Over the course of 8 races, a runner's finishing time was 5 seconds

Official solution from Veritas Prep.

This word problem is essentially a sequence problem. You don't know what her time was for each individual race, but you can use the same variable (for example, x) and relate each race to the others. If her first race took her x seconds, then the second was (x - 5), the third was (x - 10), etc. So for 8 total races, that means that the sequence would go:
x, (x - 5), (x - 10), ..., (x - 35)

You can then sum the times quickly by recognizing that each of 8 races has its own x, so that means 8x. And then you'll subtract from all but the first race. That means that 7 races will involve subtraction, with the smallest number subtracted being 5 and the largest being 35. Since this is an evenly-spaced set, you can multiply those 7 terms by the average of 20, and know that her overall time can be represented as 8x - 140.

You know that \(8x - 140 = 2660\) (NOTE: You'll calculate 2660 by multiplying 44 minutes by 60 and adding the remaining 20 seconds. 40 times 60 is 2400, plus 4 times 60 is another 240, plus 20 gives you 2660).

That then means that \(8x = 2800\), which means that x (the time of her first race) equals 350 seconds. 350 seconds is 5:50, but keep in mind that that is the time of her FIRST race, and the question asks for her second. Subtracting 5, that means that her second race took her 5:45.

This approach, which avoids the time-consuming step-by-step processes and is especially handy if you face a question involving larger numbers (say, if the number of races was 80 i/o 8), employs the formula for the summation of an arithmetic series:

Sn = (n/2){2a +(n-1)d} (where Sn is the sum of 'n' terms of the series, 'a' is the 1st term, 'n' is the number of terms and 'd' is the Common Difference between consecutive terms)
In this case: Sn=Total time for all 8 races=2660; 'a'=Time for 1st race=Unknown Qty; n=Total number of races=8; and 'd'=Time difference between each race and the next=(-5).

2660 = 4(2a - 35)
a = 350 seconds
Thus, time taken to run the 2nd race was (350-5=345 seconds=5:45 minutes. Ans: D

Firstly, 44:20 mins means (44*60)+20=2660 seconds.
Let's say, that to finish the first race, x seconds are needed. For the second race, the runner was 5 seconds faster so (x-5) seconds are required. For 3rd race, (x-10) seconds will be required and so on.
Therefore, the series becomes x+(x-5)+(x-10)+(x-15)+.... This is an AP series with n=8 and common difference d=(x-5)-x=(-5).
Therefore, \((8/2)*(2x+(8-1)*(-5)) = 2660\). Solving we get x=350. So, for the second race, 350-5=345 seconds or, 5 mins 45 seconds are required. Option (D) is correct.