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

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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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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?
$$A.\ x-1$$
$$B.\ x+1$$
$$C.\ \frac{x-1}{60}$$
$$D.\ \frac{x+1}{60}$$
$$E.\ \frac{x+3}{60}$$
Let x = 5 seconds for the first lap, implying 10 seconds for the second lap, 3 seconds for the third lap, and a total of 18 seconds for all 3 laps.
(18 seconds)/(3 laps) = (6 seconds)/(1 lap) = 1/10 minute per lap.

The correct answer must yield a value of 1/10 when x=5.
Only D works:
(x+1)/60 = (5+1)/60 = 6/60 = 1/10.

The correct answer is D.

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Hi All,

We're told that 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. We're asked for Carlos’s average (arithmetic mean) number of MINUTES per LAP, in terms of X. This question can be solved rather easily by TESTing VALUES (as Mitch has shown). It can also be solved Algebraically, although you have to make sure to do the extra 'step' of converting seconds to minutes.

To start, we know that the 3 times/lap are : X seconds, (X + 5) seconds and (X - 2) seconds, so the AVERAGE time per lap (in seconds) is:

(X + X + 5 + X - 2)/3 = (3X + 3)/3 = (X + 1) seconds

Since 1 minute = 60 seconds, to convert an answer "in seconds" to an answer "in minutes", we need to divide by 60...

(X + 1)/60

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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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?
$$A.\ x-1\,\,\,\,\,\,\,B.\ x+1\,\,\,\,\,\,\,C.\ \frac{x-1}{60}\,\,\,\,\,\,\,D.\ \frac{x+1}{60}\,\,\,\,\,\,\,E.\ \frac{x+3}{60}$$
$$?\,\,\,:\,\,\,\,\,\,f\left( x \right)\,\,\,{{\min } \over {{\rm{lap}}}}$$
$$\eqalign{
& {1^{{\rm{st}}}}\,{\rm{lap}}\,\,\,\, \to \,\,\,\,\,\,\,x\,\,\sec \cr
& {2^{{\rm{nd}}}}\,{\rm{lap}}\,\,\,\, \to \,\,\,\,\,\left( {x + 5} \right)\,\,\sec \,\,\,\,\, \cr
& {3^{{\rm{rd}}}}\,{\rm{lap}}\,\,\,\, \to \,\,\,\,\,\left( {x - 2} \right)\,\,\sec \,\, \cr} $$

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

\[?\,\,\, = \,\,\,\,\frac{{\left( {3x + 3} \right)\,\,\sec }}{{3\,\,{\text{laps}}\,}}\,\,\,\left( {\frac{{1\,\,\min }}{{60\,\,\sec }}} \right)\,\,\,\, = \,\,\,\,\frac{{x + 1}}{{60}}\,\,\,\,\frac{{\min }}{{{\text{lap}}}}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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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 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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Scott Woodbury-Stewart Founder and CEO

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