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18 Mar 2011, 17:20
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this can also be found by plugging in different numbers
1. Given x>0
x = 2 => (2^3)>(2^2) = (x^3)>(x^2)
x =0.5 => X^3<X^2
two contradicting results. so not sufficient.
2. X^2 >x
x =2 => x^2>x. also X^3>x^2
x = -2 => x^2>x. but X^3>x^2
So not sufficient
together its says x^2 > x and x>0 (so no negative numbers and no fractions between 0 &1)
sufficient to say x^3 > x^2. Hence answer is C.
1. Given x>0
x = 2 => (2^3)>(2^2) = (x^3)>(x^2)
x =0.5 => X^3<X^2
two contradicting results. so not sufficient.
2. X^2 >x
x =2 => x^2>x. also X^3>x^2
x = -2 => x^2>x. but X^3>x^2
So not sufficient
together its says x^2 > x and x>0 (so no negative numbers and no fractions between 0 &1)
sufficient to say x^3 > x^2. Hence answer is C.
23 Apr 2015, 07:05
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Hi Guys,
The range of inequality can also be found out by using the 'Wavy Line' method. It's a powerful method of solving inequalities especially when you have trouble in finding the range algebraically. Let me illustrate briefly how to use this method.
Finding Range of \(x^3 - x^2 > 0\) through 'Wavy Line' method.
We can simplify the expression to \(x^2 (x -1) > 0\). Now, the LHS of the inequality becomes 0 at 0 and 1. So, {0, 1} are the zero points of the inequality.
Plot these points on the number line and start from the top right corner and draw a Wavy Line. Be ready to alternate the region of the wave based on how many times a a point is root to the equation. If the power of a term is odd, the wave simply passes through the point into the other region. If the power of a term is even, the wave bounces back to the same region. In this case, power of the term was 2 for the zero point 0, hence the wave bounced back.
The range of the inequality will be +ve for region above the number line and -ve for region below the number line.
We observe from the graph that the range is +ve when \(x > 1\). So, all we need is to find from the statements- I & II is if \(x >1\) or not.
Statement-I
St-I tells us that \(x > 0\). This does not tell us for sure if \(x >1\). Hence, the statement is insufficient.
Statement-II- Finding Range of \(x^2 -x >0\) through Wavy Line method
The expression in st-II can be simplified to \(x(x-1) > 0\). Let's find the range by the Wavy Line method. The zero points of the expression are { 0,1}.
The Wavy Line graph would be represented in the manner shown below:
We see that the value of the inequality is +ve for \(x < 0\) and \(x > 1\). But the inequality in the question statement is +ve only when \(x > 1\). Hence, Statement-II is insufficient.
Combining Statement-I & II
If we combine both the statements, we can infer that \(x > 1\), which is what we needed. Hence, combining both the statements is sufficient to answer the question.
You can practice a similar question at is-x-196185.html. Try solving it by Wavy Line method.
Also, an article for you on the Wavy Line method inequalities-trick-91482-80.html#p1465609
Hope it helps!
Regards
Harsh
The range of inequality can also be found out by using the 'Wavy Line' method. It's a powerful method of solving inequalities especially when you have trouble in finding the range algebraically. Let me illustrate briefly how to use this method.
Finding Range of \(x^3 - x^2 > 0\) through 'Wavy Line' method.
We can simplify the expression to \(x^2 (x -1) > 0\). Now, the LHS of the inequality becomes 0 at 0 and 1. So, {0, 1} are the zero points of the inequality.
Plot these points on the number line and start from the top right corner and draw a Wavy Line. Be ready to alternate the region of the wave based on how many times a a point is root to the equation. If the power of a term is odd, the wave simply passes through the point into the other region. If the power of a term is even, the wave bounces back to the same region. In this case, power of the term was 2 for the zero point 0, hence the wave bounced back.
The range of the inequality will be +ve for region above the number line and -ve for region below the number line.
We observe from the graph that the range is +ve when \(x > 1\). So, all we need is to find from the statements- I & II is if \(x >1\) or not.
Statement-I
St-I tells us that \(x > 0\). This does not tell us for sure if \(x >1\). Hence, the statement is insufficient.
Statement-II- Finding Range of \(x^2 -x >0\) through Wavy Line method
The expression in st-II can be simplified to \(x(x-1) > 0\). Let's find the range by the Wavy Line method. The zero points of the expression are { 0,1}.
The Wavy Line graph would be represented in the manner shown below:
We see that the value of the inequality is +ve for \(x < 0\) and \(x > 1\). But the inequality in the question statement is +ve only when \(x > 1\). Hence, Statement-II is insufficient.
Combining Statement-I & II
If we combine both the statements, we can infer that \(x > 1\), which is what we needed. Hence, combining both the statements is sufficient to answer the question.
You can practice a similar question at is-x-196185.html. Try solving it by Wavy Line method.
Also, an article for you on the Wavy Line method inequalities-trick-91482-80.html#p1465609
Hope it helps!
Regards
Harsh
23 Apr 2015, 15:13
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Hi All,
This question can be solved by TESTing VALUES and taking advantage of the built-in Number Properties that exist in this question.
We're asked if X^3 > X^2. This is a YES/NO question.
Fact 1: X > 0
IF...
X = 1
1^1 is NOT > 1^2 and the answer to the question is NO
IF....
X = 2
2^3 IS > 2^2 and the answer to the question is YES
Fact 1 is SUFFICIENT
Fact 2: X^2 > X
IF....
X = 2
2^3 IS > 2^2 and the answer to the question is YES
IF....
X = -1
(-1)^3 is NOT > (-1)^2 and the answer to the question is NO
Fact 2 is INSUFFICIENT
Combined, we know....
X > 0
X^2 > X
The combined Facts really 'limit' the possible values of X:
Since X > 0, X MUST be positive.
Since X^2 > X, X CANNOT be a positive fraction (between 0 and 1). X also CANNOT = 1.
By extension, X MUST be > 1. In ALL of these situations, X^3 will be greater than X^2, so the answer to the question is ALWAYS YES.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This question can be solved by TESTing VALUES and taking advantage of the built-in Number Properties that exist in this question.
We're asked if X^3 > X^2. This is a YES/NO question.
Fact 1: X > 0
IF...
X = 1
1^1 is NOT > 1^2 and the answer to the question is NO
IF....
X = 2
2^3 IS > 2^2 and the answer to the question is YES
Fact 1 is SUFFICIENT
Fact 2: X^2 > X
IF....
X = 2
2^3 IS > 2^2 and the answer to the question is YES
IF....
X = -1
(-1)^3 is NOT > (-1)^2 and the answer to the question is NO
Fact 2 is INSUFFICIENT
Combined, we know....
X > 0
X^2 > X
The combined Facts really 'limit' the possible values of X:
Since X > 0, X MUST be positive.
Since X^2 > X, X CANNOT be a positive fraction (between 0 and 1). X also CANNOT = 1.
By extension, X MUST be > 1. In ALL of these situations, X^3 will be greater than X^2, so the answer to the question is ALWAYS YES.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
