ksatish wrote:Is x > y?
1. x^2 > y^2+1.
2. The product of xy is negative.
you can also approach the problem like this:
when you look at statement 1, the important part is that x^2 is bigger than y^2. (the +1 is of no consequence, at least in this problem.)
what does this mean?
remember that squares of positives are positive, but that squares of negatives are also positive. so, if x^2 is bigger than y^2, then we know that the "size" of x is more than the "size" of y -- but we don't know the signs.
this can be written as "|x| > |y|", but to me, at least, that's not a particularly meaningful statement.
here's how i like to put it:
if we know that x^2 > y^2, then
x = ±BIG
y = ±small (or 0, if the problem allows).
so... now that we have this, let's kill this problem.
(1)
if x = -BIG, then either value of y = ±small will be greater than x.
if x = +BIG, then that's greater than either value of y = ±small.
not sufficient.
(2)
in this case, one of the following is true:
x = positive, y = negative
or
x = negative, y = positive
the first is always a yes, the second always a no. not sufficient.
(together)
here there are still two possibilities:
x = -BIG and y = +small
or
x = +BIG and y = -small
the first is a no, the second a yes. not sufficient.
so, (e).
--
mitch's method of
testing numbers is also valuable, although you shouldn't memorize a specific set of numbers to test -- too limiting. (for instance, if i change y^2 + 1 to y^2 + 10, then the essential character of the problem doesn't change, but none of those numbers will work anymore.)
