If x is not equal to 0, is 1 x 1 ?

Question

If x is not equal to 0, is –1 ≤ x ≤ 1 ?

(1)  x^3 < x^2

(2)  x^2 < x

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shameekv1989 · Answer

If x is not equal to 0, is –1 ≤ x ≤ 1 ?

(1) \(x^3 < x^2\)

(2) \(x^2 < x\)

Statement i)
\(x^3 - x^2 < 0\) => \(x^2 (x-1) < 0\) => \(x<1 \) - x can be in the range and then it may not

Insufficient

Statement ii)
x^2 - x < 0 => x(x-1) < 0 => 0<x<1 => as x can be only in this range it will definitely be in the range mentioned. Hence Sufficient

Answer - B

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • x is not equal to zero. 
We need to find; whether -1 ≤ x ≤ 1 or not.
 Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:   x^3 < x^2 
 •  x^3 - x^2 < 0 
•  x^2(x-1) < 0  
We know that  x^2  is greater than 0.
 • Therefore,  x - 1 < 0 , and  x < 1
 
• However, we don't know whether x > -1 or not. 
Hence, statement 1 is not sufficient, we can eliminate answer options A and D.
 Statement 2:    x^2 < x 
 • We know that  x^2  is always positive
 o Thus, x is also positive because  x^2 < x  
• This is only possible when x is between 0 and 1 exclusively.
 o  0 < x < 1 
  It means  -1 ≤ x ≤ 1    
Hence, statement 2 is sufficient, the correct answer is  Option B .

Archit3110 · Answer

#1
  x^3 < x^2 
would be valid when either x= fraction value b/w 0 to 1 or x is -ve integer insufficient
#2
 x^2 < x 
valid when 0>x<1 sufficient
IMO B

minustark · Answer

If x is not equal to 0, is –1 ≤ x ≤ 1 ?

(1)  x^3 < x^2

(2)  x^2 < x

For any negative value of x, x^3 will be lower than x^2. For example, if x= -1/2, x^2 =1/2 and x^3 =-1/8. Again when x = -2, x^2 =4 and x^3 =-8. Not sufficient.

For value from 0 to 1, x^2 < x. e.g: x =1/4, x^2 = 1/16 or x =1/3, x^2=1/9. So, x is between 0 and 1. Sufficient.

B is the answer.

GunjanMadan1 · Answer

In 2nd condition, how it can be true for x=+-1

Efstathios · Answer

1) x^3-x^2<0 =>x^2(x-1)<0 => x<1, since x^2>0 (x not equal to 0) => not sufficient
2) x^2<x then x>0, since x>x^2 and x^2>0(again x not equal to 0) and we can divide by x without flipping the inequality, so x<1. Finally x>0 and x<1 => sufficient
Therefore B.

harshittahuja · Answer

In B how one can make sure the X lies between -1to+1 . 
from my point of view answer is option E.

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If x is not equal to 0, is 1 x 1 ?

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