rahul.s wrote:If x and y are distinct integers, is (x + 7y)^3 > 0?

(1) x^2 > 49y^2
(2) x < 7y

OA: C
Source: Knewton

this is a pretty cool problem.

REPHRASE THE QUESTION
cubes don't affect positive/negative, so take the cube root:
is x + 7y > 0?
that's your question.

statement (1)
this is equivalent to |x| > |7y|. in other words:
x = +/- BIG NUMBER
7y = +/- SMALL NUMBER
that's all we know.
in this situation, x + 7y could have either sign, so, insufficient.

statement (2)
rearranges to x - 7y < 0.
the sign of x - 7y is not going to help us find the sign of x + 7y.
insufficient.

together
if x < 7y, then 7y must be POSITIVE BIG NUMBER. (see the possibilities outlined in #1)
this means that x + 7y is POSITIVE (since x, even if it is negative, is not big enough to turn this sum negative overall.)
therefore, sufficient.

(c).