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Is x^3 > x^2?

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15 Mar 2011, 03:24
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Is x^3 > x^2?

(1) x > 0
(2) x^2 > x
[Reveal] Spoiler: OA
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15 Mar 2011, 03:24
I'm confused because MGMAT starts off the answer this way:

X^3 > X^2 equals:
X^3 - X^2 > 0
X^2 (X-1) > 0
Therefore we have an inequality in the form xy>0. After this they start heading to statement 1 and 2.

But on page 89 (Chapter 6 - Inequalities strategy of the same book) it says "Note that we should never subtract or divide two inequalities."
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15 Mar 2011, 03:25
Inequalities are definitely my nemesis...It's worse when they are combined with DS. I HATE YOU DS!
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15 Mar 2011, 03:54
This is how the answer is presented in MGMAT:
Question: X^3>X^2?
Rephrase: X^3 - X^2>0?
Rephrase again: X^2(X-1)>0?

Both terms must be positive or negative, because we have an inequality in the form of xy>0. X^2 can never be negative, so we just need to know that both X^2>0 (meaning X is non zero) and X>1. Since every number greater than 1 is non-zero, we can rephrase the question to:
X>1?

Statement 1 tells us: X is positive. Insufficient.
Statement 2 tells us: X^2>X, so X<0 OR X>1. Insufficient.

Combining the statements, X>1. SOLVED...

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15 Mar 2011, 03:56
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2
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Moved to DS subforum.

thesfactor wrote:
I have been trying to understand inequalities by reading MGMAT's VIC book and am confused. I was trying to solve a few problems and had some questions. I'd appreciate your help...

MGMAT VIC book (Inequalities : Advanced Set)
Page 187, Problem 9:

Is X^3 > X^2?
1) X>0
2) X^2 > X

1.
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

2. You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Also note that in your example there is only one inequality and MGMAT solution just manipulates within it.

Back to the original question:

Is x^3 > x^2?

Is $$x^3 > x^2$$? --> is $$x^3-x^2>0$$? --> is $$x^2(x-1)>0$$? --> this inequality holds true for $$x>1$$. So the question basically asks whether $$x>1$$.

(1) x > 0. Not sufficient.
(2) x^2 > x --> $$x(x-1)>0$$ --> $$x<0$$ or $$x>1$$. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) gives: $$x>1$$. Sufficient.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Similar questions:
inequality-110191.html
friday-algebra-ds-108395.html
last-one-for-today-ds-algebra-108217.html
one-more-algebra-ds-108207.html
algebra-ds-108110.html
ds-algebra-107401.html
inequalities-concept-based-ds-107397.html
quant-review-2nd-edition-ds-104028.html
inequality-98674.html
inequality-ds-100086.html

DS questions on inequalities to practice: search.php?search_id=tag&tag_id=184
PS questions on inequalities to practice: search.php?search_id=tag&tag_id=189

Hope it helps.
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15 Mar 2011, 04:12
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Is X^3 > X^2?
1) X>0
2) X^2 > X

I have seen addition and subtraction of two inequalities. It helps us solve many inequality question. Usually, the inequalities can be subtracted when two inequalities have opposite sign;

1<2 ---A
100>50 ----B

Subtract B from A because the two inequalities have opposite signs; first inequality has "<" sign and second inequality has ">".
1-100<2-50 ; The sign "<" will remain that of A's
-99<-48

Perhaps you will need to read gmatclub's Number properties.

Coming to the question:
Is X^3 > X^2?
1) X>0
2) X^2 > X

X^3>X^2
X^3-X^2>0
X^2(X-1)>0
X>1

Question is reduced to:
Is X>1?

1. X>0.
So, X can be greater than 1 or smaller than 1.
Not sufficient.

2. X^2 > X
X^2-X>0
X(X-1)>0
X>1 or X<0
Not Sufficient.

Combining both;
X>1.
Sufficient.

Ans: "C"
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15 Mar 2011, 06:36
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X^3 - x^2 > 0

x^2(x-1) > 0

So either x^2 > 0 and x-1 > 0, because x^2 > 0 is always true therefore x > 1 has to be true

Note that the other part is not possible, i.e, x^2 < 0 and x-1 < 0

Now, (1) is insuff as we can't infer if x > 1 from this.

