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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If x and y are positive, is x > y^2? tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## If x and y are positive, is x > y^2? $$If\ \ x\ \ and\ \ y\ \ are\ \ positive,\ \ is\ \ x>y^2?$$ $$\left(1\right)\ \ y\ >\ x^2$$ $$\left(2\right)\ \ y=x+1$$ The OA is B. I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 LUANDATO wrote: $$If\ \ x\ \ and\ \ y\ \ are\ \ positive,\ \ is\ \ x>y^2?$$ $$\left(1\right)\ \ y\ >\ x^2$$ $$\left(2\right)\ \ y=x+1$$ The OA is B. I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced. Statement 1: Pick some easy numbers Case 1: x = 1, y = 2. In this case, 1 is not greater than 2^2, so we have a NO Case 2: x = 1/2, y = 1/2. (These numbers are eligible to test because 1/2 > (1/2)^2, and therefore y > x^2.) In this case, 1/2, or x, is greater than (1/2)^2, or y^2, so we have a YES. Because we can answer NO or YES to the original question, this statement alone is not sufficient. (Put another way, anytime you have a fraction between 0 and 1, when you square that fraction, the result will be smaller than the original value.) Statement 2: Pick more number. Case 1 x = 1, y = 2. In this case, 1 is not greater than 2^2, so we have a N0. Case 2: x = 2, y = 3; 2 is not greater than 3^2, so we have a NO. Case 3: x = 1/2, y = 3/2. 1/2 is not greater than (3/2)^2, so we have a. NO No matter what we pick, the answer to the original question is NO, so this statement alone is sufficient. The answer is B _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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LUANDATO wrote:
$$If\ \ x\ \ and\ \ y\ \ are\ \ positive,\ \ is\ \ x>y^2?$$

$$\left(1\right)\ \ y\ >\ x^2$$
$$\left(2\right)\ \ y=x+1$$

The OA is B.

I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
We can also address statement 2 algebraically. If y = x + 1, we can rephrase the initial questions from "Is x > y^2 to Is x > (x +1)^2?
Is x > (x +1)^2?
Is x > x^2 + 2x + 1?
Is -x > x^2 + 1?

Notice that the value on the left side of the equation must be negative, as x is positive. Moreover, the value on the right side of the equation must be positive, as x^2 must be positive and we're adding 1 to it. Put another way, the answer to "Is a negative > a positive?" is always NO. Therefore this statement alone is sufficient to answer the question.

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