Can you please explain how the sunny time speed 2 miles/hour [ i.e. s miles/h ] and cloudy time speed 3 miles /hour [ i.e. (s+1 ) miles/h ] corroborate to the average speed of 2.8miles /h in the entire distance travelled ?
kind regards,

Sorry, I don't understand your question.
What does h stand for?
Also, I'm not sure where you got 2/h and 3/h. What do these expressions represent?

Cheers,
Brent

if he only has two running speeds: s and s + 1 then the range of the running speeds is 1.

We also know that his sunny walking speed is an integer. (there would be infinite solutions if this wasn't the case)

Since the average speed is 2.8, we know that s = 2, otherwise we wouldn't be able to get the average 2.8

Let's set up the problem a weighted averages problem:
(2)(t) + (2+1)(1-t) where t is bounded on the interval [0,1]

we want the weighted average to equal 2.8

2t + 3 - 3t = 2.8
-t = -0.2
so t = .2

This means that he spent 1/5 the total time going 2 mph and 4/5 the total time going 3 mph.

In one hour he travels 1/5 * 2 in the sun = 2/5
and one hour 4/5 * 3 partially sunny = 12/5

total distance traveled is 14/5 = 2.8 miles (consistent with the given information)

d_partiallyCloudy / d_total = fraction walked in clouds
(2/5) / (14/5) = 2/14 = 1/7

------------------------

I think pranab01's way is better but I guess here's another perspective haha

Here's another approach that doesn't assign a "nice value" for the distance traveled...

Let's first start with Derek's regular speed and increased speed (in mph)
We know that these two speeds differ by 1 mph, and we know that Derek's average speed is 2.8 mph.

This tells us that Derek walked 2 mph when it was sunny, and he walked 3 mph when it was cloudy.

Okay now let's deal with the rest of the problem.....

Let's say d = the distance Derek travel

Time = distance/speed
So, Derek's travel time = d/2.8

Since Derek's TOTAL travel time = d/2.8, let's say that:
t = number of hours walking while sunny
So d/2.8 - t = number of hours walking while cloudy

We'll begin with a word equation: (distance traveled while sunny) + (distance traveled while cloudy) = TOTAL distance traveled
Since distance = (speed)(time), we can now write:

(2)(t) + (3)(d/2.8 - t) = d
Expand: 2t + 3d/2.8 - 3t = d
Simplify: 3d/2.8 - t = d
Multiply both sides by 2.8 to get: 3d - 2.8t = 2.8d
Rearrange to get: 0.2d = 2.8t
Solve: t = 0.2d/2.8 = 2d/28 = d/14
In other words, Derek walked for d/14 hours while sunny.

Distance = (speed)(time)
At a walking speed of 2 mph, the distance Derek walked while sunny = (2)( d/14) = 2d/14 = d/7
So, Derek walked d/7 miles of the total distance while the sun was shining on him.

In other words, Derek walked 1/7 of the total distance while the sun was shining on him.

Answer: 1/7

ASIDE: As you can see the problem is much harder to solve when we don't have signed a nice value for the distance in this case

Cheers,
Brent

the harmonic mean of 2 and 3 is 2.4 and not 2.8

I think it is by logic/reasoning that avg speed should between the speeds and as the speeds are integers they must be 2 and 3

What I think is important to understand is being able to find that value of \(s\).

If we have two different segments of travel with two arbitrary speeds, call the slower one \(x\) and faster one \(y\), then the average speed, call it \(r_{average}\), is between those two speeds.

In other words:

\(x <= r_{average} <= y\)

And this works for any arbitrary time for the speeds were travelling at, and for any number of segments greater than 1.

So for example, if for one segment I travel 20mph for 1 minute, and the next segment I travel 30mph for 1 hour, my average speed for the whole trip would be very close to 30 since I spent more time travelling at that speed (in fact, it would be 29.83mph).

But here's the important part: No matter how I manipulate the times spent at those speeds, the average speed would never exceed 30mph, or be less than 20mph.


And that's the logic we need to figure out \(s\).

Since the average speed is 2.8, and we're given that \(s\) is an integer, then we know two things:

1) \(s+1\) is an integer.

2) Using the formula above: \(s <= 2.8 <= s+1 \), so it follows that \(s\) must be 2 and \(s+1\) must be 3.

why did you chose 2 and 3 why not 20 and 21

I'll answer that question by way of example.
If I spend part of a trip walking 5 miles per hour, and the other part of the trip walking 6 miles per hour, then my average speed will be between 5 and 6 mph.

In this particular question, we're told Derek's average speed is 2.8 miles per hour.
Since s and s+1 are integer values (Derek's speeds), it must be the case that s = 2 and s+1 = 3 (so that Derek's average speed of 2.8 mph is BETWEEN 2 and 3 mph.

Does that help?

The average speed -- 2.8 miles per hour -- must be BETWEEN the two individual rates (s and s+1).
Thus, s = 2 miles per hour and s+1 = 3 miles per hour.

This is a MIXTURE problem
A rate of 2 miles per hour is being combined with a rate of 3 miles per hour to yield an average speed of 2.8 miles per hour.
To determine how much WEIGHT must be given to each rate, we can use ALLIGATION:

Step 1: Plot the 3 rates on a number line, with the two individual rates (2 miles per hour and 3 miles per hour) on the ends and the average speed for the whole trip (2.8) in the middle.
2---------------2.8-------------3

Step 2: Calculate the distances between the rates.
2-----0.8-----2.8-----0.2-----3

Step 3: Determine the ratio of the rates.
The required ratio is equal to the RECIPROCAL of the distances in red.
(2 miles per hour) : (3 miles per hour) = 0.2 : 0.8 = 2:8 = 1:4

In rate problems, the weight for each rate is the amount of TIME spent at each rate.
The ratio above implies the following:
For every 1 hour spent traveling at 2 miles per hour, 4 hours must be spent traveling at 3 miles per hour.

Distance traveled in 1 hour at rate of 2 miles per hour = r*t = 2*1 = 2 miles.
Distance traveled in 4 hours at a rate of 3 miles per hour = r*t = 3*4 = 12 miles.
Of the total distance, the fraction traveled at 2 miles per hour \(= \frac{2}{(2+12)} = \frac{2}{14} = \frac{1}{7}\)

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.