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If 1 < x < 1 and x is not equal to 0, then which of the following mu Updated on: 10 Feb 2026, 12:37
Originally posted by EMPOWERgmatRichC on 22 Oct 2015, 20:08.
Last edited by Bunuel on 10 Feb 2026, 12:37, edited 2 times in total.
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55% (hard)

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70% (02:14) correct 30% (02:17) wrong based on 1187 sessions

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If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. \(|x| > x^2\)

II. \(x – x^2 > x^3\)

III. \(|1 – x| = |x – 1|\)

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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QUANT 4-PACK SERIES Problem Solving Pack 1 Question 4 IF -1 < x < 1 and ...

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If 1 < x < 1 and x is not equal to 0, then which of the following mu 22 Oct 2015, 21:02
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RANDOM 4-PACK SERIES Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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After glancing at the answer choices, I observed that there is no answer choice that includes all 3 statements. Statement 3 is clearly true because the absolute distance from x to 1 will certainly be equal to the absolute distance from 1 to x. Then I noticed that there is no answer choice for Statement 3 only. This eliminates answer choices A,B, and C. Now it is between D and E. Now either statement 1 is true or statement 2 is true, but both can't be true. It is easier to disprove something with a counterexample then proving something must be true. If you plug in \(x=-\frac{1}{2}\) into statement 2, you will find it is not true. This eliminates answer choice E and leaves me with answer choice D.
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If 1 < x < 1 and x is not equal to 0, then which of the following mu 22 Oct 2015, 21:55
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RANDOM 4-PACK SERIES Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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My solution:

Let's take two values of x as -0.5 and 0.5, then

I) Any value inside the mode is positive then whether we take 0.5 or -0.5 this will remain positive. We also know that squaring a fraction gives us a value less then the original fraction(if the original fraction is positve) and always positive. So statement (I) is always true as 0.5 > 0.125.

II) Not always true: If x is -0.5 then the expression becomes 0.625 = 0.625

III) Always holds true as anything inside the mode is positive. If x is -0.5 then mode 1-(-0.5) = mode -0.5-1 becomes 1.5 =1.5 (Don't know how to use mode symbol) :(

Option D
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RANDOM 4-PACK SERIES Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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When -1<x<1: |x| > \(x^{2}\) => I must be true
x – \(x^{2}\) > \(x^{3}\) => Not sure it is true with all values of x
|1 – x| = |x – 1| true with all the value of x => it is also true with all x: -1<x<1: III must be true

Ans: D
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If 1 < x < 1 and x is not equal to 0, then which of the following mu 23 Oct 2015, 04:21
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RANDOM 4-PACK SERIES Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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Given, \(- 1 < x < 1\) and \(x \neq 0\),

For a Must be true question, ALL cases posisble must satisfy the given conditions.

i) \(|x| > x^2\) ---> 1 positive quantity > another positive squared quantity , can only happen for fractions between -1 and 1. As this is the given range of x, this has to be true for all cases.

Additionally, algebraically,

when \(x<0\) --> \(|x|=-x\) ---> \(-x>x^2\) --->\(x^2+x<0\) --->\(-1<x<0\) ...(1)
when \(x\geq 0\)-->\(|x|=x\)---> \(x>x^2\) ---> \(x^2-x<0\) ---> \(0<x<1\) ...(2)

Both (1) and (2) lie in the given ranges and hence must be true for all cases. Eliminate B and E.

ii) \(x – x^2 > x^3\) ---> \(x^3+x^2-x <0\) ---> \(x(x^2+x-1)<0\) ---> \(x(x-a)(x-b)<0\) ---> where \(a<0\) and \(b>0\)--->\(x<a\) and \(0<x<b\).

As \(a=\frac{-1-\sqrt{5}}{2}\) (= a quantity <-1) and \(b=\frac{-1+\sqrt{5}}{2}\), clearly see that \(x<a\) is out of the given range of -\(1<x<\)1, hence making this statement NOT a must be true statement. Eliminate C.

iii) \(|1 – x| = |x – 1|\), \(|x-1| > 0\) for all values \(-1<x<1\) and as \(|x-1| > 0\) for \(-1<x<0\) and as this range lies within the given range, this is a MUST BE TRUE statement, making D as the correct answer.
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If 1 < x < 1 and x is not equal to 0, then which of the following mu 26 Oct 2015, 16:42
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My solution:

Let's take two values of x as -0.5 and 0.5, then

I) Any value inside the mode is positive then whether we take 0.5 or -0.5 this will remain positive. We also know that squaring a fraction gives us a value less then the original fraction(if the original fraction is positve) and always positive. So statement (I) is always true as 0.5 > 0.125.

