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DS: Inequalities

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GMAT Kolaveri Really wants to Beat The GMAT!
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DS: Inequalities Post Sun Apr 29, 2012 6:56 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Is p a negative number?

    (1) p^3(1 - p^2) < 0

    (2) p^2 - 1 < 0

    Source: grockit. OA: C

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    aneesh.kg GMAT Destroyer!
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    Post Sun Apr 29, 2012 7:45 pm
    This question can be solved very easily using 'Critical Points Method'. I have explained the method in this post:
    http://www.beatthegmat.com/critical-points-method-for-inequalities-explained-useful-t110450.html

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    shantanu86 Rising GMAT Star Default Avatar
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    Post Sun Apr 29, 2012 9:28 pm
    GMAT Kolaveri wrote:
    Is p a negative number?

    (1) p^3(1 - p^2) < 0

    (2) p^2 - 1 < 0

    Source: grockit. OA: C
    IMO its [C]

    Lets analyze-

    Quote:
    (1) p^3(1-p^2)<0
    => either p^3 or 1-p^2 is negative and the other one is positive.

    case 1. p^3 <0 & (1-p^2)>0
    => -1< p < 0

    case 2. p^3 >0 & (1-p^2)<0
    => p > 1

    Thus combining both cases its not guaranteed that p< 0. Hence Insufficient

    Quote:
    (2) p^2 - 1 < 0
    => p^2 < 1
    => -1<p < 1

    Thus insufficient

    Quote:
    (1) & (2) together
    p^3(1-p^2)<0 .. i
    p^2 - 1 < 0 .. ii
    from .. ii
    1-p^2 >0
    substituting in i
    p^3 < 0
    => p<0

    Hence, sufficient

    Thus answer is [C]

    Hope it helps!!

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    Post Wed May 09, 2012 12:25 am
    I tried with p = 0.5 and -0.5 So, for -1 < p < 1 , the answers can be + ve or -ve.. Please explain.
    In statement 1:

    p^3(1-p)(1+p)<0

    readjusting the equations,

    p^3(p-1)(1+p)>0

    aneesh.kg GMAT Destroyer!
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    Post Wed May 09, 2012 12:42 am
    Great.

    Now
    Identify the critical points.
    Mark them on the number line.
    Draw the curve.
    Select the required region.

    If there's a doubt, read the post on Critical Points again: http://www.beatthegmat.com/critical-points-method-for-inequalities-explained-useful-t110450.html
    I would like to see you attempt it properly before giving away the solution.

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    Post Wed May 09, 2012 4:12 am
    GMAT Kolaveri wrote:
    Is p a negative number?

    (1) p^3(1 - p^2) < 0

    (2) p^2 - 1 < 0

    Source: grockit. OA: C
    The critical points method is great.

    For some test-takers, plugging in numbers might be easier and more efficient.
    Typically, we need to consider the following types of numbers:
    Zero
    One (positive and negative)
    Integers (greater than positive 1 and less than negative 1)
    Fractions (positive and negative)

    Statement 1: p^3(1 - p^2) < 0
    p=0 and p=±1 make the left-hand side equal to 0, so none of them works here.

    p=-2:
    (-2)³(1-(-2)²) < 0
    (neg)(neg) < 0.
    Doesn't work, implying that p cannot be less than -1.

    p=-1/2:
    (-1/2)³(1-(-1/2)²) < 0
    (neg)(pos) < 0.
    This works, implying that p can be a negative fraction.

    p=2:
    (2)³(1-2²) < 0
    (pos)(neg) < 0.
    This works, implying that p can be greater than 1.

    p=1/2:
    (1/2)³(1-(1/2)²) < 0
    (pos)(pos) < 0.
    Doesn't work, implying that p cannot be a positive fraction.

    Since p=-1/2 and p=2 both work, INSUFFICIENT.

    Statement 2, rephrased: p²<1.
    Thus, p can be a negative fraction, 0, or a positive fraction.
    INSUFFICIENT.

    Statements 1 and 2 combined:
    The only type of value that satisfies both statements is a negative fraction such as p=-1/2.
    SUFFICIENT.

    The correct answer is C.

    An alternate way to combine the statements:
    Since p²-1<0 (statement 2), 1-p²>0.
    Thus, statement 1 becomes:
    p³(positive) < 0.
    p³<0, implying that p is a negative number.
    SUFFICIENT.

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    krishna239455 Really wants to Beat The GMAT! Default Avatar
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    Post Wed May 09, 2012 9:50 am
    Dear Aneesh
    I tried to solve this problem with the trick mentioned by you.
    Statement 1: Critical points considered are 0,1,-1.
    Two inequalities satisfying the function are P>1 and -1<P<0
    Hence P can be both +ve and -ve , Not suffeceint.

    Statement 2: Critical points considered are 1 and -1.
    Only One inequality satisfying the function is -1<P<1
    Again P can be both +ve and -ve, Not suffeceint

    After combining both the statements -1<P<0 and -1<P<1, -1<P<0 is more restrictive than other and hence can help us to conclude that P is -ve.

    But what about P>1 that we got in statement 1

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    Post Wed May 09, 2012 11:23 am
    Good job Krishna!

    When we COMBINE two statements, we look for those values for which both Statement(1) and Statement(2) hold true. In other words, we look for the COMMON solution.

    P > 1 is a solution from Statement(1) but not from Statement(2), so we do not include it in the common solution. When you plot these values on the number line, you will see that -1 < P < 0, as you mentioned, is the only common solution to the two statements.

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    krishna239455 Really wants to Beat The GMAT! Default Avatar
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    Post Thu May 10, 2012 5:30 am
    Thanks Aneesh !!!

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