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Absolute Inequalities

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Source: — Data Sufficiency |
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by goyalsau » Fri Dec 17, 2010 10:48 pm
gmatusa2010 wrote:Is X>y?

1) sqrt (X) >y

2) X^3>y
Ist Statement -

sqrt ( x ) > y { In Gmat we don't consider negative value under sqrt so x must be positive }

Now we know x is positive , sqrt of x will always be less than x . Like sqrt ( 4 ) = 2 ,,, 2 is less than 4 ,

so x will definitely be greater than y

Sufficient

IInd Statement

X^ 3 > y

Try some values for x x = 2 , y = 3

x^ 3 > y , but x < y

Let x = 9 , y = 3

Even x^3 > y ,,, But x > y

Not sufficient

Answer should be A
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by Night reader » Fri Dec 17, 2010 11:02 pm
goyalsau wrote:
gmatusa2010 wrote:Is X>y?

1) sqrt (X) >y

2) X^3>y
Ist Statement -

sqrt ( x ) > y { In Gmat we don't consider negative value under sqrt so x must be positive }

Now we know x is positive , sqrt of x will always be less than x . Like sqrt ( 4 ) = 2 ,,, 2 is less than 4 ,

so x will definitely be greater than y

Sufficient
false reasoning, sqrt of x will be less than x if x>1 e.g. => x=0.16 sqrt (0.16) > 0.16 hence sqrt x > x;
statement (1) defines that sqrt x > y Answer A is possible only if x>1, which we don't know
and I agree with B is Not sufficient for x can be greater +ve or -ve

IOM E
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by Rahul@gurome » Fri Dec 17, 2010 11:16 pm
gmatusa2010 wrote:Is X>y?

1) sqrt (X) >y

2) X^3>y
To answer this question we must have a clear idea about the relation of x with x³ and √x. Note the followings:
  • 1. √x ≥ x for (0 ≤ x ≤ 1)
    2. √x ≤ x for (1 ≤ x)
    3. x³ ≤ x for (x ≤ -1) and (0 ≤ x ≤ 1)
    4. x³ ≥ x for (-1 ≤ x ≤ 0) and (1 ≤ x)
Now analyze the problem:

Statement 1: √x > y
For 0 ≤ x ≤ 1, √x ≥ x : In this case x may or may not be greater than y.
For x ≥ 1, √x ≤ x : In this case x is definitely greater than y.

Not sufficient.

Statement 2: x³ > y
For x ≤ -1 and 0 ≤ x ≤ 1, x³ ≤ x : In this case x is definitely greater than y.
For 1 ≤ x and -1 ≤ x ≤ 0, x³ ≥ x : In this case x may or may not be greater than y.

Not sufficient.

1 & 2 Together: Negative values of x are discarded due to √x. Only two regions of concern: 0 ≤ x ≤ 1 and x ≥ 1
  • 1. For 0 ≤ x ≤ 1 : √x ≥ x ≥ x³ > y => x > y
    2. For x ≥ 1 : x³ ≥ x ≥ √x > y => x > y
Sufficient.

The correct answer is C.
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by prachich1987 » Mon Dec 20, 2010 2:50 am
gmatusa2010 wrote:Is X>y?

1) sqrt (X) >y

2) X^3>y
Please post the OA
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by [email protected] » Fri Mar 16, 2012 9:53 pm
Is X>y?

1) sqrt (X) >y

2) X^3>y


A very important question. A must do question actually before you write any test. This gives you an idea of overall inequalities chapter as well as the chapter on Number System...

Understand this question and the next questions on number systems become damn easier. This is like a concept question...
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