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

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Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post 19 Aug 2018, 09:43
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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) \(\frac{x+3}{60}\)

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Re: Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post 19 Aug 2018, 09:50
Bunuel wrote:
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) \(\frac{x+3}{60}\)



lap 1 => x seconds
lap 2 => x+5 seconds
lap 3 = > x-2 seconds

mean = (3x+3)/3 = x+1 seconds = (x+1)/60 minutes
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Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post Updated on: 13 Jan 2019, 12:06
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, but regardless of how you approach this question, you have to make sure to do the extra 'step' of converting seconds to minutes.

IF.... X=3, then the three lap times are...
3 seconds, 8 seconds and 1 second...

Thus, the average time/lap is (3+8+1)/3 = 12/3 = 4 seconds

Since there are 60 seconds in a minute, 4 seconds = 4/60 = 1/15 of a minute. Thus, we're looking for an answer that equals 1/15 when we plug X=3 into it. There's only one answer that matches.

Final Answer:

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Originally posted by EMPOWERgmatRichC on 12 Jan 2019, 12:47.
Last edited by EMPOWERgmatRichC on 13 Jan 2019, 12:06, edited 1 time in total.
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Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post 13 Jan 2019, 04:35
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EMPOWERgmatRichC wrote:
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, but regardless of how you approach this question, you have to make sure to do the extra 'step' of converting seconds to minutes.

IF.... X=3, then the three lap times are...
3 seconds, 8 seconds and 1 second...

Thus, the average time/lap is (3+8+1)/3 = 12/3 = 4 seconds

Since there are 60 seconds in a minute, 4 seconds = 4/60 = 1/15 of a minute. Thus, we're looking for an answer that equals 1/15 when we plug X=3 into it. There's only one answer that matches.

Final Answer: D

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


Hi Rich,

Hi Rich,

The final answer is clear. I can see it only when I reply to you but not in the original post.
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Re: Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post 13 Jan 2019, 06:49
Bunuel wrote:
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!

\(?\,\,\, = \,\,\,\,{{\left( {3x + 3} \right)\,\,\sec } \over {3\,\,{\rm{laps}}\,}}\,\,\,\left( {{{1\,\,\min } \over {60\,\,\sec }}} \right)\,\,\,\, = \,\,\,\,{{x + 1} \over {60}}\,\,\,\,{{\min } \over {{\rm{lap}}}}\)


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

Regards,
Fabio.
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Re: Carlos runs a lap around the track in x seconds. His second lap is fiv  [#permalink]

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New post 14 Jan 2019, 19:06
Bunuel wrote:
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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Re: Carlos runs a lap around the track in x seconds. His second lap is fiv   [#permalink] 14 Jan 2019, 19:06
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Carlos runs a lap around the track in x seconds. His second lap is fiv

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