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Is |x| < 1 [#permalink] New post 05 Oct 2011, 05:00
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Is |x| < 1

(1) x^5 > x^3
(2) x^2 > x^3
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Nov 2013, 05:30, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Modulus question [#permalink] New post 05 Oct 2011, 05:40
Question is basically asking if X is a fraction or 0.

A is insufficient because we can see by testing numbers that it can be either a negative fraction or a positive integer and be true. For example:
If X = -1/2 that gives us -1/32 > -1/8 which is true.
If X = 2, the equation is 32 > 8 which is also true.

B is insufficient because it can be any fraction or negative number and still fulfill this prompt. For example:

If X = -1/2 then we get 1/4 > -1/8 which is true
If X = 1/2 then we get 1/4 > 1/8 which is true
If X = -2 then we get 4 > -8 which is true

The answer is C because together we know that X must be a negative fraction.
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Re: Modulus question [#permalink] New post 05 Oct 2011, 05:55
gmatblues wrote:

is |x| < 1

A. x^5 > x^3
B. X^2 > X^3


There is a way to simplify both statements.
We know that x^2 is positive.
Divide both sides by x^2 in both statements
simplified statements

1. x^3>x
2. 1>x

1. x can be a negative fraction or a positive number greater than 1.
Insufficient

2. x can be negative or a positive fraction

1+2
x is a negative fraction.
Hence sufficient.
C

Is this approach clear?
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Re: Modulus question [#permalink] New post 05 Oct 2011, 06:43
Good question.

If we draw the graph, we can figure out that individually both the option doesn't satisfy but if we join together, the value of x satisfy between the region - (-1< x < 0)

Hence, C is the answer.

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Re: Modulus question [#permalink] New post 05 Oct 2011, 06:55
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Ajay369 wrote:
Good question.

If we draw the graph, we can figure out that individually both the option doesn't satisfy but if we join together, the value of x satisfy between the region - (-1< x < 0)

Hence, C is the answer.

Cheers!


Good thing to draw a number line to solve this type of questions.
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Re: Modulus question [#permalink] New post 05 Oct 2011, 07:26
gmatblues wrote:
is |x| < 1

1) x^5 > x^3
2) X^2 > X^3


Is -1<x<1?

1)
x^5-x^3>0
x^3(x^2-1)>0
x(x^2-1)>0
(x-1)x(x+1)>0
x>1 OR -1<x<0
So, x can be 2 OR -0.5
Not Sufficient.

2)
x^3-x^2<0
x^2(x-1)<0
x<1 but x != 0
x can be 0.5 or -2.
Not Sufficient.

Combining both;
-1<x<0
Sufficient.

Ans: "C"
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Re: Modulus question [#permalink] New post 05 Oct 2011, 10:40
Thanks for the explanations. I found drawing the graph helped me understanding the solution. @blink005 : the approach is simple and clear.
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Re: Modulus question [#permalink] New post 05 Oct 2011, 11:11
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is |x| < 1

A. x^5 > x^3
B. X^2 > X^3


A X^5 > X^3

If X = 1/2 X^5 = 1/32;X^3 = 1/8 satisfies if it is negative then it does not so A not enough

B X^2 > X^3

If X = 1/2 then satisfies X = -1/2 satisfies and then if X = 2 then it does n't and if X = -2 then it does


putting together we are getting

C
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Re: Modulus question [#permalink] New post 28 Nov 2013, 04:36
fluke wrote:
gmatblues wrote:
is |x| < 1

1) x^5 > x^3
2) X^2 > X^3


Is -1<x<1?

1)
x^5-x^3>0
x^3(x^2-1)>0
x(x^2-1)>0
(x-1)x(x+1)>0
x>1 OR -1<x<0
So, x can be 2 OR -0.5
Not Sufficient.

2)
x^3-x^2<0
x^2(x-1)<0
x<1 but x != 0
x can be 0.5 or -2.
Not Sufficient.

Combining both;
-1<x<0
Sufficient.

Ans: "C"


I did this question with key points but I'm having a small doubt here

In 1, I did the same way as fluke
Key points were -1,0 and 1 so I ended up with ranges -1<x<0 and x>1
Therefore not sufficient

In 2, I ended up with x^2(x-1)<0
Now what I did was use 0 and 1 as key points but actually what I got was

- infintive---(1)---0---(2)-----1---(3)-----+ Infinite

In range (1), the expression is negative too so it is part of the range
In range (2), the expression is negative also so it is part of the range
In range (3), the expression is positive, so it does not belong to the range

Now even though I got x<1 for the range the same as fluke
I'm wondering if I did the key points correctly because we should have +,-,+,- in that order for the ranges.
I am guessing it has something to do with the fact that x^2 will always yield a positive number

Bunuel would you please help me out on this one

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Re: Is |x| < 1 [#permalink] New post 28 Nov 2013, 05:40
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Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif [ 1.5 KiB | Viewed 933 times ]
Sufficient.

