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).
Ron has been teaching various standardized tests for 20 years.
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