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Kaplan CAT 4 question

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duartemarchand Just gettin' started! Default Avatar
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Kaplan CAT 4 question Post Tue Mar 20, 2012 4:40 pm
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    Hi guys

    While reviewing a CAT exam by Kaplan I faced this question:

    What is the value of (x + y)2?

    (1) x2 - xy = 28 and 3xy + y2 = 72.

    (2) (x + y)4 = 10,000

    Statement 1 is clearly enough to solve the issue, however in statement 2 kaplan explanation states that because (x + y)4 is the square of (x + y)2 than the square root of 10,000 will be equal to (x + y)2.

    My question is: shouldn't we consider that both 100 and -100 are possible answers to statemenmt 2. Is there anything I'm missing?

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    Post Tue Mar 20, 2012 6:02 pm
    duartemarchand wrote:
    Hi guys

    While reviewing a CAT exam by Kaplan I faced this question:

    What is the value of (x + y)2?

    (1) x2 - xy = 28 and 3xy + y2 = 72.

    (2) (x + y)4 = 10,000

    Statement 1 is clearly enough to solve the issue, however in statement 2 kaplan explanation states that because (x + y)4 is the square of (x + y)2 than the square root of 10,000 will be equal to (x + y)2.

    My question is: shouldn't we consider that both 100 and -100 are possible answers to statemenmt 2. Is there anything I'm missing?
    Even though (x + y)^4 = 10,000, we cannot say that (x + y)^2 = 100 or -100
    The reason for this is that (x+y)^2 must be greater than or equal to zero (since we have something squared)
    In other words (x + y)^2 cannot equal -100, which means (x + y)^2 must equal 100

    Cheers,
    Brent

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    Post Tue Mar 20, 2012 6:03 pm
    duartemarchand wrote:
    Hi guys

    While reviewing a CAT exam by Kaplan I faced this question:

    What is the value of (x + y)2?

    (1) x2 - xy = 28 and 3xy + y2 = 72.

    (2) (x + y)4 = 10,000

    Statement 1 is clearly enough to solve the issue, however in statement 2 kaplan explanation states that because (x + y)4 is the square of (x + y)2 than the square root of 10,000 will be equal to (x + y)2.

    My question is: shouldn't we consider that both 100 and -100 are possible answers to statemenmt 2. Is there anything I'm missing?
    Statement 1: x² - xy = 28, 3xy + y² = 72.
    Adding the two equations, we get
    (x² - xy) + (3xy + y²) = 28+72.
    x² + 2xy + y² = 100
    (x+y)² = 100.
    SUFFICIENT.

    Statement 2: (x+y)⁴ = 10,000
    Thus, (x+y)² = 100.
    SUFFICIENT.

    The correct answer is D.

    It is not possible that (x+y)² = -100.
    The square of a value can never be negative.

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    Brent@GMATPrepNow GMAT Instructor
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    Post Tue Mar 20, 2012 6:10 pm
    duartemarchand wrote:
    What is the value of (x + y)^2?

    (1) x^2 - xy = 28 and 3xy + y^2 = 72.

    (2) (x + y)^4 = 10,000
    Let's solve this one.

    Statement 1
    We are given two equations:
    x^2 - xy = 28
    y^2 + 3xy = 72

    If we add the left sides and right sides, we get: x^2 + 2xy + y^2 = 100
    Now factor to get: (x+y)^2 = 100
    So, statement 1 provides SUFFICIENT information to answer the target question.

    Statement 2: (x + y)^4 = 10,000
    Normally, we might conclude that, if (x+y)^4 = 10,000, then (x+y)^2 = -100 or 100
    However, we can eliminate the possibility that (x+y)^2 = -100, since any number to the power of 2 is always greater than or equal to zero.
    So, it MUST be the case that (x+y)^2 = 100
    So, statement 2 provides SUFFICIENT information to answer the target question.

    The correct answer is D.

    Cheers,
    Brent

    _________________
    Brent Hanneson - GMAT Prep Now instructor
    - Check out GMAT Prep Now’s online course at http://www.gmatprepnow.com/
    - Use our video course in conjunction with Beat The GMAT's free 60-Day Study Guide
    - Watch hours of free videos on DS, RC and AWA
    - Our top 3 free videos:
    1) The Double Matrix method
    2) Calculating combinations in your head
    3) Managing your time on the GMAT

    Study Smart! Use Beat The GMAT’s FREE 60-Day Study Guide in conjunction with GMAT Prep Now’s video course and reach your target score in 2 months! With two money-back guarantees, you can try us out risk-free.
    duartemarchand Just gettin' started! Default Avatar
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    Post Wed Mar 21, 2012 5:08 am
    Thanks for both for you answers.

    Cheers

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