Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 20 May 2013, 20:41

# Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0

Author Message
TAGS:
SVP
Joined: 05 Jul 2006
Posts: 1564
Followers: 4

Kudos [?]: 63 [1] , given: 34

Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  15 Aug 2009, 14:58
1
KUDOS
00:00

Question Stats:

58% (02:28) correct 41% (01:52) wrong based on 10 sessions
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Mar 2012, 21:00, edited 1 time in total.
Edited the question and added the OA
Director
Joined: 01 Apr 2008
Posts: 921
Followers: 8

Kudos [?]: 123 [0], given: 18

Re: Is |x| < 1? [#permalink]  15 Aug 2009, 22:37
C

1)
if x>1, solving the equality we get x=3.
if -1<x<1, we get x=1/3.
if x<-1, we get x=3.

Not suff
2)
we get x is not equal to 3.

Combining, x can only be 1/3
OA?

Last edited by Economist on 15 Aug 2009, 23:01, edited 1 time in total.
Intern
Joined: 28 Mar 2009
Posts: 12
Followers: 0

Kudos [?]: 2 [2] , given: 0

Re: Is |x| < 1? [#permalink]  15 Aug 2009, 22:51
2
KUDOS
is -1<x<1 ?

1) |x + 1| = 2|x - 1|

two possibilities:
a) x+1=2(x-1) or x=3 (consider x+1 and x-1 to be of same sign)
b) x+1=2(1-x) or x=1/3 (consider x+1 and x-1 to be of different signs)
not sufficient, x may or may not be between -1 and 1

2) |x - 3| ≠ 0
=> x≠3
not sufficient, x may or may not be between -1 and 1

together, x≠3 so x is 1/3 which is between -1 and 1
sufficient

hence, C
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [9] , given: 826

Re: Is |x| < 1? [#permalink]  23 Oct 2009, 20:18
9
KUDOS
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1|
Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

So we got TWO values of x (two solutions): \frac{1}{3} and 3, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) |x - 3|\neq{0}
Just says that x\neq{3}. But we don't know whether x is in the range (-1,1) or not.

(1)+(2) x=\frac{1}{3} or x=3 AND x\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Hope it helps.
_________________
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 319
Followers: 2

Kudos [?]: 76 [0], given: 37

Re: Is |x| < 1? [#permalink]  23 Oct 2009, 20:23
I got C too substituting values and evaluating each of those.

thanks Bunuel and Economist for elaborations
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

CEO
Joined: 17 Nov 2007
Posts: 3594
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 230

Kudos [?]: 1299 [6] , given: 346

Re: Is |x|< 1? [#permalink]  09 Nov 2009, 22:26
6
KUDOS
|x|<1 means x e (-1,1)

1)|x+1|=2|x-1|
There are 2 key points (x=-1 and x=1) where one of the expressions under modules passes 0 and change its sign.
x<-1: -(x+1)=-2(x-1) --> -x-1=-2x+2 ---> x = 3. But x=3 does not satisfy x<-1 condition.
-1<=x<1: x+1 = -2(x-1) ---> x+1= -2x+2 --> x=1/3. It satisfies the condition.
x>=1: x+1=2x-2 ---> x=3. It satisfies the condition.
Insufficient

2) |x - 3| ≠ 0
x≠3
Insufficient

1&2) Only x=1/3 satisfies both statements. Sufficient.

If you find it useful/useless, let me know
_________________

iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free* (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.

SVP
Joined: 30 Apr 2008
Posts: 1900
Location: Oklahoma City
Schools: Hard Knocks
Followers: 25

Kudos [?]: 339 [0], given: 32

Re: Is |x|< 1? [#permalink]  09 Nov 2009, 22:32
Answer is C - you beat me to it walker

For statement 1, you need to actually break it down to 2 separate statements to account for both scenarios:

Option #1 ->
x + 1 = 2(x-1)
x + 1 = 2x -2
x = 2x -3
3 + x = 2x
3 = x {or also Option #2}

Option #2 -> x + 1 = -2(x-1)
x + 1 = -2x +2
x = -2x + 1
3x = 1
x = 1/3

Statement (1) is insufficient because |x| could be 3 or 1/3. So the answer is sometimes yes, sometimes no which means insufficient.

Statement (2) is insufficient because we are told that |x -3| ≠ 0. This means that as long as x ≠ 3, then we're ok. There are far too many options so we certainly have a sometimes yes, and sometimes no.

