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Doubt in Data Sufficiency

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anveshan_reddy_k
 Post subject: Doubt in Data Sufficiency
 Post Posted: Sun Dec 04, 2011 9:58 am 
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Students


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If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x
(2) |x| > x


I feel 2nd stmt alone should be sufficient since from second statement, we knew x is negative. So if it is negative, it should be less than 1..But answer says both are required....Can anybody help here..
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stud.jatt
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Sun Dec 04, 2011 12:16 pm 
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Well you have to consider that the equation in the question includes that modulus function. so the question being asked by |x| < 1 isn't just whether x < 1 but whether -1 < x < 1 (i.e. does x lie between -1 and 1), because only in that case will |x| < 1

Statement 1 gives us two ranges for x
-1 < x < 0 OR x > 0

Statement 2 gives us that x < 0 but we don't know if x lies between -1 and zero

Upon combining the two we have -1 < x < 0 a range that lies between -1 and 1 and hence the answer is C
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jaspreetp
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Fri Dec 09, 2011 8:36 am 
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Hi stud.jatt

Can you please explain how you got the two ranges for Statement 1 ( x -1 < x < 0 OR x > 0)

Thanks
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akunduru4
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Sat Dec 10, 2011 4:07 am 
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Students


Posts: 1
stud.jatt wrote:
Well you have to consider that the equation in the question includes that modulus function. so the question being asked by |x| < 1 isn't just whether x < 1 but whether -1 < x < 1 (i.e. does x lie between -1 and 1), because only in that case will |x| < 1

Statement 1 gives us two ranges for x
-1 < x < 0 OR x > 0

Statement 2 gives us that x < 0 but we don't know if x lies between -1 and zero

Upon combining the two we have -1 < x < 0 a range that lies between -1 and 1 and hence the answer is C



Thanks
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stud.jatt
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Sat Dec 10, 2011 12:46 pm 
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Students


Posts: 21
The definition of a modulus function is

|y| = y if y > 0 and
|y| = -y if y < 0

From Statement 1 we have x/|x| < x

Now x/|x| can have only two values:

1 if x > 0 and
-1 if x < 0

Substituting these values in the above equation we have

when x is +ve, 1 < x
and
when x is -ve, -1 < x; and since this is the case for x < 0, instead of an infinite range we get the limited range -1 < x < 0

Combining the two cases, the set of possible values of x which satisfies x/|x| < x is -1 < x < 0 and x > 1; plug any value of x from this range to check

Also be careful that you may be tempted to simplify this equation by eliminating the x on both sides leading to 1/|x| < 1 or 1 < |x|, however this would give a faulty solution as we don't know whether x is +ve or -ve and hence what the effect on the inequality sign would be when dividing both sides by x.

jaspreetp wrote:
Hi stud.jatt

Can you please explain how you got the two ranges for Statement 1 ( x -1 < x < 0 OR x > 0)

Thanks
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lj_neary
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Tue Dec 13, 2011 6:24 pm 
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Can someone please post a step by step explanation of how statement 2 is simplified into x<0. Thanks.
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jnelson0612
 Post subject: Re: Doubt in Data Sufficiency
 Post Posted: Mon Dec 26, 2011 12:06 am 
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ManhattanGMAT Staff


Posts: 1857
lj_neary wrote:
Can someone please post a step by step explanation of how statement 2 is simplified into x<0. Thanks.


Sure. Here's the statement:
(2) |x| > x

Let's think about this. The absolute value of x, |x|, is always the positive distance from x to zero on the number line. So unless x itself is zero, |x| will always be positive.

Can x itself be positive? Can x be something such as 3? Well, no, because |3|=3, and thus |3| is not greater than 3, as our statement says.

Can x be zero? No, because the two sides would both be zero; the left is not greater than the right.

Can x be negative? Yes! If x is something such as -2, note that |-2|=2, and 2 > -2. The absolute value transforms the left to a positive number, but the right is left as a negative number. Note that this works for any negative number.

I hope this helps!

_________________
Jamie Nelson
ManhattanGMAT Instructor
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