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abhinav008
Is x > 0

(1) x^2 > x
(2) x^3 > x

I can't understand why is first option insufficient.

x^2>x
x^2-x>0
x(x-1)>0
x>0 or x>1 which gives me the answer. but this is not the correct answer. can some one please explain me
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ravinder050484
abhinav008
Is x > 0

(1) x^2 > x
(2) x^3 > x

I can't understand why is first option insufficient.

x^2>x
x^2-x>0
x(x-1)>0
x>0 or x>1 which gives me the answer. but this is not the correct answer. can some one please explain me

When u substitute x> 1 say x = 1.1
1.1 ( 1.1-1 ) is greater than 0

When substitute X between 0 and 1 you can see that 0.5 ( 0.5 -1 ) is less than 0

When substitute value x< 0 say -0.5
-0.5 ( -0.5 - 1 ) we get value > 0

So the equation satisfies all values except values between 0 and 1

So X> 0 i.e 1.1 or less than 0 i.e. -0.5 both these values when substituted in x ( x-1 ) gives answer > 0
So X can be > 1 or x can be < 0
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ravinder050484
abhinav008
Is x > 0

(1) x^2 > x
(2) x^3 > x

I can't understand why is first option insufficient.

x^2>x
x^2-x>0
x(x-1)>0
x>0 or x>1 which gives me the answer. but this is not the correct answer. can some one please explain me

Hi ravinder050484,

For the inequality x(x-1) > 0, there are a couple of ways to find out the range of x.

Method-1
You can interpret this inequality as product of two numbers being greater than 0 i.e. it tells us about the same sign of the numbers. So, either

a. x & x-1 both are positive which would be true for range x > the largest zero point i.e. x > 1 or

b. x & x-1 both are negative which would be true for range x < the smallest zero point i.e. x < 0.

Combining these two we can say that x > 1 or x < 0. Thus statement-I is insufficient to answer the question.

Method-2
You can find the range of this inequality using the Wavy line method. Refer the following diagram for the range of x .



We can see from the wavy line diagram that the inequality is positive when x > 1 or x < 0.
Thus we can't say for sure if x > 0. Hence, statement-I is insufficient to answer the question.

You can read more about the Wavy Line method here

Hope its clear!

Regards
Harsh
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Hi abhinav008,

This DS question can be solved with a combination of Number Properties and TESTing VALUES.

We're asked if X > 0. This is a YES/NO question.

Fact 1: X^2 > X

IF...
X = 2
the answer to the question is YES.

IF...
X = ANY NEGATIVE
the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: X^3 > X

IF....
X = 2
the answer to the question is YES.

IF...
X = -1/2
the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know...
X^2 > X
X^3 > X

From our prior work, we have VALUES that 'fit' BOTH Facts, so we know....

IF....
X = 2
the answer to the question is YES.

IF...
X = -1/2
the answer to the question is NO.
Combined, INSUFFICIENT

Final Answer:
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Rich
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