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xcise_science
- Master | Next Rank: 500 Posts
- Posts: 106
- Joined: Tue Oct 09, 2007 12:25 pm
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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