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Hi Learning4mU,

Your logic IS correct, but you have to be careful about your calculations. \((.5)^{2}\) = .25 (NOT .125). Working through this prompt, your logic is sound, so the miscalculation wouldn't hurt you. In other types of questions though (especially DS), a miscalculation could cost you the points.

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If 1 < x < 1 and x is not equal to 0, then which of the following mu 26 Oct 2015, 17:06
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Hi All,

In Roman Numeral questions, the arrangement of the answer choices often provides a clue as to how you could deal with the prompt in the most efficient way possible (and avoid some of the work). From these answer choices, we know that only 1 or 2 of the 3 Roman Numerals MUST be true, so we can deal with the Roman Numerals 'out of order' if it will save us some effort.

We're given a range of possible values (-1 < x < 1) and that X CANNOT be 0. We can prove/disprove the various Roman Numerals by TESTing VALUES and looking for patterns.

I. I. |x| > \(x^{2}\)

|x| will be positive for any value of X that we can choose. Since we'll be SQUARING a fraction (regardless of whether it's a positive fraction or a negative fraction), the result WILL be smaller than the |x|.

eg.
IF....
X = 1/2
\((1/2)^{2}\) = 1/4

IF...
X = -1/3
\((-1/3)^{2}\) = 1/9

Thus, Roman Numeral 1 is ALWAYS TRUE.

II. x – \(x^{2}\) > \(x^{3}\)

This Roman Numeral is probably the 'scariest looking', but it can also be dealt with by TESTing VALUES.

While....
(1/2) - \((1/2)^{2}\) > \((1/2)^{3}\)
1/2 - 1/4 > 1/8
1/4 > 1/8
Is TRUE

(-1/2) - \((-1/2)^{2}\) > \((-1/2)^{3}\)
-1/2 - 1/4 > -1/8
-3/4 > -1/8
is NOT TRUE

Thus, Roman Numeral 2 is NOT always true.

III. |1 – x| = |x – 1|

This Roman Numeral is actually a math 'truism'; it's always true for any rational value of X. You can use as many different values as you like, the resulting calculations are ALWAYS equal.

Thus, Roman Numeral 3 is ALWAYS TRUE.

Final Answer:
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D

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Learning4mU


My solution:

Let's take two values of x as -0.5 and 0.5, then

I) Any value inside the mode is positive then whether we take 0.5 or -0.5 this will remain positive. We also know that squaring a fraction gives us a value less then the original fraction(if the original fraction is positve) and always positive. So statement (I) is always true as 0.5 > 0.125.


Hi Learning4mU,

Your logic IS correct, but you have to be careful about your calculations. \((.5)^{2}\) = .25 (NOT .125). Working through this prompt, your logic is sound, so the miscalculation wouldn't hurt you. In other types of questions though (especially DS), a miscalculation could cost you the points.

GMAT assassins aren't born, they're made,
Rich
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Thanks a lot Rich for pointing out this silly error of mine. Surely it has worked here but some other day it can be a blunder and cost me heavy points. I ll certainly take care of this.
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RANDOM 4-PACK SERIES Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

48 Hour Window Answer & Explanation Window
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.


Given, \(- 1 < x < 1\) and \(x \neq 0\),

For a Must be true question, ALL cases posisble must satisfy the given conditions.

i) \(|x| > x^2\) ---> 1 positive quantity > another positive squared quantity , can only happen for fractions between -1 and 1. As this is the given range of x, this has to be true for all cases.

