.

sambam · Answer

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

VeritasPrepRon · Answer

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.

mau5 · Answer

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.

Rock750 · Answer

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

or  s>1.8  and  s<2.8 .

Regards

mau5 · Answer

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 0n2;lets assume that this average speed will be always between n1 & n2 or n20. 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

swarman · Answer

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

KarishmaB · Answer

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/

SpotlessMind · Answer

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

chechaxo · Answer

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

WholeLottaLove · Answer

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

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!!

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

Thanks!!

VeritasPrepRon · Answer

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

SVaidyaraman · Answer

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.

WholeLottaLove · Answer

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

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

m3equals333 · Answer

Is there an algebraic way to solve for s (or even solve directly for the distance ratio) if it wasn't immediately apparent to us that 2.8 is between s and s+1 and s is an integer so it has to be 2?

I attempted to do this and it not only turned into a mess, but also didn't seem possible? Knowing that s=2 is definitely the key, but I was curious if there is a feasible workaround, thanks.

KarishmaB · Answer

I wrote this question and I will tell you what my intent was in writing it - I wanted to make people use logic. I wanted them to use deduction on the given data. Use of logic greatly simplifies many high level GMAT questions. So I tried to force them to deduce this vital piece of information - the two speeds must be 2 and 3. You can't put everything given to you in terms of an equation and that is what I tried to exploit.

Even if you don't start with it, at some point you will require some deduction.

For example, say you use weighted averages. 2.8 is the average speed. Weights in case of speeds are the time taken in the two cases.

Ratio of time spent while its sunny and time spent while its cloudy =  \frac{Ts}{Tc} = \frac{(s+1) - 2.8}{2.8 - s}

\frac{Ts}{Tc}= \frac{s - 1.8}{2.8 - s}

Now what can you say about s?

s must be greater than 1.8 since the fraction cannot be negative. It must also be less than 2.8 since the fraction cannot be negative.

Since s must be an integer, it must be 2. So (s+1) must be 3.

raks38 · Answer

sunny speed (s) < 2.8< (s+1) cloudy speed
As S is integer. The only possible value is S=2.

As avg speed is 2.8 .
i.e total distance/ total time = 2.8
Let the total distance is 28 and time taken 10 hrs.

Out of 10 hours X hours in Sunny and 10-X hours in cloudy

So total distance S*X + (S+1) * (10-x) = 28
2x + 30 - 3x = 28 ( substituted s = 2)
x = 2

In sunny at speed 2 miles/hr , the person walked for 2 hrs. Hence distance covered = 2*2 = 4 miles
Ration = 4miles ( distance in sunny) / 28miles( total distance) = 1/7

Ans E.

manishtank1988 · Answer

My 2 cents: My solution based on the veritas article: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/0 ... -the-gmat/

BrentGMATPrepNow · Answer

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

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

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

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615 to 715 in 15 Days (Ria's Strategies)

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