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

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

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New post 08 Feb 2017, 02:02
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A
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C
D
E

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  95% (hard)

Question Stats:

50% (02:56) correct 50% (03:19) wrong based on 82 sessions

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Re: Over the course of 8 races, a runner's finishing time was 5 seconds  [#permalink]

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New post 08 Feb 2017, 07:58
We know that the combined time for the races is 44:20 or 2660 seconds.

Since the runner's times decrease by 5 seconds per race, we can think of the final race as x seconds, the next to last race as x+5 seconds, the one before it as x+10, etc. These can be added together and rewritten as follows:

\(8x+5(1+2+3+4+5+6+7)=2660\)
\(8x=2520\)
\(x=315\)

So we know the final race was 315 seconds. We also know that the second race was 30 seconds longer than the final race. 315+30=345. We convert this to minutes to get 5:45. Answer is D.
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Re: Over the course of 8 races, a runner's finishing time was 5 seconds  [#permalink]

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New post 13 Feb 2017, 08:39
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Bunuel wrote:
Over the course of 8 races, a runner's finishing time was 5 seconds faster each race than the race prior. If her combined total for all 8 races was 44:20, what was her finishing time for the second race?

A. 5:30
B. 5:35
C. 5:40
D. 5:45
E. 5:50


Since in each race, the runner was 5 seconds faster (i.e., 5 seconds less) than in the race prior, we can let the time of each race be as follows:

Race 1 = x

Race 2 = x - 5

Race 3 = x - 10

Race 4 = x - 15

Race 5 = x - 20

Race 6 = x - 25

Race 7 = x - 30

Race 8 = x - 35

Thus, the sum of the race times is 8x - 140.

We are given that the total was 44 minutes and 20 seconds. Converting to seconds gives us:

44 minutes and 20 seconds = 44 x 60 + 20 seconds = 2660 seconds

We can now determine x:

8x - 140 = 2660

8x = 2800

x = 350 seconds

350 seconds = 350/60 = 5 and 50/60 minutes = 5 minutes and 50 seconds.

Thus, the time of the second race is 5 minutes 50 seconds - 5 seconds = 5 minutes and 45 seconds.

Answer: D
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Re: Over the course of 8 races, a runner's finishing time was 5 seconds  [#permalink]

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New post 21 Feb 2017, 07:09
Bunuel wrote:
Over the course of 8 races, a runner's finishing time was 5 seconds faster each race than the race prior. If her combined total for all 8 races was 44:20, what was her finishing time for the second race?

A. 5:30
B. 5:35
C. 5:40
D. 5:45
E. 5:50


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.
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Re: Over the course of 8 races, a runner's finishing time was 5 seconds  [#permalink]

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New post 08 Aug 2019, 00:41
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
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Re: Over the course of 8 races, a runner's finishing time was 5 seconds   [#permalink] 08 Aug 2019, 00:41
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Over the course of 8 races, a runner's finishing time was 5 seconds

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