But together, from Statement (1) we have either 3 or 1/3 and from statement (2) we have everything BUT 3. So the only overlap we have for possible values to make both statements true is 1/3. So the answer is Yes, |x| IS less than 1 and we have enough info to answer the question.

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

Find out what's new at GMAT Club - latest features and updates

SVP
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1756
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 50

Kudos [?]: 145 [0], given: 108

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  16 Jan 2012, 08:48
+1 C

However, I used using other methods: a) Squaring both sides of the equation because |x| = (x^2)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: my-ir-logbook-diary-133264.html

Find out what's new at GMAT Club - latest features and updates

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [0], given: 826

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  16 Jan 2012, 14:12
metallicafan wrote:
+1 C

However, I used using other methods: a) Squaring both sides of the equation because |x| = (x^2)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?

It's about expanding the absolute value in different ranges. Discussed in Walker's post on absolute values: math-absolute-value-modulus-86462.html

If you're more comfortable with "squaring" method (note that it's not always applicable) then you can apply it to the first statement too (and not only to the stem).

Is |x| < 1?

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

(1) |x + 1| = 2|x - 1| --> square both sides: (x+1)^2=4(x-1)^2 --> x^2+2x+1=4x^2-8x+4 --> 3x^2-10x+3=0 --> x=\frac{1}{3} or x=3 --> first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) |x - 3|\neq{0} --> just says that x\neq{3}. But we don't know whether x is in the range (-1,1) or not.

(1)+(2) x=\frac{1}{3} or x=3 AND x\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Hope it's clear.
_________________
Manager
Joined: 14 Feb 2012
Posts: 226
Followers: 1

Kudos [?]: 6 [0], given: 7

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  24 Apr 2012, 04:04
Hi Bunuel ,
I also used the squaring method....
In what all situations is the squaring method not applicable ???
Can you please explain why ?
_________________

The Best Way to Keep me ON is to give Me KUDOS !!!
If you Like My posts please Consider giving Kudos

Shikhar

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [1] , given: 826

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  24 Apr 2012, 05:43
1
KUDOS
shikhar wrote:
Hi Bunuel ,
I also used the squaring method....
In what all situations is the squaring method not applicable ???
Can you please explain why ?

Sure, for example it's not always applicable for inequalities.

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
2<4 --> we can square both sides and write: 2^2<4^2;
0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;

But if either of side is negative then raising to even power doesn't always work.
For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
-2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3;
x<y --> we can raise both sides to third power and write: x^3<y^3.

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.
_________________
Manager
Joined: 16 Mar 2010
Posts: 59
Followers: 0

Kudos [?]: 4 [0], given: 4

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  28 Apr 2012, 10:42
Nice explaination Bunuel, Thanks! 1+
Intern
Joined: 20 Mar 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: Is |x| < 1? [#permalink]  21 May 2012, 22:35
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1|
Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

So we got TWO values of x (two solutions): \frac{1}{3} and 3, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) |x - 3|\neq{0}
Just says that x\neq{3}. But we don't know whether x is in the range (-1,1) or not.

(1)+(2) x=\frac{1}{3} or x=3 AND x\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Hope it helps.

How is the sign applied for different range?can someone help please
X<-1
-x-1=2(-x+1)

-1<=x<=1
x+1=2(-x+1)

X>1
x+1=2(x-1)
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [0], given: 826

Re: Is |x| < 1? [#permalink]  21 May 2012, 23:18
sdpkind wrote:
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1|
Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

So we got TWO values of x (two solutions): \frac{1}{3} and 3, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) |x - 3|\neq{0}
Just says that x\neq{3}. But we don't know whether x is in the range (-1,1) or not.

(1)+(2) x=\frac{1}{3} or x=3 AND x\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Hope it helps.

How is the sign applied for different range?can someone help please
X<-1
-x-1=2(-x+1)

-1<=x<=1
x+1=2(-x+1)

X>1
x+1=2(x-1)

Absolute value properties:
When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(some \ expression)}. For example: |-5|=5=-(-5);

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|\leq{some \ expression}. For example: |5|=5;

Applying this to |x + 1| = 2|x - 1|.

If x<-1 then x+1<0 and x-1<0 so |x + 1| =-(x+1) and 2|x - 1|=-2(x-1) which means that |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3.

Similarly for other ranges.

