Averages | OG 12

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Averages | OG 12

by [email protected] » Tue Oct 02, 2012 12:37 pm
During a 6-day local trade show, the least number of
people registered in a single day was 80. Was the
average (arithmetic mean) number of people
registered per day for the 6 days greater than 90 ?

(1) For the 4 days with the greatest number of
people registered, the average (arithmetic mean)
number registered per day was 100.
(2) For the 3 days with the smallest number of
people registered, the average (arithmetic mean)
number registered per day was 85.

Not able to understand the OG Explanation for 2nd statement.

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by GMATGuruNY » Tue Oct 02, 2012 7:11 pm
During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90?

(1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.

(2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.
Statement 1: For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.
For 4 of the days, the average per day = 100.
For the other 2 days, the least possible average per day = 80.
Since there are MORE days with an average of 100 per day than with an average of 80 per day, the average for all 6 days must be CLOSER to 100 than to 80.
Thus, the average for all 6 days must be greater than 90.
SUFFICIENT.

Statement 2: For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.
If there are 3 days with an average of 85 per day and 3 days with an average of 86 per day, then the average for all 6 days will be less than 90.
If there are 3 days with an average of 85 per day and 3 days with an average of 1000 per day, then the average for all 6 days will be greater than 90.
INSUFFICIENT.

The correct answer is A.
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