Is x > y^2? (1) x > y + 5 (2) x^2 - y^2 = 0

Is \(x \gt y^2\)?

(1) \(x \gt y+5\).

Rewrite as \(x-y \gt 5\). This is 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. We have two different answers, so this is not sufficient.

(2) \(x^2-y^2=0\).

Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) or \(x+y=0\). This is also insufficient: if \(x=1\) and \(y=1\), then the answer is NO, but if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. We have two different answers, so this is not sufficient.

(1)+(2) Since we know from (1) that \(x-y \gt 5\), we can conclude that \(x-y \neq 0\). Therefore, based on (2), we must have \(x+y=0\). This means that \(y=-x\), and we can rewrite the question as: is \(x \gt x^2\)?

Now, let's substitute \(y=-x\) in (1): \(x \gt -x + 5\). From this inequality, we can infer that \(x \gt \frac{5}{2}\). Since \(x \gt \frac{5}{2}\), it implies that \(x^2 \gt x\) (because for any \(x\) greater than 1, \(x^2 \gt x\)). Therefore, the answer to the question of whether \(x \gt y^2\) is NO. Sufficient.


Answer: C

Hope it's clear.
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Hi, there! I'm happy to help with this.

The question: is x > y^2

Statement #1: x > y + 5

This doesn't necessarily tell us anything. If y = 1, and x = 7, then x > y^2, but if y = -6 and x = 0, then x < y^2. But itself, Statement #1 is not sufficient.

Statement #2: x^2 - y^2 = 0

This means that x^2 = y^2, which means that x = ±y. Same absolute value, but both could be positive, both could be negative, or either one could be positive and the other negative. We know that y^2 will be positive, but the x can be positive or negative, so by itself, Statement #2 is insufficient.

Combined
Now, we know that x^2 - y^2 = 0 ---> x = ±y, AND we know that x > y + 5. This leads immediate to a few conclusions
(a) x is positive and y is negative --- that's the only way they could have the same absolute value, but with x bigger than y + 5
(b) x and y must have an absolute value greater than 2.5, so that the different between positive x and negative y is more than 5

So we are comparing a positive number x, greater than 2.5, to the square of the negative number with the same absolute value. Of course, x^2 and y^2 are equal, so the question really boils down to: given that x > 2.5, is x > x^2?

For all x greater than one, the square of x is greater than x. That's because, squaring is multiplying a number by itself, and when you multiply anything by a number greater than one, it gets bigger.

Thus, if x > 2.5, when we square it, it will get bigger. Therefore, x^2 = y^2 > x for all values of x > 2.5.

Thus, combined, the statements are sufficient together. Answer = C

Does that make sense? Please let me know if you have any questions.

Mike
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Statement 1:
@y= 1 x= 7
7 > 1²
@y=20 x= 27
27 < 20²
Not Sufficient

Statement 2:
X² - y² = 0
I.e. IxI = IyI
I.e. for integer values of x and y
X < y²
For 0<x<1
x> y²
Not Sufficient

Combining
Now we can only take values of x out of the range 0 and 1

For all values of x and y now,
X is not greater than y^2
hence

Sufficient

C

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