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14 Feb 2019, 02:31
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
84% (01:46) correct
16%
(02:08)
wrong
based on 58
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
A. 2
B. 5
C. 6
D. 10
E. 12
14 Feb 2019, 04:29
Kudos
Bookmarks
SajjadAhmad
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
14 Feb 2019, 05:58
Kudos
Bookmarks
SajjadAhmad
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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... etic-mean/
