SOLUTION

Is x between 0 and 1?

(1) x^2 is less than x --> \(x^2<x\) --> \(x(x-1)<0\):

Multiples must have opposite signs:
\(x<0\) and \(x-1>0\), or \(x>1\) --> no solution (\(x\) cannot be simultaneously less than zero and more than 1);
\(x>0\) and \(x-1<0\), or \(x<1\) --> \(0<x<1\);

So \(x(x-1)<0\) holds true when \(0<x<1\). Sufficient.

For alternate approach check "How to solve quadratic inequalities": http://gmatclub.com/forum/x2-4x-94661.html#p731476

(2) x^3 is positive --> \(x^3>0\) just tells us that x is positive. Not sufficient.

Answer: A.
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\(x(x-1) < 0\) gives you the solution 0 < x < 1.

check out the link below for the explanation:
http://gmatclub.com/forum/inequalities-trick-91482.html#p804990
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Hmm i think i am more confused after reading that the first time through....

I think i am missing the reason why the inequality sign for \(x < 0\) should actually be\(x > 0\). I determined x\(< 1\) because i set the inequality of \(x - 1 < 0\) and after subtracting from both sides give me \(x < 1\) . What is different about \(x < 0\) becoming \(x > 0\) ?

x(x - 1) < 0 is not the same as x <0 and (x - 1)< 0

When I multiply two terms, the result is negative if and only if one of them is negative and the other is positive. When I multiply x with (x - 1), the result x(x - 1) will be negative (less than 0) in two cases:

Case I: x < 0 (x is negative) but (x - 1) > 0 (x - 1 is positive)
(x - 1) > 0 implies x > 1
But this is not possible. x cannot be less than 0 and greater than 1 at the same time.

Case II: x > 0 (x is positive) but (x - 1) < 0 (x - 1 is negative)
(x - 1) < 0 implies x < 1
This will happen when x lies between 0 and 1. i.e. when 0 < x < 1.

The link gives you the shortcut of solving inequalities of this type.
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is x between 0 and 1?
1. x^2 is less than x
2. x^3 is positive

I answer C considering x could be negative or positive but option 2 ensures x is positive. Please help what is the wrong with me.
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Merging similar topics.



In (1) as x^2<x then x can not be negative, because if it is then we would have x^2>0>x.
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Hi All,

We can also solve this question using the Wavy Line method.

Given:
We are asked if \(0< x <1\). For finding the range of \(x\), we need to solve the inequality given in the statements.

Statement-I:
St-I tells us that \(x^2 < x\) i.e. \(x(x -1) < 0\). Let's find the range of this inequality using the Wavy Line method.

The zero points of this inequality are {1,0}.



Plotting them on the number line using the Wavy Line method we see that the inequality is negative for the range \(0 < x < 1\).

Hence, statement-I is sufficient to answer the question.

Statement-II:
St-II tells us that \(x^3 > 0\). Let's draw a Wavy line for this inequality. The zero point of the inequality is {0}



Plotting it on the number line we see that the inequality is positive for the range \(x > 0\). Thus, we can't say for sure if \(0 < x < 1\).

Hence, statement-II is not sufficient to answer the question.

For more reading on the Wavy Line method try out inequalities-trick-91482-80.html#p1465609

Hope its clear!


Regards
Harsh
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Target question: Is x between 0 and 1 ?

Statement 1:x² is less than x
In other words, x² < x
We can apply some inequality rules here.
Since x² must be POSITIVE here, we can take x² < x and divide both sides by x²
We get: 1 < 1/x
Since 1/x is greater than 1, we can conclude that 1/x is positive, which means x is POSITIVE (i.e., x > 0)
Since x is POSITIVE, we can take 1 < 1/x and multiply both sides by x to get: x < 1
When we combine our two inequalities, we get 0 < x < 1
In other words, x IS between 0 and 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³ is positive
There are several values of x that satisfy statement 2. Here are two:
Case a: x = 1/2 (so, x³ = (1/2)³ = 1/8). In this case, x IS between 0 and 1
Case b: x = 2 (so, x³ = 2³ = 8). In this case, x is NOT between 0 and 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO FROM OUR COURSE

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Sir, i am having one query . In this question condition is not given , that X is negative. How can we say that multiples must have opposite signs to find 0<X<1.

Using statement 1,
If X were negative, X^2 would be positive and certainly greater than X. Hence X cannot be negative.
So X must be positive.
If X is 0 or 1, X^2 = 0 or 1 too.
If X is greater than 1, X^2 would be greater than X.
If X is less than 1, X^2 will be less than X.

So 0 < X < 1
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Hi for statement 1

if I take x= 0.5
x^2 = 0.25 (satisfies st 1)

but even if I take x= -0.5
x^2 = 0.25 (satisifies st 1)
here x is negative but and doesnt fall under 0<x<1

Isn't statement 1 giving us -1<x<1 ?

could someone pls point out what I am missing? Bunuel VeritasKarishma GMATPrepNow

(1) says that x^2 < x.

If x = -0.5, then (x^2 = 0.25) > (x=-05). For any negative x, (x^2 = positive) > (x = negative).
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The question asks us if x is a proper fraction.

From statement I alone, \(x^2\) is less than x.
This is true only when x is a proper fraction. Statement I alone is sufficient to answer the question with a Yes.
Answer options B, C and E can be eliminated.

From statement II alone, \(x^3\) is positive.
This means x is positive or x>0. However, this is insufficient to find if x is between 0 and 1. Statement II alone is insufficient to answer the question.
Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!
Aravind BT
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Solution:

We need to determine whether x is between 0 and 1.

Statement One Alone:

Since x^2 is less than x, then x is between 0 and 1 (notice that if x ≥ 1, x^2 ≥ x, and if x ≤ 0, x^2 ≥ x). Statement one alone is sufficient.

Statement Two Alone:

x^3 is positive only means x is positive. However, there is no way we can determine if x is between 0 and 1 OR if it’s greater than or equal to 1. Statement two alone is not sufficient.

Answer: A
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