For more check Absolute Value chapter for Math Book: math-absolute-value-modulus-86462.html

Hope it helps.
_________________
Intern
Joined: 27 Dec 2011
Posts: 37
Followers: 0

Kudos [?]: 1 [0], given: 5

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  22 May 2012, 20:28
Hi Bunuel,

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)"
Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here?
How are you deciding the sign and coming here: x+1=2(-x+1)
I couldn't understand this

I have seen this post: http://gmatclub.com/forum/math-absolute-value-modulus-86462.html as you referred above but ctill couldn't understand it.

I would really appreciate if you can throw a little light on this.

Thanks,
Intern
Joined: 01 Apr 2012
Posts: 26
Location: United States
Concentration: Technology, Economics
GMAT Date: 05-13-2012
WE: Consulting (Computer Software)
Followers: 0

Kudos [?]: 10 [0], given: 18

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  22 May 2012, 23:05
hey kartik,

i also had the same issue initially...then i went thru the link Bunuel has given and it toook me a little time to get it....

so seee.....
the Question states

|x+1| = 2|x-1|
if we take the number line

--------------{-1}----------------{1}---------------

|x+1| if u see will be negative till -1 and after that it will be positive
|x-1| will be negative till 1 and will be positive after that.....

Now we have 3 areas where we need to check

x<-1
-1<=x<=1
x>1

If u check the number line till {-1} both the expression's value will be negative......so when we solve x<-1 we take both of them as Negative

but when we solve -1<=x<=1 the expression |x+1| is positive as its sign changed on x=-1 but the other one |x-1| is still Negative as its sign would change on {1}.
therefore when we solve of this range we take |x+1|= x+1 and 2|x-1| = -2(x-1) = 2(-x+1)

in the third one both are Positive os both the expressions will be positive....

I hope this help you...as it also took me a lot of time to get this concept in my head.........
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [0], given: 826

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  22 May 2012, 23:30
kartik222 wrote:
Hi Bunuel,

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)"
Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here?
How are you deciding the sign and coming here: x+1=2(-x+1)
I couldn't understand this

I have seen this post: http://gmatclub.com/forum/math-absolute-value-modulus-86462.html as you referred above but ctill couldn't understand it.

I would really appreciate if you can throw a little light on this.

Thanks,

This issue is discussed here: is-x-1-1-x-1-2-x-1-2-x-82478.html#p1088086

Absolute value properties:
When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(some \ expression)}. For example: |-5|=5=-(-5);

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|\leq{some \ expression}. For example: |5|=5;

Applying this to |x + 1| = 2|x - 1|.

If -1\leq{x}\leq{1} then x+1>0 and x-1<0 so |x + 1| =x+1 and 2|x - 1|=-2(x-1) which means that |x + 1| = 2|x - 1| becomes: x+1=2(-x+1).

Hope it's clear.
_________________
Manager
Status: exam is close ... dont know if i ll hit that number
Joined: 06 Jun 2011
Posts: 209
Location: India
GMAT Date: 10-09-2012
GPA: 3.2
Followers: 2

Kudos [?]: 6 [0], given: 1

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  30 Aug 2012, 20:12
good explanation sir thanks
a lot learnt new concept in this
_________________

just one more month for exam...

Manager
Joined: 27 May 2010
Posts: 205
Followers: 2

Kudos [?]: 7 [0], given: 3

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  31 Aug 2012, 02:36
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11522
Followers: 1795

Kudos [?]: 9549 [0], given: 826

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  31 Aug 2012, 02:42
qweert wrote:
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.

Check our Question Banks: viewforumtags.php

PS questions on Absolute Values: search.php?search_id=tag&tag_id=58
DS questions on Absolute Values: search.php?search_id=tag&tag_id=37

Tough inequality and absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Theory on absolute values: math-absolute-value-modulus-86462.html

Hope it helps.
_________________
Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0   [#permalink] 31 Aug 2012, 02:42
Similar topics Replies Last post
Similar
Topics:
Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 6 01 Dec 2005, 00:32
Is |x| < 1 1) |x+1| = 2 |x-1| 2) |x-3| not equal to 0 If 3 07 Dec 2005, 21:12
Is |x| < 1 (1) |x + 1| = 2|x -1| (2) |x - 3| <> 0 1 05 Aug 2007, 09:06
Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 A. 2 09 Feb 2008, 21:13
Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 4 19 Mar 2008, 23:47
Display posts from previous: Sort by