From (2)

x^2 - x > 0

x(x-1) > 0

For this to be true, either x > 0 and x-1 > 0, (which cannot be inferred)

or x < 0 and x < 1, which is contrary to what should happen, hence insufficient

But on taking (1) and (2) together, x > 0, so x > 1, hence sufficient, answer is C.
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15 Mar 2011, 11:42
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Step 1) The first thing I notice is that one exponent is odd and the other is even. So I immediately know what it's testing.

Step 2) I know I need to check positive and negative...but I also need to check fractions as well.

Check positive/negative:
If x = 2, then of course x^3 > x^2 [GOOD]
If x= -2, then x^3 is NOT > x^2 [NO GOOD]

So (A) and (D) are no good since there's inconsistency here with GOOD/NO GOOD.
For now, I don't need to check fractions. Just evaluate (2) for now.

Step 3) Evaluating (2): x^2 > x
Genearlly, squaring a number means it gets bigger. BUT--I know there's an exception--FRACTIONS!
If x = 1/2, then (1/2)^2 = (1/4)
So when you square a fraction, it actually gets smaller!
OK, so (2) is telling us: "Let's use only non-fractions."

Obviously, if we plug in a normal number like 3, it's gonna be a [TRUE] statement.
What happens if we plugin a negative number to x^3 > x^2?

Try x=-2. Well, then you'd get -8>+4 [FALSE].
So with x=+2, you get TRUE, but with x=-2, you get FALSE. .

So since you sometimes get FALSE and sometimes get TRUE---then you know this conflict means (B) is NO GOOD!

Step 4) But what about if you combine (1) and (2) together.

Well, (1) basically tells us we can restrict x to be positive.
(2) basically tells us we can restrict x to be only non-fractions.

So what if we only use x values that are positive AND non-fractions. That means x>=1.

Can we answer this original question definitively?

Is x^3 > X^2?

Well, when we only use x>1, then this statement is ALWAYS true. So voila!

When we combine (1) and (2) we can restrict the scope of possible X's to only those that are POSITIVE and NON-FRACTIONS--which means x>1.

Turns out this works out great. So we can choose (C) is the final answer.

http://www.gmatpill.com/practice-questi ... exponents/
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15 Mar 2011, 12:56
Thank you all for the awesome explanations. It makes more sense now.

"You can only apply subtraction when their signs are in the opposite directions:

If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from).
Example: 3<4 and 5>1 --> 3-5<4-1."

This was definitely news to me. Glad to know...
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15 Mar 2011, 13:26
Bunuel wrote:
Moved to DS subforum.

thesfactor wrote:
I have been trying to understand inequalities by reading MGMAT's VIC book and am confused. I was trying to solve a few problems and had some questions. I'd appreciate your help...

MGMAT VIC book (Inequalities : Advanced Set)
Page 187, Problem 9:

Is X^3 > X^2?
1) X>0
2) X^2 > X

1.
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

2. You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Also note that in your example there is only one inequality and MGMAT solution just manipulates within it.

Back to the original question:

Is x^3 > x^2?

Is $$x^3 > x^2$$? --> is $$x^3-x^2>0$$? --> is $$x^2(x-1)>0$$? --> this inequality holds true for $$x>1$$. So the question basically asks whether $$x>1$$.

(1) x > 0. Not sufficient.
(2) x^2 > x --> $$x(x-1)>0$$ --> $$x<0$$ or $$x>1$$. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) gives: $$x>1$$. Sufficient.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Similar questions:
inequality-110191.html
friday-algebra-ds-108395.html
last-one-for-today-ds-algebra-108217.html
one-more-algebra-ds-108207.html
algebra-ds-108110.html
ds-algebra-107401.html
inequalities-concept-based-ds-107397.html
quant-review-2nd-edition-ds-104028.html
inequality-98674.html
inequality-ds-100086.html

DS questions on inequalities to practice: search.php?search_id=tag&tag_id=184
PS questions on inequalities to practice: search.php?search_id=tag&tag_id=189

Hope it helps.

Thanks for the detailed post. I apologize for posting this in the wrong forum - duly noted.

