During a 6·day local trade show, the least number of people

The question says that the least number of people registered in a single day was 80. So do we have to consider 80 to be the least value on only one day or all 6 days ?
The solution given in OG takes it for one day, 80+a+b+c+d+e/6.

Please guide me. I think I am a little lost in the wording of the question.

It actually doesn't matter! All you know is that the lowest number of people to ever register was 80. So, there was at least one day where exactly 80 people registered, and there were no days where fewer than 80 people registered. The other days could have had 80 registrants, or any number of registrants greater than 80. That's why the other five days are represented as variables - you can't really say much about them, until you get to the statements.

You could argue about whether 'least' implies that there was exactly one day with 80 registrants, but it isn't important to the solution.

And if they wanted to specifically say that exactly 80 people registered on every day, they'd have to state that clearly. Since they don't, you can't make that assumption.

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!

I chose C at first and then when I redid the question after knowing the answer now I understand.
The least day = 80
Four greatest days = 400 with a mean of 100

Therefore, the one day that left (let's call it x) must vary between 80 and 100 (if it's more than 100 than it must be one of those 4 greatest days, while we are now looking for the second least day).

In scenario when x = 80, you see the result in other comments in this topic
In scenario when x = 100, the mean of 6 days is (400+100+80)/6 = 580/6 = 8x (I stopped calculating when I see the tenth number being 8).

Then in all scenario, the mean is always less than 90.

Hope this helps