IMO C.

From st2, 9x < 16y - 1 or, 16y>9x+1 (here x & y can't be 0)

Combining we get, x & y both as +ve or y +ve/x -ve.

E.g. lets say y=-2, x=-3. So x<y but st2 doesnt hold -32>-26.

ankurmit wrote:If x and y are integers, is 3x < 4y?

(1) x < y
(2) 9x < 16y - 1


I could not solve 2nd equation

Stmnt 1 : x < y
for all positive values equation holds true
Now y = -5 x = -6
-18 < -20 Hence not suff

Stmnt 2 : 16y > 9x + 1
y = 2 and x= 1 32 > 10
y = -2 x = -1 -32 > -8 Insuff

combining y > x
and 16y > 9x + 1
to satisfy 2nd condition Y shud be +ve x can be + or -ve
Hence C

OA : B

as in 1st intergers can be negative also.

Statement 1 doesn't provide enough information. If x = -4 and y = -3, 3x = 4y, and the answer is No. On the other hand, if x = - 1,000,000 and y = 1,000,000, the answer is Yes. Insufficient. (Notice that I chose extremes when plugging in numbers here. When plugging in questions, you want to invalidate your original relationship, and the best way to do this is to look at extreme cases).

Statement 2 is tricky. When I see 9x and 16y, I ask myself how I could rephrase the original question to look like the inequality in Statement 2. We can re-write the original question as: Is 9x < 16y? Statement 2 tells us that even if you take 1 away from 16y, it will still be greater than 9x. Thus 16y itself must be greater than 9x. Statment 2 is sufficient. The answer is B.

ankurmit wrote:If x and y are integers, is 3x < 4y?

(1) x < y
(2) 9x < 16y - 1


I could not solve 2nd equation