Gurpinder wrote:Stmt 1: Insuff
All we know is that 8 copies were sold on Thursday. We know nothing about the rest of the 82 copies.
Stmt 2: Insuff
Again, now all we know is that the greatest number of copies sold were 38. We know nothing about the rest!
Together:
8 on Thurs
38 on Sat - LARGEST
So now we have 46-90=44 copies left and 5 days in the week left. The greatest copies sold in a day CANNOT exceed 38. And the copies sold on any 2 given days is to be different! And friday must have the 2nd largest copies sold!
So lets distribute:
fri --11
sun --10
mon --9
tues--8
wed--6
in this case, store S sold exactly 11 copies.
lets try a different distribution:
fri --10
sun --9
mon --8
tues--7
wed--10 <<<< doesent work.
so the only way this works is if we choose the last distribution.
Therefore the answer is (C). And with that we know that store S did not sell more copies than 11
The most important part of this question is to assume five different non-negative integers each less than 11, which could add up to 90 - (38 + 11) = 41 or more. We cannot be repetitive at the same time. If we can find such a set of 5, then Friday sales could have been 11 too; but if we cannot find such a set of 5, then Friday sales got to be more than 11.
So let's try our max, take 10 + 9 + 8 + 7 + 6 = 40, just falling short of 41, so yes, Friday sales got to be more than 11.
[spoiler]
B[/spoiler]
