Is
x>y^2?
(1)
x>y+5 -->
x-y>5. Clearly insufficient, for example: if
x=1 and
y=-10 then the answer is NO, but if
x=10 and
y=1 then the answer is YES. Two different answers, hence not sufficient.
(2)
x^2-y^2=0 -->
(x-y)(x+y)=0 --> so either
x-y=0 or
x+y=0. Also insufficient: if
x=1 and
y=1, then answer is NO, buy if
x=\frac{1}{2} and
y=\frac{1}{2}, then the answer is YES. Two different answers, hence not sufficient.
(1)+(2) As from (1)
x-y>5\neq{0}, then from (2) must be true that
x+y=0 --> so
x=-y --> substitute
x in (1) -->
-y-y>5 -->
y<-\frac{5}{2}<0, as
x=-y, then
x>\frac{5}{2}>0, so
y^2 (or which is the same
x^2) will always be more than
x, thus the answer to the question "Is
x>y^2" is NO. Sufficient.
To elaborate more as
x=-y>0, the only chance for
x>y^2 to hold true (or which is the same for
x>x^2 to hold true) would be if
x is fraction (
0<x<1). For example if
x=\frac{1}{2} and
y=-\frac{1}{2} then
x=\frac{1}{2}>y^2=\frac{1}{4}. But the fact that
x>\frac{5}{2}>0 rules out this option.
Answer: C.
Hope it's clear.
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-y-2-1-x-y-5-2-x-2-y-97142.html
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