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

Can someone please explain why that approach is not correct?

Ds = distance sunny time
Dc = distance cloudy time

Average speed = 2.8 = 14/5

and we know, Average speed = total distance / total time
then, why can't we simply express that with Ds and Dc?

I mean : \(14/5 = \frac{(Ds+Dc)}{(Ds/2)+(Dc/3)}\)

and then try to obtain the ratio asked in the question

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

This implies that the ratio of time for s=2 is 1/4 of the total time
I didn't understand the Bold part...
i am missing something here.. how come we know this??
Please explain. may be i am missing some concept here..

this is what I missed here..
Thank you got it now...

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

I should admit that I could not solve this mathematically. I assumed total distance as 2.8 miles. Then I tried from the options to get 's' as an integer. Option E gives an integer value for s. Therefore, that should be the 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.

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

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.

Could you explain the deduction in the bolded part?