Doubt in Data Sufficiency

[anveshan_reddy_k]

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..

[stud.jatt]

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

[jaspreetp]

Hi stud.jatt

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

Thanks

[akunduru4]

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

[stud.jatt]

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.

Hi stud.jatt

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

Thanks

[lj_neary]

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

[jnelson0612]

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!