DS Absolute value

Question

Please see the attachment

crackgmat750 · Answer

C.

Combining both IxI has to be decimal value less than one. x square will be less than x in that case.

anandnk · Answer

(A) for me.

a) x^2/|x| = sign of (x) * |x|
if x < 1 for example x = 0.99999 then x^2/|x| = 0.9999
if x = -0.5 then value is -0.5
if x = 100 then value is -100

there fore (A) is sufficient.

Dookie · Answer

If x=-10 the result is 100/10 =10, which is greater than 1.
if x=1/2 the result is (1/4)/(1/2)=1/2.

Therefore A is insufficient. 

Answer is C,

Paul · Answer

should be C
x = 1/2 then x^2/ lxl = 2/4 = 1/2 which is less than 1
x = -2 then x^2 / lxl = = 4/2 = 2 which is greater than 1
A is not sufficient

We need to know that both x and y are b/w -1 and 1 in order to draw some conclusion.

smandalika · Answer

anand,
 what if x= -100 the n the function will evalute to 100 which is greater than 0.
 so it has to lie between -1 and 1 hence ans is C

twixt · Answer

I think answer is A

When you pick numbers, you consider x = -2 ... -100 while statement 2 refers to "x > -1"

rahul · Answer

I will go with A

1 . x can take any value less than 1

x = 0.5 res = 0.25/0.5 < 1 - true
x = -0.5 res = - (o.25/0.5 ) < 1 true

x = -10 res = - (100/10) < 1 true
x = 10 we cannot take this value as x < 1 .

so for any value x < 1 , the equation is true

on the other hand x > -1

x = - 0.5 res = -(o.25/0.5 ) < 1 and > -1 - true

x = 10 , res = 100 /10 = 10 > 1 so 2 is insufficient .

gmatmath · Answer

Rahul,
if x = 10, the result is a positive value. coz (x)^2 = (-10)^2 = 100
|x| = |-10| = 10
=> result = 100/10 = 10

hardworker_indian · Answer

C.

The question boils down to:
x^2/|x| < 1
x^2 < |x|
|x| > x^2

(x is a proper fraction, if 1<x<1 & x!=0, like -2/3, 2/3, 5/7, etc.)

In general, if you see
1. x > x^odd powers, then x is a proper fraction or negative.
2. x > x^even powers, then x is a positive proper fraction.
3. |x| > x^2 only when x is a proper fraction (can be negative or positive)

Dharmin · Answer

We have to combine the two options because x^2 and |x| both change the sign of the negative numbers to positive and to make |x| > x^2, x must be between -1 and 1.

"C" should be it.

saurya_s · Answer

Why the answer is not B?
When x=-0.5, we get answer as No.
When x=1, answer is No
When x=5, answer is still No.
So is B sufficent?

DS Absolute value

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