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.
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