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

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

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/

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.

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.

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
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!

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

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

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

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!!

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

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

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!

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......

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