chitrangada:
chitrangada.maitra wrote:
Is 1/3y> 1 ?
If we investigate this carefully, we find that if y is an nonzero integer, 1/3y is never greater than 1. Try y = 2 and y = -2, In both cases 1/3y is less than 1.
I absolutely agree..
chitrangada.maitra wrote:
My question is:
If we plug the fraction 1/9 for Y, we get, 1/3y >1, which renders the first statement insufficient.
However,how can you plug a fractional value here, when the question stem specifies x and y to be non zero
integers?
chitrangada.maitra wrote:
Alternatively, if we simplify and rephrase the question like this:
Is y/3 > y^2, and plug 1/10 (or 1/ 9) for Y, we get, Y/3 > y^2. However, for integar values of Y, we get, Y^2 > y/3
Therefore, statement 1 is insufficient.
How did you arrive at this?You can rephrase the stem as:
xy/x+y>xy
Using Statement 1(x = 2y), we can further reduce it to
2(y^2)/3y>2(y^2) =>Is 1/3y>1? Whatever integer value you substitute will result in a
No as an answer. Hence "A" is sufficient.
I Hope you are clear with this...