Suppose 6 days are represented by {x1, x2, x3, x4, x5, x6} each >=80

Stmt1: For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.

Let x3+x4+x5+x6/4 = 100.

x3+x4+x5+x6=400

Lets suppose x1 and x2 to be minimum i.e 80.

So, x1+x2+x3+x4+x5+x6/6 = 80+80+400/6 = 560/6 = 93.3 > 90

Now, Lets suppose x1 and x2 to be more than minimum, 90

So, x1+x2+x3+x4+x5+x6/6 = 90+90+400/6 = 580/6 = 96.6 > 90 Sufficient.

Stmt2: For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.

Let x1+x2+x3/3 = 85

x1+x2+x3=255

Lets suppose x4,x5 and x6 to be minimum. i.e 80

x1+x2+x3+x4+x5+x6/6= 255+80+80+80/6=495/6=82.5 < 90

Lets suppose x4,x5 and x6 to be more than minimum, 100

x1+x2+x3+x4+x5+x6/6= 255+100+100+100/6=/6=92.5 < 90

Hence two answers. Insufficient.

OA A.

_________________

My dad once said to me: Son, nothing succeeds like success.