Data Sufficiency

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 number registered per day was 100.

2) For the 3 days with the smallest number of people registered, the average number registered per day was 85.

In general when a DS Q asks about average, you should think in terms of total. We want to know whether the sum of all registrations is greater than 90*6 = 540. The answer is A. I go through the question in detail in the full solution below (taken from the GMATFix App).



-Patrick

BlueDragon2010 wrote: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 number registered per day was 100.

2) For the 3 days with the smallest number of people registered, the average number registered per day was 85.

This is a weighted average problem.
If we understand how weighted averages work, almost no math is required.

Statement 1: For the 4 days with the greatest number of people registered, the average number registered per day was 100.
Since the average for more than 1/2 of the 6 days is 100, and the LEAST possible average for the remaining days is 80, the average for all 6 days must be weighted CLOSER TO 100 THAN TO 80.
Thus, the average for all 6 days must be greater than 90 (the halfway point between 80 and 100).
SUFFICIENT.

Statement 2: For the 3 days with the smallest number of people registered, the average number registered per day was 85.
If the average for the other 3 days is 86, then the average for all 6 days will be less than 90.
If the average for the other 3 days is 1000, then the average for all 6 days will be greater than 90.
INSUFFICIENT.

The correct answer is A.