You solved (2) x^2 > x --> x(x-1)>0 --> x<0 or x>1. Not sufficient.
What did you do to arrive at this conclusion. The way I see it, x(x-1)>0 is an xy inequality...where x can be -ve and y +ve OR vice versa
Therefore
X>0 and X<1
OR
X<0 and X>1

What's wrong with my logic here?

Finally, how do we get (1)+(2) Intersection of the ranges from (1) and (2) gives: x>1. Sufficient.
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15 Mar 2011, 13:29
fluke wrote:
Is X^3 > X^2?
1) X>0
2) X^2 > X

I have seen addition and subtraction of two inequalities. It helps us solve many inequality question. Usually, the inequalities can be subtracted when two inequalities have opposite sign;

1<2 ---A
100>50 ----B

Subtract B from A because the two inequalities have opposite signs; first inequality has "<" sign and second inequality has ">".
1-100<2-50 ; The sign "<" will remain that of A's
-99<-48

Perhaps you will need to read gmatclub's Number properties.

Coming to the question:
Is X^3 > X^2?
1) X>0
2) X^2 > X

X^3>X^2
X^3-X^2>0
X^2(X-1)>0
X>1

Question is reduced to:
Is X>1?

1. X>0.
So, X can be greater than 1 or smaller than 1.
Not sufficient.

2. X^2 > X
X^2-X>0
X(X-1)>0
X>1 or X<0
Not Sufficient.

Combining both;
X>1.
Sufficient.

Ans: "C"

Thanks for the answer, I have a question here too...how did you arrive at this conclusion:
X^3>X^2
X^3-X^2>0
X^2(X-1)>0
X>1

The way I see it: X^2(X-1)>0 is an xy>0 solution where either x>0 and y<0 OR x<0 and y>0
Therefore in this case, X^2>0 and X-1<0 which means X>0 and X<1 ----- (A)
OR
X^2<0 and X-1>0 which means X<0 and X>1 ----- (B)

When we combine statements 1 and 2 we know that X>0 in (1)
Therefore (B) is not possible because X cannot be less than 0...what am I doing wrong here...
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15 Mar 2011, 13:41
thesfactor wrote:

Finally, how do we get (1)+(2) Intersection of the ranges from (1) and (2) gives: x>1. Sufficient.

I'll stay away from the top part of your question. As for the bottom:
[quote2gmatpill]Well, (1) basically tells us we can restrict x to be positive. (x>0)
(2) basically tells us we can restrict x to be only values outside of the -1 to 1 range.

So what if we only use x values that are positive AND non-fractions. That means x>1.[/quote2]

Hope that helps! By the way, be careful with all these symbols and calculations you are doing. The GMAT tests your REASONING skills and your ability to "make sense" of it. I may be wrong--but I get the sense you're getting too bogged down on the nitty gritty details yet don't really understand how to translate the math symbols into logic and reasoning.
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16 Mar 2011, 02:10
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thesfactor wrote:
Thanks for the answer, I have a question here too...how did you arrive at this conclusion:
X^3>X^2
X^3-X^2>0
X^2(X-1)>0
X>1

The way I see it: X^2(X-1)>0 is an xy>0 solution where either x>0 and y<0 OR x<0 and y>0
Therefore in this case, X^2>0 and X-1<0 which means X>0 and X<1 ----- (A)
OR
X^2<0 and X-1>0 which means X<0 and X>1 ----- (B)

When we combine statements 1 and 2 we know that X>0 in (1)
Therefore (B) is not possible because X cannot be less than 0...what am I doing wrong here...

X^2(X-1)>0 is an xy>0 solution where either x>0 and y<0 OR x<0 and y>0

This is not correct.

If a*b > 0; It means either a and b are both +ve OR a and b are both -ve.
If a*b < 0; It means either a is +ve and b is -ve OR a is -ve and b is +ve.

X^2(X-1)>0

Here; either X^2 and (X-1) are both negative OR X^2 and (X-1) are both +ve

X^2 can't be -ve.
Note:
A square can never be a negative value. It can be 0 or +ve.
X^2 is also not 0. Because if X=0; then X^2=0; That will make X^2(X-1) = 0; but X^2(X-1)>0. Thus, X is not 0.

