redng wrote:2. For the past x laps around the track, Steven's average time per lap was 51 seconds. If a lap of 39 seconds would reduce his average time per lap to 49 seconds, what is the value of x?
- (A) 2
(B) 5
(C) 6
(D) 10
(E) 12
answer:
B
I don't even know how to approach this. Thanks!
We can plug in the answers, which represent the x number of laps.
Answer choice C: 6 laps
Time for the first 6 laps = 6*51 = 306.
Average time for all 7 laps = (306+39)/7 = 345/7 = 49 2/7.
Incorrect: the average time must be 49.
In order to reduce the average time just a bit more -- to 49 seconds per lap -- the time spent on the faster lap must be a slightly higher fraction of the total amount of time.
Here is the reasoning: If there are FEWER original laps, then the one additional faster lap will have a GREATER impact on the average.
Thus, the original number of laps must be just a bit less than 6.
The correct answer is
B.
Answer choice B: 5 laps
Time for the first 5 laps = 5*51 = 255.
Average time for all 6 laps = (255+39)/6 = 294/6 = 49.
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