Additionally, algebraically,

when \(x<0\) --> \(|x|=-x\) ---> \(-x>x^2\) --->\(x^2+x<0\) --->\(-1<x<0\) ...(1)
when \(x\geq 0\)-->\(|x|=x\)---> \(x>x^2\) ---> \(x^2-x<0\) ---> \(0<x<1\) ...(2)

Both (1) and (2) lie in the given ranges and hence must be true for all cases. Eliminate B and E.

ii) \(x – x^2 > x^3\) ---> \(x^3+x^2-x <0\) ---> \(x(x^2+x-1)<0\) ---> \(x(x-a)(x-b)<0\) ---> where \(a<0\) and \(b>0\)--->\(x<a\) and \(0<x<b\).

As \(a=\frac{-1-\sqrt{5}}{2}\) (= a quantity <-1) and \(b=\frac{-1+\sqrt{5}}{2}\), clearly see that \(x<a\) is out of the given range of -\(1<x<\)1, hence making this statement NOT a must be true statement. Eliminate C.

iii) \(|1 – x| = |x – 1|\), \(|x-1| > 0\) for all values \(-1<x<1\) and as \(|x-1| > 0\) for \(-1<x<0\) and as this range lies within the given range, this is a MUST BE TRUE statement, making D as the correct answer.
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I have been making this mistake for a lot of modulus problems...please help me...

when x<0
we have x^2+x<0
that is x(x+1)<0
so x<0 and x+1<0
hence X<0 and x<-1
how is it that x>-1 according to your solution? You said when x<0 the range is -1<x<0

please explain
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QUANT 4-PACK SERIES Problem Solving Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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x is a +ve or -ve fraction <1 but not equal to 0

I. |x| > \(x^{2}\)

x^2 and |x| will always be +ve
and squaring +ve fraction always decrease the value.
hence |x| > \(x^{2}\)

II. x – \(x^{2}\) > \(x^{3}\)

If we take x= 1/2 or -1/2 and put the value in the given equation, it might or might not be right.

III. |1 – x| = |x – 1|
If we put 1/2 or -1/2 in the given equation, the result will be same. True

D) I and III only is the answer.
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QUANT 4-PACK SERIES Problem Solving Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. |x| > \(x^{2}\)
II. x – \(x^{2}\) > \(x^{3}\)
III. |1 – x| = |x – 1|

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

48 Hour Window Answer & Explanation Window


Earn KUDOS! Post your answer and explanation.
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x is a +ve or -ve fraction <1 but not equal to 0

I. |x| > \(x^{2}\)

x^2 and |x| will always be +ve
and squaring +ve fraction always decrease the value.
hence |x| > \(x^{2}\)

II. x – \(x^{2}\) > \(x^{3}\)

If we take x= 1/2 or -1/2 and put the value in the given equation, it might or might not be right.

III. |1 – x| = |x – 1|
If we put 1/2 or -1/2 in the given equation, the result will be same. True

D) I and III only is the answer.
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hello disha
can u answer my question above?
please tell me what is the mistake in that approach...
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I have been making this mistake for a lot of modulus problems...please help me...

when x<0
we have x^2+x<0
that is x(x+1)<0
so x<0 and x+1<0
hence X<0 and x<-1
how is it that x>-1 according to your solution? You said when x<0 the range is -1<x<0

please explain
Show more

Hi mup05ro,

When dealing with that inequality, you CANNOT "break it into pieces" the way that you did.

When you wrote that X < 0, you have to consider how that impacts the other part of the calculation. For example, If X = -2, how does THAT value impact the overall inequality?

IF X = -2...

X(X+1) =
(-2)(-2 + 1) =
(-2)(-1) = +2.... but that is NOT less than 0, so X CANNOT be -2. In that same way, while X can be negative, X cannot be less than -1.

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mup05ro


I have been making this mistake for a lot of modulus problems...please help me...

when x<0
we have x^2+x<0
that is x(x+1)<0
so x<0 and x+1<0
hence X<0 and x<-1
how is it that x>-1 according to your solution? You said when x<0 the range is -1<x<0

please explain

Hi mup05ro,

When dealing with that inequality, you CANNOT "break it into pieces" the way that you did.

When you wrote that X < 0, you have to consider how that impacts the other part of the calculation. For example, If X = -2, how does THAT value impact the overall inequality?

IF X = -2...

