Bunuel
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 ?
The question basically asks whether more than 6*90=540 people registered during the show.
(1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100 --> in these 4 days total of 4*100=400 people registered. Since the minimum number of people registered on the other two days is 80*2=160, then the total number of people registered is at least 400+160=560. Sufficient.
(2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85. Clearly insufficient. Consider: {85, 85, 85, 86, 86, 86} and {85, 85, 85, 1,000, 1,000, 1,000}.
Answer: A.
Expanding on Statement 2:
The average of number of people registered per day = 85, but the question stems says the minimum number in any single day was 80. So the least value for one out of the 3 days can be 80.
In that case : Average of 3 days = 3x85 = 255, so the sum of number of people in the 2 remaining days = 255-80 = 175.
This remaining number of 175 people for Day 2 and 3 can be split in 2 ways:
Min possible number of people registered for Day 2 = 80, so Day 3= 95
Max possible number of people registered for Day 2 = (175/2) rounded down to 87, so Day 3 = 88
So the range of possible cases for the 3 days with the smallest number of people registered becomes:
{80, 80, 95} and {80, 87, 88}
Remaining Day 4 to Day 6 will have at least as many people as Day 3. So the minimum possible number of people registered for Day 4, 5 and 6 will be at least 88.
(80, 87, 88, 88, 88, 88}
This gives a total of = 519. So the minimum possible number of people registered for all 6 days >= 519
As
Bunuel rightly pointed out, the question basically asks whether more than 540 people registered during the show.
Statement 2 is not sufficient!