So, X^2 is +ve and X-1 is also +ve.
X-1>0
X>1 and X>0
Combing both; X>1.

So; the ultimate question becomes;
Is X>1

1. This statement tells us that X>0.
X can be 0.5<1 or 2>1.
Thus not sufficient.

2. X^2>X
X^2-X>0
X(X-1)>0

Again a*b>0; a +ve and b+ve OR a-ve and b-ve

X(X-1)>0
X>0 and X-1>0 i.e. X>1
Combing both; X>1
OR
X<0 and X-1<0 i.e. X<1
Combing both; X<0

We have X>1 or X<0
Not sufficient.

Combing both;
Statement 1 tells us that X>0
Statement 2 tells us that X could be less than 0 or greater than 1. Since St 1 tells us that X>0; we can ignore the St2 condition where it says X<0.
Thus, X>0 and X>1. Means X>1

C.
***************************************************

These questions become easier if you knew how to find proper ranges of X in the inequality. Get yourself acquainted with the following trick:
http://gmatclub.com/forum/inequalities-trick-91482.html

This was little cumbersome for me in the beginning. But, with adequate practice and guidance from Bunuel, I am getting hold of it. Yet to become an exponent though. It's really proven fruitful for me.

Good luck!!
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16 Mar 2011, 02:12
thesfactor wrote:
Thanks for the answer, I have a question here too...how did you arrive at this conclusion:
X^3>X^2
X^3-X^2>0
X^2(X-1)>0
X>1

The way I see it: X^2(X-1)>0 is an xy>0 solution where either x>0 and y<0 OR x<0 and y>0
Therefore in this case, X^2>0 and X-1<0 which means X>0 and X<1 ----- (A)
OR
X^2<0 and X-1>0 which means X<0 and X>1 ----- (B)

When we combine statements 1 and 2 we know that X>0 in (1)
Therefore (B) is not possible because X cannot be less than 0...what am I doing wrong here...

X^2(X-1)>0 is an xy>0 solution where either x>0 and y<0 OR x<0 and y>0

This is not correct.

If a*b > 0; It means either a and b are both +ve OR a and b are both -ve.
If a*b < 0; It means either a is +ve and b is -ve OR a is -ve and b is +ve.

X^2(X-1)>0

Here; either X^2 and (X-1) are both negative OR X^2 and (X-1) are both +ve

X^2 can't be -ve.
Note:
A square can never be a negative value. It can be 0 or +ve.
X^2 is also not 0. Because if X=0; then X^2=0; That will make X^2(X-1) = 0; but X^2(X-1)>0. Thus, X is not 0.

So, X^2 is +ve and X-1 is also +ve.
X-1>0
X>1 and X>0
Combing both; X>1.

So; the ultimate question becomes;
Is X>1

1. This statement tells us that X>0.
X can be 0.5<1 or 2>1.
Thus not sufficient.

2. X^2>X
X^2-X>0
X(X-1)>0

Again a*b>0; a +ve and b+ve OR a-ve and b-ve

X(X-1)>0
X>0 and X-1>0 i.e. X>1
Combing both; X>1
OR
X<0 and X-1<0 i.e. X<1
Combing both; X<0

We have X>1 or X<0
Not sufficient.

Combing both;
Statement 1 tells us that X>0
Statement 2 tells us that X could be less than 0 or greater than 1. Since St 1 tells us that X>0; we can ignore the St2 condition where it says X<0.
Thus, X>0 and X>1. Means X>1

C.
***************************************************

These questions become easier if you knew how to find proper ranges of X in the inequality. Get yourself acquainted with the following trick:
http://gmatclub.com/forum/inequalities-trick-91482.html

This was little cumbersome for me in the beginning. But, with adequate practice and guidance from Bunuel, I am getting hold of it. Yet to become an exponent though. It's really proven fruitful for me.

Good luck!!
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17 Mar 2011, 22:03
Thanks for the post. You're right, I might be getting bogged down in the details. I need to see through the haze...
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18 Mar 2011, 17:20
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.
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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
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Re: Is x^3 > x^2? [#permalink]

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23 Apr 2015, 15:13
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

[Reveal] Spoiler:
C

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Re: Is x^3 > x^2?   [#permalink] 23 Apr 2015, 15:13

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