Carlos runs a lap around the track in x seconds. His second

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Source: Manhattan Prep

Carlos runs a lap around the track in x seconds. His second lap is five seconds slower than the first lap, but the third lap is two seconds faster than the first lap. What is Carlos's average (arithmetic mean) number of minutes per lap, in terms of x?

A. \(x-1\)
B. \(x+1\)
C. \(\frac{x-1}{60}\)
D. \(\frac{x+1}{60}\)
E. \(\frac{x+3}{60}\)

The OA is D

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by ceilidh.erickson » Sat May 25, 2019 10:39 am
We're simply looking for the average time per lap (not an average rate or anything more complicated). His lap times in seconds are:
1) x
2) x + 5
3) x - 2

The average of these is:
$$\frac{\left(x\right)+\left(x+5\right)+\left(x-2\right)}{3}=\frac{3x+3}{3}=x+1$$

Bear in mind that this is the average in SECONDS, but the question is asking for the average in minutes. Because each second is 1/60 of a minute, simply divide by 60:
$$\frac{x+1}{60}$$

The answer is D.
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Harvard Graduate School of Education

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by [email protected] » Tue May 28, 2019 4:45 pm
BTGmoderatorLU wrote:Source: Manhattan Prep

Carlos runs a lap around the track in x seconds. His second lap is five seconds slower than the first lap, but the third lap is two seconds faster than the first lap. What is Carlos's average (arithmetic mean) number of minutes per lap, in terms of x?

A. \(x-1\)
B. \(x+1\)
C. \(\frac{x-1}{60}\)
D. \(\frac{x+1}{60}\)
E. \(\frac{x+3}{60}\)

The OA is D
The total number of seconds it takes Carlos to run the 3 laps is x + (x + 5) + (x - 2) = 3x + 3. So the number of seconds per lap is (3x + 3)/3 = x + 1, and hence, the number of minutes per lap is (x + 1)/60.

Answer: D

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