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    Within my grasp! nazar's Avatar
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    Interesting DS_Please Explain!!

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    If x and y are real numbers, is (x^2 -y^2)^5>0?
    (1) (x+y)^2 =25
    (2) x-y>0

    OA: Later

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    An Urch Guru Pundit Swami Sage Makumajon's Avatar
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    LHS = (x+y)^5*(x-y)^5

    (1) x+y=+5 or -5. And do not know which one is greater: x or y? Insufficient.
    (2) x>y. Do not know whether they are (+, +) or (-,-). Insufficient.

    Combining: Insufficient. Suppose x=3, y=2. LHS>0.
    Suppose x=-2, y = -3. LHS<0.

    E

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    (x^2 -y^2)^5 = ((x-y)(x+y))^5=(x-y)^5*(x+y)^5

    Statement 1 Alone not enough
    Statement 2 Alone not enough
    Togetherx-y)^5>0 Also (x+y)^4>0 But we do not know if the remaining term (x+y)>0 since x+y can +0r -5 derving from (1).
    So the answer is E.

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    Eager! rupali41's Avatar
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    E is the answer

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    GRE-GMAT Instructor Brent Hanneson's Avatar
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    Quote Originally Posted by nazar View Post
    If x and y are real numbers, is (x^2 -y^2)^5>0?
    (1) (x+y)^2 = 25
    (2) x-y>0
    r
    Target question:Is (x^2 - y^2)^5 > 0?

    IMPORTANT: (x^2 - y^2)^5 will be positive (i.e., > 0) if and only if (x^2 - y^2) > 0.
    So, we can REPHRASE the target question....

    REPHRASED target question:Is x^2 - y^2 > 0?

    Statement 1: (x+y)^2 = 25
    This tells us that EITHER x+y = 5 or x+y = -5
    There are several possible values for x and y that satisfy this condition. Here are two:
    case a: x = 3 and y = 2, in which case x^2 - y^2 > 0
    case b: x = 0 and y = -5, in which case x^2 - y^2 < 0
    Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

    Statement 2: x - y > 0
    In other words, x is GREATER than y
    There are several possible values for x and y that satisfy this condition. Here are two:
    case a
    : x = 3 and y = 2, in which case x^2 - y^2 > 0
    case b: x = 0 and y = -5, in which case x^2 - y^2 < 0
    Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

    Statements 1 and 2 combined
    There are still several possible values for x and y that satisfy BOTH conditions. Here are two:
    case a
    : x = 3 and y = 2, in which case x^2 - y^2 > 0
    case b: x = 0 and y = -5, in which case x^2 - y^2 < 0
    Since we cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT

    Answer = E

    Cheers,
    Brent

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