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605-655 Level|   Algebra|   Inequalities|                           
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
General Discussion
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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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AbdurRakib
What is the value of x if \(x^3 < x^2\) ?

(1) -2<x<2

(2) x is an integer greater than -2

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

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


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

(1) –2< x < 2
(2) x is an integer greater than –2.

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
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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:
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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?
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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.

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

See this question approach using this method.

Attachment:
WhatsApp Image 2018-05-05 at 09.38.19.jpeg
WhatsApp Image 2018-05-05 at 09.38.19.jpeg [ 56.64 KiB | Viewed 87303 times ]
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AbdurRakib
What is the value of x if x³ < x²?

(1) –2< x < 2
(2) x is an integer greater than –2.

Target question: What is the value of x?

Given: x³ < x²
If we do a little bit of work, we'll see that this given information tells us A LOT about x
x² must be POSITIVE here (since we can see that x ≠ 0, otherwise we can't have x³ < x²). So, we can safely divide both sides of the inequality by x² to get: x < 1
So, x < 1 AND x ≠ 0

Statement 1: –2< x < 2
There are several values of x that satisfy statement 1 (and the given information). Here are two:
Case a: x = -1
Case b: x = 0.5
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x is an integer greater than –2
So, x is an INTEGER that's less than 1, but greater than -2 AND x ≠ 0
There's only one x-value (x = -1) that satisfies these conditions. So, x must equal -1
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer:
RELATED VIDEO

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 ? :-)
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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
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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?
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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.

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

(1) –2< x < 2
(2) x is an integer greater than –2.

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.

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

(1) –2< x < 2
(2) x is an integer greater than –2.

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

See this question approach using this method.


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
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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.
Attachment:
06_takeaway-copy-e1504685518406.png
06_takeaway-copy-e1504685518406.png [ 40.07 KiB | Viewed 24727 times ]

578vishnu
gmatbusters
Wavy Curve method for inequality is a very important tool
Wavy curve method

See this question approach using this method.


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
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