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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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22 May 2012, 23:05
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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 = 2x1 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 x1 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 x1 is still Negative as its sign would change on {1}. therefore when we solve of this range we take x+1= x+1 and 2x1 = 2(x1) = 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.........



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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22 May 2012, 23:30
kartik222 wrote: Hi Bunuel, A. x<1 (blue range) > x + 1 = 2x  1 becomes: x1=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 = 2x  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 = 2x  1 becomes: x+1=2(x1) > 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 "x1=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/mathabsolutevaluemodulus86462.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: isx11x12x12x82478.html#p1088086Absolute 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 \(x + 1 = 2x  1\). If \(1\leq{x}\leq{1}\) then \(x+1>0\) and \(x1<0\) so \(x + 1 =x+1\) and \(2x  1=2(x1)\) which means that \(x + 1 = 2x  1\) becomes: \(x+1=2(x+1)\). Hope it's clear.
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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30 Aug 2012, 20:12
good explanation sir thanks a lot learnt new concept in this
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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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.



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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31 Aug 2012, 02:42



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 > 0 [#permalink]
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02 Nov 2012, 02:00



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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08 Jul 2013, 01:24
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yezz wrote: Is x < 1?
(1) x + 1 = 2x  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(x1)^2 On simplifying we get 3x^210x3= 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 = 2x  1 (2) x  3 0 [#permalink]
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08 Jul 2013, 16:27
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Is x < 1?
(1) x + 1 = 2x  1 (My first problem was to decide how to solve this. I had recently tackled a problem to the effect of xz=xy 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 = 2x  1 x=1, x=1 so, x<1, 1<x<1, x>1 x<1: (x+1) = 2 (x1) x1 = 2(x+1) x1 = 2x+2 x=3 INVALID as 3 does not fall within the range of x<1
1<x<1 (x+1) = 2 (x1) 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(x1) x+1=2x2 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 = 2x  1 (2) x  3 0 [#permalink]
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03 Sep 2013, 06:27
From (1) Taking sqaure of both sides ((X+1)^2)/((X1)^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 = 2x  1 (2) x  3 0 [#permalink]
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03 Sep 2013, 06:54
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 = 2x  1 (2) x  3 0 [#permalink]
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03 Sep 2013, 06:58



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Re: Is x < 1? [#permalink]
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30 Jan 2014, 07:18
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 J



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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05 Jul 2014, 01:21
This is very good question X<1 x1=2(x+1)
1<=x<=1 x+1=2(x+1)
X>1 x+1=2(x1)



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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02 Sep 2014, 04:17
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) \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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. Answer: C. Hope it helps. Your explanation always help!! Thanks



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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20 Oct 2014, 23:57
yezz wrote: Is x < 1?
(1) x + 1 = 2x  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 = 2x  1 (2) x  3 0 [#permalink]
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21 Oct 2014, 04:46
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) \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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. Answer: C. 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 = 2x  1 (2) x  3 0 [#permalink]
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21 Oct 2014, 06:25
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) \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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. Answer: C. 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: mathabsolutevaluemodulus86462.htmlAbsolute value tips: tipsandhintsforspecificquanttopicswithexamples172096.html#p1381430
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Re: Is x < 1 ? (1) x + 1 = 2x – 1 (2) x – 3 > 0 [#permalink]
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13 Jul 2015, 06:51
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apple08 wrote: Need help for a simple solution to get the answer as I m confuse with this question , many thankss Is x < 1 ?
(1) x + 1 = 2x – 1
(2) x – 3 > 0
Really appreciate it , Many Thanks Good question. For absolute values, you need to take into consideration all possible values. For example, if I say x <1, then 1<x<1. We get this by looking at 2 cases : 1. When x>= 0, x=x > x<1. This value of x<1 does satisfy x>=0. So keep this. 2. When x<0 , x = x > x<1 > x>1 . This value of x>1 (think x= 0.7 or 0.5) still satisfies x<0. So keep this value as well. Thus, from 1 and 2 , x <1 is nothing but a fancy way of saying 1<x<1. You can also visualise this from numberline: x is the distance of x from 0 (as x = x0) on the number line. Thus , when I say x <1 this means that distance of x from 0 is less than 1 on either side (as we can approach 0 from 2 directions, 1 from positive and 1 from the negative!). So, the question actually asks is x <1 > is 1<x<1. Now lets look at the 2 statements, Per statement 1, x + 1 = 2x – 1 Now, your points of analysis are x+1 =0 > x=1 and x1=0> x=1. Thus you will have 3 regions to look at: 1. x<1 2. 1<x<1 3. x>1 When you look at the above 3 regions, you will see that case (1) is not going to satisfy while case (2) will give you x=1/3 and case 3 will give you x = 3. So we see that 1<x<1 is true if x=1/3 but is false if x=3. Thus this statement is not sufficient. Per statement 2, x3 >0 > this is true for all x as if you substitute x = 1 or 2 or 0.75 or 7 or 10,x3 will always greater than 0 as x is DISTANCE of x from 0 (as x = x0). As such x3 is also distance of x from 3 and as distance is always > 0, this is true for all values of x. Note, that we will not take x = 3 as we have been given x3 > 0 and not x3 > = 0. Thus this statement is not sufficient. Combining 1 and 2 we get, x=1/3 or x=3 Now, x3 >0 will not be true if x=3 as 33 = 0 and NOT >0. Thus only 1 value of x is possible > x = 1/3. Hence both the statements together are sufficient. C is the correct answer.



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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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17 Aug 2016, 23:48
Bunuel wrote: 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) \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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. Answer: C. Hope it helps. I don't understand the statement 1, B part. (I am bad with inequalities) so if 1<=x<=1 then the equation becomes x+1=2(x+1) and that gives x = 1/3; but what about the possibility of the equation (x+1)=2(x1) ? When do we consider that inequality? I am confused? PS: I am a new user
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