Answer: C.
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Re: Is |x| < 1 [#permalink] New post 28 Nov 2013, 19:13
Bunuel wrote:
Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
Sufficient.

Answer: C.


How do you do it
(x+1)x(x-1)>0 --> [m]-1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:
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Re: Is |x| < 1 [#permalink] New post 28 Nov 2013, 23:20
Expert's post
rango wrote:
Bunuel wrote:
Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
Sufficient.

Answer: C.


How do you do it
(x+1)x(x-1)>0 --> [m]-1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:


One statement says "x must be either greater than 1 OR between -1 and 0"
Another says "x must be less than 1"

So what values can you give x such that both statements are satisfied? You cannot have x greater than 1 since second statement says x must be less than 1. So x must be between -1 and 0. This satisfies both the statements.
-1 < x < 0
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Re: Is |x| < 1 [#permalink] New post 29 Nov 2013, 00:51
VeritasPrepKarishma wrote:
rango wrote:
Bunuel wrote:
Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
Sufficient.

Answer: C.


How do you do it
(x+1)x(x-1)>0 --> [m]-1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:


One statement says "x must be either greater than 1 OR between -1 and 0"
Another says "x must be less than 1"

So what values can you give x such that both statements are satisfied? You cannot have x greater than 1 since second statement says x must be less than 1. So x must be between -1 and 0. This satisfies both the statements.
-1 < x < 0


Thanks , Sorry to say BUT Do not got the point.

:shock:
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Re: Is |x| < 1 [#permalink] New post 29 Nov 2013, 01:09
Expert's post
1
This post was
BOOKMARKED
rango wrote:
Bunuel wrote:
Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
Sufficient.

Answer: C.


How do you do it
(x+1)x(x-1)>0 --> -1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:


I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1;
-1<x<0;
0<x<1;
x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

Theory on 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
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is |x| < 1 [#permalink] New post 29 Nov 2013, 05:11
Bunuel wrote:
rango wrote:
Bunuel wrote:
Is |x| < 1

Is |x| < 1? --> is -1<x<1?

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.
Attachment:
MSP824621h8d31791668agg000062b1e415iai4e4i2.gif
Sufficient.

Answer: C.


How do you do it
(x+1)x(x-1)>0 --> -1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:


I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1;
-1<x<0;
0<x<1;
x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

Theory on 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
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.


Yes i was asking for the first sentence; i wonder about the inequality x(x+1) (x-1) >0
x>0
x+1>0; x>-1
x-1>0; x>+1
To satisfy the Q Is [X] <1 ; so the above range will be displayed as two options -1<x<0 / 0<x<+1.

Is that what you have suggested in previous equations ?

But i am still skeptical about the Inequality.
By the way thanks for help; Appreciate .

:lol:
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Re: Is |x| < 1 [#permalink] New post 29 Nov 2013, 05:14
Expert's post
rango wrote:
Bunuel wrote:
rango wrote:
How do you do it
(x+1)x(x-1)>0 --> -1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link


:shock:


I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1;
-1<x<0;
0<x<1;
x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

Theory on 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
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.


Yes i was asking for the first sentence; i wonder about the inequality x(x+1) (x-1) >0
x>0
x+1>0; x>-1
x-1>0; x>+1
To satisfy the Q Is [X] <1 ; so the above range will be displayed as two options -1<x<0 / 0<x<+1.

Is that what you have suggested in previous equations ?

But i am still skeptical about the Inequality.
By the way thanks for help; Appreciate .

:lol:


Please follow the links provided in my previous post.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is |x| < 1 [#permalink] New post 17 Dec 2013, 23:20
For second option I tried to solve it with handy trick.
x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

You can refer trick on link : inequalities-trick-91482.html#p804990

I am not sure where I am making mistake with this trick.
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Re: Is |x| < 1 [#permalink] New post 18 Dec 2013, 00:09
Expert's post
PiyushK wrote:
For second option I tried to solve it with handy trick.
x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

You can refer trick on link : inequalities-trick-91482.html#p804990

I am not sure where I am making mistake with this trick.


The squared term, x^2, should be omitted because it's always positive and thus doesn't affect the sign of the left side.
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Re: Is |x| < 1 [#permalink] New post 18 Dec 2013, 03:30
Ohh Thanks Bunuel I understood and can relate it to this trick.
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Re: Is |x| < 1 [#permalink] New post 01 Feb 2014, 07:22
Bunuel wrote:
PiyushK wrote:
For second option I tried to solve it with handy trick.
x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

You can refer trick on link : inequalities-trick-91482.html#p804990

I am not sure where I am making mistake with this trick.


The squared term, x^2, should be omitted because it's always positive and thus doesn't affect the sign of the left side.


Ahhh now I get it, that's awesome thanks. Still I don't quite know what's wrong with including x^2 in the key points, I know it will give you a different valid range but I would like to understando the logic behind it.

Cheers
J
Re: Is |x| < 1   [#permalink] 01 Feb 2014, 07:22
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