18

The approach is to try and pick numbers <11> 11

From stem Total = 90

Sat = Max
Friday = second max


From stmt 1

M 2
T 4
W 7
T 8
F 20
S 30
S 19

Cud ne one possible combination ..where F > 11

M 5
T 6
W 7
T 8
F 10
S 50
S 4

Cud ne another combination. F <11> 11

Now we will try to take a value < 11

M
T
W
T
F 10
S 38
S


YOu will notice that the value of F needs to be more that 11 for the conditions to be satisfied that F is second max ,a value below that will not satisfy the condition.


All values M T w T S have to be less that 10 if F is 10,this will never be possible

Hope this helps

Hence B


32

From stem n = n(n+5)

From stmt 1 n+ 5 = 20 Hence we know n .Hence sufficient

From stmt 2 a=s/n
i.e 17.5 = 6n+15/6

We can find n ..Hence sufficient
Hence D

Hi,
I'm sorry, I'm not clear on your explanation for Q18 because this scenario
seems to work for F < 11:
M: 2
T: 1
W: 4
TH: 8
F:10
Sa:60
Su: 5
Total: 90

Where the question says: and it sold different numbers of copies on any two of the days.... does that mean a different number each day?

M: 2
T: 1
W: 4
TH: 8
F:10
Sa:60
Su: 5
Total: 90

In this example we need Sat = 38 ..and if that is put then F has to be greater than > 11

As far as this is concerned
Where the question says: and it sold different numbers of copies on any two of the days.... does that mean a different number each day?


All the days should have unique number of books sold
then only will any two days have different number of copies.

Hope this helps