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Is x < 1? [#permalink]
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Updated on: 27 Jul 2015, 15:11
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55% (01:40) correct 45% (02:03) wrong based on 290 sessions
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Is x < 1? (1) x + 1 = 2x  1 (2) x  3 ≠ 0 My soln
we have to find 1<x<1 ....
From 1 , we have x + 1 = 2x  1 Assume 1 < x < 1 so the expression on RHS will be ve while LHS will be +ve => x+1 =2(1x) solving we get x=1/3 so we can easily say that x is <1
From 2 we cannot find whether x <1
PLease let me know where I am wrong
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Originally posted by boros2203 on 12 Feb 2010, 04:20.
Last edited by Bunuel on 27 Jul 2015, 15:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: Is x < 1? [#permalink]
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12 Feb 2010, 04:52
IMO C you have done right, but do consider the cases x>1 and x<1 also and they will give x=3 and using 2nd equation this isnt possible. Thus both taken together are sufficient.
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Re: Is x < 1? [#permalink]
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12 Feb 2010, 10:26
Thanks Gurpreet , i was missing this second part



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Re: Is x < 1? [#permalink]
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12 Feb 2010, 15:23
IMO A
(1) x + 1 = 2x  1 when x + 1>=0, you get x+1=2(x1) x+1=2x2 x=3
when x + 1<=0, you get x+1=2(x1) x+1=2x+2 x=1/3 which does not satisfy the inequality
so, only answer is x=3 and you can answer the question
(2)x  3 ≠ 0 => x is all but 3 This is not sufficient to answer
So, A  Statement 1 alone is sufficient



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Re: Is x < 1? [#permalink]
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12 Feb 2010, 15:52
Is \(x < 1\), means is \(x\) in the range (1,1)? or is \(1<x<1\) true? (1) \(x + 1 = 2x  1\) There will be two checkpoints (1 and 1), thus three ranges to test: A. \(x<1\) > \(x1=2(x+1)\) > \(x=3\) not good, as \(x<1\); B. \(1\leq{x}\leq{1}\) > \(x+1=2(x+1)\) > \(x=\frac{1}{3}\) is in the range \(1<x<1\). OK. C. \(x>1\) > \(x+1=2(x1)\) > \(x=3\) is in the range \(x>1\). OK So we got TWO values of \(x\), \(\frac{1}{3}\) and \(3\). One (\(\frac{1}{3}\)), is in the range (1,1) and another (\(3\)), is out of this range. Not sufficient. (2) \(x  3\neq0\) Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (1,1) or not. (1)+(2) From (1): \(x=\frac{1}{3}\) or \(x=3\) and from (2) \(x\neq{3}\) > means \(x\) can have only one value \(\frac{1}{3}\), which IS in the range (1,1). Sufficient. Answer: C.
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Is x < 1? [#permalink]
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Updated on: 06 Mar 2016, 11:49
testprep2010 wrote: IMO A
(1) x + 1 = 2x  1 when x + 1>=0, you get x+1=2(x1) x+1=2x2 x=3
when x + 1<=0, you get x+1=2(x1) x+1=2x+2 x=1/3 which does not satisfy the inequality
so, only answer is x=3 and you can answer the question
(2)x  3 ≠ 0 => x is all but 3 This is not sufficient to answer
So, A  Statement 1 alone is sufficient I originally thought A was the correct answer. But after redrawing the graphs, I realized I misinterpreted the graphs. Condition A gives two options, one of which falls between 1 and 1 and the other, 3, doesn't. Since B is basically eliminates 3 as a candidate, we are left with one option. Hence, the answer is C.
Originally posted by jjsverbal on 06 Mar 2016, 00:39.
Last edited by jjsverbal on 06 Mar 2016, 11:49, edited 1 time in total.



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Re: Is x < 1? [#permalink]
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06 Mar 2016, 03:24
Square both sides of statement (1) and we get x=3 or x=1/3. So statement (1) is insufficient. statement (2) tells us that X is not equal to 3. Clearly, statement (2) alone is insufficient. But if we combine statement (1) and statement (2), we get x= 1/3 which is definitely less than 1. Hence C .
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Re: Is x < 1? [#permalink]
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09 Mar 2016, 09:15
Here is my approach here we need to see whether x lies in the zone => (1,1) statement 1 is a double modulus equality => Remember whenever you encounter a double modulus equality (not inequality ) we can have only two cases => one where both of the signs will be +ve and the other when one is positive and the other is negative(take any one positive and the other negative) as per this logic => x=3 and x=1/3 => 3 does not satisfy our bound but 1/3 does hence insufficient statement 2 is insufficient too as x can be 0 or 50 combining them we get x≠3 so x=1/3 => which satisfies our bound => C is sufficient
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Re: Is x < 1? [#permalink]
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09 Jun 2018, 02:07
Bunuel wrote: Is \(x < 1\), means is \(x\) in the range (1,1)? or is \(1<x<1\) true?
(1) \(x + 1 = 2x  1\) There will be two checkpoints (1 and 1), thus three ranges to test: A. \(x<1\) > \(x1=2(x+1)\) > \(x=3\) not good, as \(x<1\);
B. \(1\leq{x}\leq{1}\) > \(x+1=2(x+1)\) > \(x=\frac{1}{3}\) is in the range \(1<x<1\). OK.
C. \(x>1\) > \(x+1=2(x1)\) > \(x=3\) is in the range \(x>1\). OK
So we got TWO values of \(x\), \(\frac{1}{3}\) and \(3\). One (\(\frac{1}{3}\)), is in the range (1,1) and another (\(3\)), is out of this range. Not sufficient.
(2) \(x  3\neq0\) Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (1,1) or not.
(1)+(2) From (1): \(x=\frac{1}{3}\) or \(x=3\) and from (2) \(x\neq{3}\) > means \(x\) can have only one value \(\frac{1}{3}\), which IS in the range (1,1). Sufficient.
Answer: C. Hi Bunuel, For the 2nd statement, since it is modulus, don't we need to consider this as well: (x3)!=0 so x!=3 ? So there will two not equals, x!=3 or x!=3. Please let me know whether I am right in this logic.



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Re: Is x < 1? [#permalink]
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09 Jun 2018, 05:02
kkrrsshh wrote: For the 2nd statement, since it is modulus, don't we need to consider this as well: (x3)!=0 so x!=3? So there will two not equals, x!=3 or x!=3. Please let me know whether I am right in this logic. Hey kkrrsshh , The error is in the highlighted statement above. (x3)!=0 =>  x + 3 !=0 => x !=  3 => x != 3 Does that make sense?
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