x/y > 2 means x and y are of same sign,

Stmnt 1.

Case x, y are -ve , which clearly satisfies the 3x+2y<18

Consider case , x and y as +ve
, gives x > 2y --(1)
2 > x - y --(2)
Using (1) and (2) gives x + 2 > 2y + x - y ==> y < 2 --(3)
From (2) you get 3x-3y<6 ---(4)
From (3) you get 5y<10 ---(5)
Add eqn (4) and(5) gives 3x+2y<16 , clearly satisfies , so stmnt I is sufficient

Stmnt II

Case , x , y are -ve will satisfy for sure
Consider x and y as +ve gives
Picking numbers like x = 10 and y = 4 , doesnt satisfy 3x+2y<18
Hence it works for -ves for sure but for +ve cases we saw one case where it fails , hence not sufficient

rohan_vus wrote:x/y > 2 means x and y are of same sign,

Stmnt 1.

Case x, y are -ve , which clearly satisfies the 3x+2y<18

Consider case , x and y as +ve
, gives x > 2y --(1)
2 > x - y --(2)
Using (1) and (2) gives x + 2 > 2y + x - y ==> y < 2 --(3)
From (2) you get 3x-3y<6 ---(4)
From (3) you get 5y<10 ---(5)
Add eqn (4) and(5) gives 3x+2y<16 , clearly satisfies , so stmnt I is sufficient

Stmnt II

Case , x , y are -ve will satisfy for sure
Consider x and y as +ve gives
Picking numbers like x = 10 and y = 4 , doesnt satisfy 3x+2y<18
Hence it works for -ves for sure but for +ve cases we saw one case where it fails , hence not sufficient

nice

The idea is to get to know of the max value for each 'x' and 'y' and check is these max values satisfy the inequality 3x+2y<18

We are'nt worried if both are negative, hence assume both 'x' & 'y' > 0

Given x>2y
St1] x-y<2
Add both, we know that y<2; now put his result back to st1 because we can get rid of 'y'
x-y<2
y<2 this means x<4.
Hence the max values of 'x' & 'y' are definitely lower than 4 & 2 repectively
Put this into the testing inequality
3x+2y<18
12+4<18.
Suff.

St2] y-x is less than 2
If we apply the same tech as above we see that it does not help us to identify the max values,
Hence the best thing to do is plug extreme nos to check both outcomes of the testing inequality.