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Steven has run a certain number of laps around a track at an average (

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Steven has run a certain number of laps around a track at an average (  [#permalink]

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New post 14 Feb 2019, 02:31
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Question Stats:

85% (01:38) correct 15% (02:26) wrong based on 27 sessions

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Steven has run a certain number of laps around a track at an average (arithmetic mean) time per lap of 51 seconds. If he runs one additional lap in 39 seconds and reduces his average time per lap to 49 seconds, how many laps did he run at an average time per lap of 51 seconds?

A. 2
B. 5
C. 6
D. 10
E. 12

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Re: Steven has run a certain number of laps around a track at an average (  [#permalink]

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New post 14 Feb 2019, 04:29
SajjadAhmad wrote:
Steven has run a certain number of laps around a track at an average (arithmetic mean) time per lap of 51 seconds. If he runs one additional lap in 39 seconds and reduces his average time per lap to 49 seconds, how many laps did he run at an average time per lap of 51 seconds?

A. 2
B. 5
C. 6
D. 10
E. 12


Total Time = 51*x for x laps

Total time after one more lap = 51x + 39

Average time now = (51x + 39) / (x+1) = 49

i.e. x = 5

Answer: Option B
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Re: Steven has run a certain number of laps around a track at an average (  [#permalink]

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New post 14 Feb 2019, 05:58
SajjadAhmad wrote:
Steven has run a certain number of laps around a track at an average (arithmetic mean) time per lap of 51 seconds. If he runs one additional lap in 39 seconds and reduces his average time per lap to 49 seconds, how many laps did he run at an average time per lap of 51 seconds?

A. 2
B. 5
C. 6
D. 10
E. 12



Average = 51 secs
One additional lap time = 39 secs, which reduces average by 2 secs (from 51 secs to 49 secs).
So the -12 secs (39 is 12 less than 51) from the average is distributed among 6 laps to give -2 to each)

So there must have been 5 laps at the average of 51 secs.

Answer (B)

This method is discussed in detail here: https://www.veritasprep.com/blog/2012/0 ... etic-mean/
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Re: Steven has run a certain number of laps around a track at an average (  [#permalink]

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New post 14 Feb 2019, 09:39
SajjadAhmad wrote:
Steven has run a certain number of laps around a track at an average (arithmetic mean) time per lap of 51 seconds. If he runs one additional lap in 39 seconds and reduces his average time per lap to 49 seconds, how many laps did he run at an average time per lap of 51 seconds?

A. 2
B. 5
C. 6
D. 10
E. 12



given
sum of laps / total laps = 51

now
sum of laps + 39 / total laps + 1 = 49

so
we can say sum of laps = 51 * total laps

so
51 * total laps + 39 = 49 * total laps + 49
total laps = 10/2 ~ 5
IMO B
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Re: Steven has run a certain number of laps around a track at an average (   [#permalink] 14 Feb 2019, 09:39
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