What is the value of x if x^3 < x^2?

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1) Not Sufficient.
2) According to the question stem x<1 and x>-2 => x = -1 so satisfy the conditional mentioned in the stem.
Hence. B. What's the OA?


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\(x^3 < x^2\)
The above condition is possible only when x is negative
For example,
if x=-5 \(-125<25\)
if x=-3\(-27<9\)

St1: x can have values in the range of \(-1<=x<=0\)
For x= -1,-0.9,-0.8 the equation holds true
No unique value NS

St2: x is int and x>-2
Then x can have only one value in this range x=-1
Since, x>0(Positive values) will not hold true; hence eliminated Suff

Option B

The above condition is possible only when x is negative
This is not quite true.
x can be any number less than 1 (but not equal to zero)
For example x = 1/2 satisfies the inequality x^3 < x^2

Cheers,
Brent

We need to determine the value of x, given that x^3 < x^2.

Statement One Alone:

–2 < x < 2

We see that x could be, for example, -1 or -½.

For either of these values, we have x^3 < x^2, since x^3 will be negative and x^2 will be positive. Since we don’t have a unique value for x, statement one alone is not sufficient.

Statement Two Alone:

x is an integer greater than –2.

We see that if x = -1, then x^3 < x^2, since x^3 = -1 and x^2 = 1.

If x = 0, then x^3 = x^2, since x^3 = 0 and x^2 = 0 (so x can’t be 0).

Similarly, if x = 1, then x^3 = x^2, since x^3 = 1 and x^2 = 1 (so x can’t be 1).

If x is an integer > 1, then x^3 will always be greater than x^2. Thus, x can’t be any integer > 1.

Therefore, we see that the only value x can be is -1. Since we have a unique value for x, statement two alone is sufficient.

Answer: B

Hi All,

We're told that X^3 is LESS than X^2. We're asked for the value of X. This question can be solved with a mix of Number Properties and TESTing VALUES.

To start, there are only certain types of values that will fit the given information that X^3 is less than X^2:
-ANY negative value
-Positive fractions (0 < X < 1)

1) -2 < X < 2

With Fact 1, we have LOTS of different possible values for X: any negative value and any positive fraction in that range.
Fact 1 is INSUFFICIENT

2) X is an INTEGER greater than -2

The information in Fact 2 eliminates most of the possibilities that we started with. Since X has to be an INTEGER, none of the positive fractions are possible and since X has to be GREATER than -2, there's only one option possible: -1
Fact 2 is SUFFICIENT

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Why when we solve x^3<x^2 algebraicly we get (0;1) as an answer, but not negative possibilities of x. I mean

x^3<x^2
x^3-x^2<0
x^2(x-1)<0
x=0 x=1 So answer is 0<x<1

Whats wrong with this solution?

Isn't there any fixed approach for this kind of inequalities or we have to think out of box/logically everytime?

Hi Mehemmed,

To start, you're treating this inequality as if it were an EQUATION - which it is NOT. The two values of X that you solved for actually do not "fit" the given inequality at all. Once you get to this point:

(X^2)(X-1)<0

You still have deal with the fact that the 'left side' of the inequality has to be be LESS than 0. Since (X^2) with either be 0 or positive, we need to focus on the other piece. With (X-1), you MUST end up with a negative.... so what values of X will make that result happen...? IF X is a positive fraction (re: 0 < X < 1) OR X is ANY negative number.

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

Wavy Curve method for inequality is a very important tool
Wavy curve method

See this question approach using this method.

hello GMATPrepNow regarding ST 2

x can be -1 in this case x³ < x² YES, but x can be any positive value if x is an integer greater than –2 ,
say X is 2, then answer is NO

Why do you limit yourself to only -1 ?

Statement 2 tells us that x is an integer greater than –2.
So, x could be -1, 0, 1, 2, 3, etc

From the given information (x³ < x²), we know that x < 1
So, if x is greater than -2 and less than 1, then x must be EITHER -1 or 0

HOWEVER, we also know that x cannot equal zero, because the inequality (x³ < x²) would not hold true.
So, it must be the case that x = -1

Does that help?

Cheers,
Brent

Hello!

-1 seems to be a logic answer but is says "what is the value of x IF x cube is LESS than x square".

I a bit of a dummie in math but isn't -1 is -1 in whichever degree?... and hence -1^3 is not less than -1^2?

Hi

when \(x = -1, x^3 = -1\), this is true for any odd power of x

and , when \(x = -1, x^2 = 1\), this is true for any even power of x

Hence, for \(x = -1, x^3 < x^2\)

hope, it is clear now, feel free to tag me in case of any doubt.

Since \(x^3 < x^2\) is true only if x is NONZERO, we can divide by \(x^2\), which must be POSITIVE:
\(\frac{x^3}{x^2} < \frac{x^2}{x^2}\)
\(x < 1\)
Question stem, rephrased:
If x is a nonzero value less than 1, what is the value of x?

Statement 1:
Here, x can be any nonzero value between -2 and 1.
INSUFFICIENT.

Statement 2:
Here, x must be a nonzero integer such that -2 < x < 1.
Thus, x=-1.
SUFFICIENT.

Answer:

Given -> x^3 < x^2, this implies x is negative or x <0

1) –2< x < 2 , or -2< x < 0

so x can take any value between -2 and 0. NOT SUFFICIENT.

2) x > -2, also we know that x < 0 , this implies -2< x < 0
Now it is also given that x is an integer and the only integer that satisfies the above inequality is -1. Hence SUFFICIENT.

Hi,

Can somebody please explain to me why gmatbusters has considered -infinity <x < 1 as the range?
During wavy line method, shouldn't the signs be alternate? Why has he considered two subsequent signs as negatives?

Please help!

chetan2u VeritasKarishma Bunuel egmat

Hi 578vishnu

When the exponent of the variable x is even, then the curve will not pass through the number line but it will just bounce back at that critical point.
Steps


The inequality we get is x^2(x-1) or (x-1)^2 * (x-1)
here, the exponent is 2 (even ) at critical point x=0, so it wil bounce back at x = 0.



Hope it is clear now.