X(X+1) =
(-2)(-2 + 1) =
(-2)(-1) = +2.... but that is NOT less than 0, so X CANNOT be -2. In that same way, while X can be negative, X cannot be less than -1.

GMAT assassins aren't born, they're made,
Rich
Show more

thank you

how do determine this for every such problem? I would do the steps that I did above and simply say x<-1, so should i always verify with the range I got and conclude? I have not done that for many problems that i got right
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Hi mup05ro,

Some of the questions that you're going to face on Test Day will test the "thoroughness" of your thinking (Data Sufficiency does this often). If there are certain question types that you find challenging (such as absolute value/modulus questions), then you should consider doing a little more work to prove that you're correct. By TESTing VALUES, you would have PROOF of whether your deductions are correct or not.

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If 1 < x < 1 and x is not equal to 0, then which of the following mu 16 Jun 2018, 12:29
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We're given that -1<x<0 or 0<x<1. To see if each statement must be true, we should test these ranges.

I. \(x^2<|x|\)
If x is negative, then |x|=-x, so expression becomes
\(x^2<|x|\)
\(x^2<-x\)
\(x>-1\) (dividing by x and reversing the inequality since x is negative)
So if x is negative, -1<x<0, which falls inside the negative part of the given range for x.

If x is positive, then |x|=x, so expression becomes
\(x^2<|x|\)
\(x^2<x\)
\(x<1\)
So if x is positive, 0<x<1, which falls inside the positive part of the given range for x.

Thus, this statement must be true.

II. \(x^3<x-x^2\)
\(x^3+x^2-x<0\)
If -1<x<0, then we have
\(x^3+x^2-x<0\)
(small negative) + (larger positive) + (even larger positive)
which cannot be less than 0

Thus, this statement must not be true.

III. |1-x|=|-(x-1)|=|x-1|
identity property. always true

Thus, this statement must be true.

Answer: D
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If 1 < x < 1 and x is not equal to 0, then which of the following mu 03 Feb 2026, 12:59
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since the given range is -1<x<1 i.e. a fraction whose absolute value is less than 1.
1. its for certain that in this range the absolute value of the fraction is greater than its square. must true
3. also the absolute distance of this fraction from 1 is same. must true
so no need to waste time on evaluating the second expression as no where in the choices is given that all the 3 choices must be true.
hence D
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If 1 < x < 1 and x is not equal to 0, then which of the following mu 10 Feb 2026, 12:18
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Hello experts,
How can I algebraically solve the statement II to get the answer
EMPOWERgmatRichC
QUANT 4-PACK SERIES Problem Solving Pack 1 Question 4 IF -1 < x < 1 and ...

If – 1 < x < 1 and x is not equal to 0, then which of the following must be true?

I. \(|x| > x^2\)

II. \(x – x^2 > x^3\)

III. \(|1 – x| = |x – 1|\)

A) I only
B) II only
C) I and II only
D) I and III only
E) II and III only

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If 1 < x < 1 and x is not equal to 0, then which of the following mu 10 Mar 2026, 07:36
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Deconstructing the Question

We are given

\(-1 < x < 1\)

and

\(x \ne 0\)

We must determine which statements must always be true.

Step-by-step

Evaluate statement I

\(|x| > x^2\)

If \(0<x<1\), then

\(|x| = x\)

and

\(x > x^2\)

since squaring a number between \(0\) and \(1\) makes it smaller.

If \(-1<x<0\)

Example

\(x=-0.5\)

\(|-0.5|=0.5\)

\((-0.5)^2=0.25\)

\(0.5>0.25\)

So statement I is always true.

Evaluate statement II

\(x-x^2 > x^3\)

Test \(x=0.5\)

\(0.5-0.25=0.25\)

\(0.5^3=0.125\)

\(0.25>0.125\)

True.

Test \(x=-0.5\)

\(-0.5-0.25=-0.75\)

\((-0.5)^3=-0.125\)

\(-0.75>-0.125\)

False.

So statement II is not always true.

Evaluate statement III

\(|1-x| = |x-1|\)

Since

\(1-x = -(x-1)\)

and

\(|a| = |-a|\)

this equality is always true.

Statements I and III must be true.

Answer D: I and III only
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