Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

GMAT TIGER, so which answer will you opt for? Since x is +ve and y is -ve and both have the same numerical value, x can be > or < than y^2, depending on whether the numerical value is <or> than 1. That's why I think the answer should be (E).

GMAT TIGER, so which answer will you opt for? Since x is +ve and y is -ve and both have the same numerical value, x can be > or < than y^2, depending on whether the numerical value is <or> than 1. That's why I think the answer should be (E).

Quote:

Is x > y^2?

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

x^2 - y^2 = 0 is possible only when either x = y or x = -y but since x > y+5, x must be +ve and therefore y is -ve.

x > y+5
x - y > 5
- y - y > 5
y < -2.5

suppose, y = - 3, x = 6, which is grater than -3+5 = 2.

2) insuff. this statement means (x+y) (x-y)=0
therefore:

x=y or
x=-y

statement 2 alone would be true if we know at least that the variables are integers, however, we weren't told that they are integers. Therefore they could be fractions, which can mess up the whole logic and give different results.

Both statement together

worst case scenario:
scenario a: if y=1/2, then x is at least 6
6>1/4 true

scenario b: if y=-1/2, then x is at least 5
5> 1/4 true

Firstly, looking at the statement: y^2 will always be positive, unless y = 0. Therefore, for x > y^2, one of the conditions is that x must be > 0.

Looking at the statements:

1) INSUFFICIENT: we cannot determine from this equation whether x is > 0. For example: -5 > -11 + 5 AND 5 > -11 + 5

2) x^2 - y^2 = 0, therefore (x-y)(x+y) = 0, therefore x = -y OR + y

INSUFFICIENT: we cannot determine from this equation whether x is > 0. All we know is that x = y OR x is equal to y * -1

1) & 2) Together:

Subsitute x = y into the equation in Statement 1: y > y + 5 This cannot be true. Therefore, x is NOT equal to y! Therefore x = -y
Substitute x = -y into the equation in Statement 1:

-y > y + 5
-2y > 5
-y > 5/2.... x > 5/2
y < -5/2 Because y is <1 and x = -y, as x gets bigger, y gets smaller. Therefore y^2 will always be greater than x. The answer to the question "Is x > y^2" is NO. Therefore it's C.

OE
when we combine the two statements, we see that the absolute values of x and y are equal and that x > y , implying that x is positive and y is negative and neither is equal to zero. Therefore, x > y2 is true.

(1) or (2) alone is not sufficient
Together we have lxl = lyl and x>y+5
so x= -y and y<-2.5
thus x>2.5 and x<x2 = y2
The answer is with (1)(2) together we have x<y2