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

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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| > 0 [#permalink]  02 Nov 2012, 01:00
Expert's post
himanshuhpr wrote:
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0

From 1st statement we know that x = 1/3 or x= 3
From 2nd statement we know that that x>3 or x<3

My question is can we combine these 2 results and say (c) is sufficient as x>3 in statement creates an ambiguity and therfore it should be E.

(2) |x - 3| > 0. Absolute value is always more than or equal to zero. Thus this statement just says that $$|x - 3|\neq{0}$$ --> $$x\neq{3}$$.

Hope it helps.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  07 Jul 2013, 23:53
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  08 Jul 2013, 00:24
1
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yezz wrote:
Is |x| < 1?

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

Questions asks if -1<x<1

From St1 we can say that since both LHS and RHS is positive therefore squaring both sides we get

(x+1)^2= 4(x-1)^2
On simplifying we get 3x^2-10x-3= 0 -----> x=3 or x= 1/3

2 ans and hence st 1 is not sufficient. Therefore Option A and D ruled out

St 2 we have |x - 3| ≠ 0 and since Modulus of any no is greater than or equal to 0 (In this case greater than 0) which means x not equal to 3

Not sufficient.

Combining both equation we see that x=1/3 which is in the Question stem range

Ans C
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  08 Jul 2013, 15:27
1
KUDOS
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(My first problem was to decide how to solve this. I had recently tackled a problem to the effect of |x-z|=|x-y| where I squared both sides to help simplify. In this case, it might be better to find the check points of the LHS and RHS as we are given a variable and an integer as opposed to all variables like the problem I had to solve before)
|x + 1| = 2|x - 1|
x=-1, x=1
so, x<-1, -1<x<1, x>1
x<-1:
-(x+1) = 2 -(x-1)
-x-1 = 2(-x+1)
-x-1 = -2x+2
x=3 INVALID as 3 does not fall within the range of x<-1

-1<x<1
(x+1) = 2 -(x-1)
x+1 = -2x+2
3x=1
x=1/3 VALID as x falls within the range of -1<x<1

x>1
(x+1)=2(x-1)
x+1=2x-2
-x=-3
x=3 VALID as x falls within the range of x>1

We have two valid values for x and therefore we cannot determine a single, correct answer.
INSUFFICIENT

(2) |x - 3| ≠ 0
Tells us nothing about the sign of x. x could be -10 or +10 and |x - 3| ≠ 0 would be valid.
INSUFFICIENT

1+2) x=1/3 OR x=3 AND |x - 3| ≠ 0

Plug in both x values into the above equation. x=3 fails as it does = 0 when the equation says it does not. Therefore, the only remaining x value is 1/3.
SUFFICIENT

(C)
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:27
From (1) Taking sqaure of both sides ((X+1)^2)/((X-1)^2 = 4, solving this you get X = 3 or 1/3, which is not sufficient.

(2) says X is not = 3

Hence both together tell us the answer (C)
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:54
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:58
Expert's post
thinktank wrote:
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?

Why does "is $$-1 < x < 1$$?" means "is x=0"? We are not told that x is an integer.
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Re: Is |x| < 1? [#permalink]  30 Jan 2014, 06:18
Economist wrote:
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?

Actually your third range is wrong I believe. If x<-1 then x cannot be 3, it isn't a valid solution but the rest is OK

Cheers
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  05 Jul 2014, 00:21
This is very good question
X<-1
-x-1=2(-x+1)

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

X>1
x+1=2(x-1)
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  02 Sep 2014, 03:17
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.

Your explanation always help!! Thanks
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  20 Oct 2014, 22:57
yezz wrote:
Is |x| < 1?

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

1 - On solving the equation, we get the following values of x - 3, 1/3
2 - There can be infinite values of x, except 3

Combining 1 & 2 = x = 1/3

Ans. C
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  21 Oct 2014, 03:46
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.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  21 Oct 2014, 05:25
Expert's post
arnabs 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.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?

It seems that you need to brush up fundamentals.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1381430
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0   [#permalink] 21 Oct 2014, 05:25

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