If xy < 3, is x < 1?
(1) y > 3. If x were \(\geq{1}\), then xy would be more than or equal to 3, which would contradict the condition that xy < 3. Thus x must be less than 1. Sufficient.
(2) x < 3. Not sufficient: consider x=y=0 and x=2 and y=0. Not sufficient.
Answer: A.I'm a bit confused after analyzing another similar question with different solution
In st1, why negative values of x are not being considered ? I came up with answer "E", considering negative values of x
However, just when I was convinced that, values of x and y need to be assumed positive in this type of prompt, I met the following question
Is xy < 6?
(1) x < 3 and y < 2.
(2) 1/2 < x < 2/3 and y^2 < 64.
Data Sufficiency, Question: 68, Page: 157, OG quant review 2nd edition
SOLUTION
Is xy < 6?
(1) x < 3 and y < 2 --> now, if both x and y are equal to zero then xy=0<6 and the answer will be YES but if both x and y are small enough negative numbers, for example -10 and -10 then xy=100>6 and the answer will be NO. Not sufficient.
(2) \frac{1}{2}<x<\frac{2}{3} and y^2<64, which is equivalent to -8<y<8 --> even if we take the boundary values of x and y to maximize their product we'll get: xy=\frac{2}{3}*8\approx{5.3}<6, so the answer to the question "is xy<6?" will always be YES. Sufficient.
Answer: B. Here, both the positive and negative values of x and y have been considered.
Bunuel can you please explain the contradiction ?