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

Inequalities are definitely my nemesis...It's worse when they are combined with DS. I HATE YOU DS!

Moved to DS subforum.



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.

Answer: C.

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

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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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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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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...

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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Thanks for the post. You're right, I might be getting bogged down in the details. I need to see through the haze...