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Is | x | < 1 ? 1) | x + 1 | = 2| x -1 | 2) | x - 3 | not [#permalink]
 11 Sep 2008, 14:13
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Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0
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Re: DS ... Absolute Value [#permalink]
11 Sep 2008, 14:27
IgnitedMind wrote: Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0 1) | x + 1 | = 2| x -1 | case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient 2) | x - 3 | not equal to 0 itself not suffcieint. From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient. C
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Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 01:08
x2suresh wrote: IgnitedMind wrote: Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0 1) | x + 1 | = 2| x -1 | case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient 2) | x - 3 | not equal to 0 itself not suffcieint. From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient. C Thanks Suresh. Thats awesome.
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Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 09:31
case III) x + 1 <0 and x -1>0 --> not possible.
Why is it not possible?
|X+1| =2|x-1| would equal
-x-1 = 2x-2
then x = 1/3 ?
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Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 09:33
bigfernhead wrote: case III) x + 1 <0 and x -1>0 --> not possible.
Why is it not possible?
|X+1| =2|x-1| would equal
-x-1 = 2x-2
then x = 1/3 ? x + 1 <0 and x -1>0 --> At the same time can x<-1 and x>1 (how come this is possible)????
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Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 22:40
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IgnitedMind wrote: Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0 There's another way to deal with absolute (in)equalities. (i) | x + 1 | = 2| x -1 | square on both sidesx²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient (ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient. combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3 so i and ii combined say x=1/3. So answer is C
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Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 23:05
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amitdgr wrote: IgnitedMind wrote: Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0 There's another way to deal with absolute (in)equalities. (i) | x + 1 | = 2| x -1 | square on both sidesx²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient (ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient. combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3 so i and ii combined say x=1/3. So answer is C Good method !!! +1 for u this saves time and actually clears the fundamental
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Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 08:00
Does this always work? Squaring both sides? absolute values, equalities, and inequalities? amitdgr wrote: IgnitedMind wrote: Is | x | < 1 ?
1) | x + 1 | = 2| x -1 |
2) | x - 3 | not equal to 0 There's another way to deal with absolute (in)equalities. (i) | x + 1 | = 2| x -1 | square on both sidesx²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient (ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient. combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3 so i and ii combined say x=1/3. So answer is C |
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Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 09:10
Just to add my late two cents perhaps sombody may find this method easier.
Statement 1: |x+1| = 2*|x-1|
The easiest way to see the various scenarios is to track the critical points of the two absolute value expressions on a number line. The critical point of the expression /x + 1/ is -1 and the critical point of the expression /x - 1/ is 1
How many distinct regions are there on the number line with the points -1 and/or 1 as boundaries?
The answer is three: (-1<x), ( x<1) and (-1 < x < 1).
Now let’s test this scenario:
SCENARIO 1: when x < -1
(x+1) is –ve in this reason. So is (x-1) so the equation becomes:
-x-1 = 2*(-x+1)
=> -x-1 = -2x+2
=> x=3
SCENARIO 2: when -1 < x < 1, (x + 1) is positive in this region and (x - 1) is negative, so the equation becomes:
(x+1) = 2*(-x+1)
=> x+1 = -2x+2
=> 3x = 1
=> X = 1/3
SCENARIO 3: when x > 1, (x + 1) is positive in this region and (x - 1) is also positive, so the equation becomes:
x+1 = 2*(x-1)
=> x+1 = 2x-2
=> x = 3
So we have 2 possible values of x. i.e. 1/3 and 3
Statement1 insufficient
Statement 2 is clearly insufficient
Combine: st.1 tells us that x is either 1/3 or 3
st. 2 tells us that x is not equal to 3
Then x must be 1/3
Sufficient
IMO C
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Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 11:43
bigfernhead wrote: Does this always work? Squaring both sides? absolute values, equalities, and inequalities?
When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check. I am no math Guru, I just happened to learn this method somewhere and this works for me 
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Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 11:51
The Critical point method is almost always the best way to solve *ANY* kind of inequality problem. Albeit, sometimes time consuming.
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Re: DS ... Absolute Value [#permalink]
16 Sep 2008, 07:15
Tracking point works, I agree, but it takes me way too long. For the squaring, do you need absolute values on both side? Or this is this the time that you choose to apply it? amitdgr wrote: bigfernhead wrote: Does this always work? Squaring both sides? absolute values, equalities, and inequalities?
When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check. I am no math Guru, I just happened to learn this method somewhere and this works for me |
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Re: DS ... Absolute Value [#permalink]
16 Sep 2008, 11:30
If squaring can remove the modulus, then run for it ... Otherwise it wont make much sense imo ...
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Re: DS ... Absolute Value
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16 Sep 2008, 11:30
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