Problem Solving

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

This math is from veritas prep practice test. I could not solve this one and i have noticed that i am facing problem regarding solving rate-distance related problems. I get confused from where to start solving this kind of problem. Even if i get clues and can solve, it takes long time. I seek help as i have my test in 2 weeks.

One more question, approx how many qs related to work rate and probability/combination usually come in the test?

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

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.

We can treat this as 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 the RECIPROCAL of the distances in red.
(2 miles per hour) : (3 miles per hour) = 0.2 : 0.8 = 1:4.

Here, the weight given to 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 = 2/(2+12) = 2/14 = 1/7.

The correct answer is E.

Thanks to both of you.

Can anyone please comment on the rest part of my question?

This math is from veritas prep practice test. I could not solve this one and i have noticed that i am facing problem regarding solving rate-distance related problems. I get confused from where to start solving this kind of problem. Even if i get clues and can solve, it takes long time. I seek help as i have my test in 2 weeks.

One more question, approx how many qs related to work rate and probability/combination usually come in the test?

What is the difficulty level according to you experts?
I took 3 minutes to solve this question. Is this reasonable?