pingala wrote:

I messed up on how i interpreted statement two, the fact that

x=y=1 would create the right condition for statement 2 to be correct, also means that when answering the prompt,

x>y, we get the answer

NO; trying another set of numbers- x=2 and y=1- where statement two meets the condition,

2>1/3, means that the answer to the question

x>y is

YES.

Bunuel explains this but I didn't quite understand the process because I wasn't registering the prompt and statement criteria clearly.

Hi

pingala,

here is how i would approach this question

we are asked that IS x>y

so is x-y>0

what could be the possible senarios

x>0, y>0 and |x|>|y|

x<0, y<0 and |y|>|x|

x>0, y<0

x<0, y>0 and |y|>|x|

in above all scenarios we would get the answer to our question.

Now statement A:

It says y>0. It could be x>0 or x<0, but still we don't about |x| and |y|

So insufficient.

Now Statement B:

3x>2y

or we can have

\(x>\frac{2}{3} y\)

if we read along it says x is greater than two-thirds of y.

We will have two senarios

y>0 even then we can't conclude that x> y. why say y>1 then x can be \(>\frac{2}{3}\) . So its possible that the range of of x is \(\frac{2}{3}<x<\infty\) and range of y would be \(1<y< \infty\) . Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now if y<0 even then we can't conclude that x> y. Then x can be \(>-\frac{2}{3}\) . So its possible that the range of of x is \(-\frac{2}{3}<x<\infty\) and range of y would be \(-\infty<y<0\). Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now even if we combine both the statements still we don't have any new information to conclusively answer if x>y

hence option E

let me know if this helps

Probus