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M24-03

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New post 16 Sep 2014, 01:20
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A
B
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D
E

Difficulty:

  55% (hard)

Question Stats:

57% (00:52) correct 43% (01:14) wrong based on 138 sessions

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New post 16 Sep 2014, 01:20
1
Official Solution:


(1) \(|1-x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A
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New post 29 Jan 2015, 04:31
Bunuel wrote:
Official Solution:


(1) \(|1−x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A

|1-x|<1
will have two parts to the solution

1-x<1 =====>x-1>-1==========> x>0
and -(1-x)<1======>x-1<1 =====> x<2

So does this mean that x lies between 0 and 2???

Or does it mean that either X is greater than 0 or X is less than 2.

Because if it means that x is less than 2 or x is greater than 0, then x is less than 2 could also signify that x is negative.
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New post 29 Jan 2015, 05:30
mayankpant wrote:
Bunuel wrote:
Official Solution:


(1) \(|1−x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A

|1-x|<1
will have two parts to the solution

1-x<1 =====>x-1>-1==========> x>0
and -(1-x)<1======>x-1<1 =====> x<2

So does this mean that x lies between 0 and 2???

Or does it mean that either X is greater than 0 or X is less than 2.

Because if it means that x is less than 2 or x is greater than 0, then x is less than 2 could also signify that x is negative.


\(|1−x| \lt 1\) translates to 0<x<2 as written above.
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New post 02 Mar 2016, 08:26
Bunuel wrote:
Official Solution:


(1) \(|1-x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A


in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!
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New post 03 Mar 2016, 07:27
nishantdoshi wrote:
Bunuel wrote:
Official Solution:


(1) \(|1-x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A


in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!


No. |x-1| is the distance between x and 1, so is |1-x|. Thus those two are the same.
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New post 06 Mar 2016, 22:10
Bunuel wrote:
nishantdoshi wrote:
Bunuel wrote:
Official Solution:


(1) \(|1-x| \lt 1\). From this statement it follows that \(-1 \lt 1-x \lt 1\). Add -1 to all parts: \(-2 \lt -x \lt 0\). Now multiply by -1: \(2 \gt x \gt 0\). So, \(x\) is positive. Sufficient.

(2) \(|1+x| \gt 1\). From this statement it follows that \(x \gt 0\) or \(x \lt -2\). So, \(x\) may or may not be positive. Not sufficient.


Answer: A


in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!


No. |x-1| is the distance between x and 1, so is |1-x|. Thus those two are the same.


so you mean

if |1-x|<1 then |x-1|<1 , it wont be |x-1|>1

please clear this doubt
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New post 07 Mar 2016, 10:48
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New post 09 Nov 2016, 22:32
Hi bunuel,

I understand that when we open a mod, we have to do things:

a. Solve without considering the mod (i.e normally)
b. Change the signs of any one side of the equation.

I believe thus same process is applicable both when = sign is used and when >= or <= signs are used.

Getting very confused here.

Request you to kindly clarify.

Many Thanks.
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New post 10 Nov 2016, 07:34
dsheth7 wrote:
Hi bunuel,

I understand that when we open a mod, we have to do things:

a. Solve without considering the mod (i.e normally)
b. Change the signs of any one side of the equation.

I believe thus same process is applicable both when = sign is used and when >= or <= signs are used.

Getting very confused here.

Request you to kindly clarify.

Many Thanks.


Here are the topics that will help you to brush up fundamentals on absolute values:
Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope it helps.
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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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New post 10 Apr 2018, 03:04
+1 for option A
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New post 22 Jul 2018, 01:42
dsheth7, yes it works the same way.
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M24-03 &nbs [#permalink] 22 Jul 2018, 01:42
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