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Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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Is x < 1? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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Last edited by Bunuel on 02 Mar 2012, 20:00, edited 1 time in total.
Edited the question and added the OA



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Re: Is x < 1? [#permalink]
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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?
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Re: Is x < 1? [#permalink]
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is 1<x<1 ?
1) x + 1 = 2x  1
two possibilities: a) x+1=2(x1) or x=3 (consider x+1 and x1 to be of same sign) b) x+1=2(1x) or x=1/3 (consider x+1 and x1 to be of different signs) not sufficient, x may or may not be between 1 and 1
2) x  3 ≠ 0 => x≠3 not sufficient, x may or may not be between 1 and 1
together, x≠3 so x is 1/3 which is between 1 and 1 sufficient
hence, C



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Re: Is x < 1? [#permalink]
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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.
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Re: Is x < 1? [#permalink]
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23 Oct 2009, 19:23
I got C too substituting values and evaluating each of those. thanks Bunuel and Economist for elaborations
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Re: Is x< 1? [#permalink]
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x<1 means x e (1,1) 1)x+1=2x1 There are 2 key points (x=1 and x=1) where one of the expressions under modules passes 0 and change its sign. x<1: (x+1)=2(x1) > x1=2x+2 > x = 3. But x=3 does not satisfy x<1 condition. 1<=x<1: x+1 = 2(x1) > x+1= 2x+2 > x=1/3. It satisfies the condition. x>=1: x+1=2x2 > x=3. It satisfies the condition. Insufficient 2) x  3 ≠ 0 x≠3 Insufficient 1&2) Only x=1/3 satisfies both statements. Sufficient. Jade3, check this post, please: mathabsolutevaluemodulus86462.htmlIf you find it useful/useless, let me know
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Re: Is x< 1? [#permalink]
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Answer is C  you beat me to it walker For statement 1, you need to actually break it down to 2 separate statements to account for both scenarios: Option #1 > x + 1 = 2(x1) x + 1 = 2x 2 x = 2x 3 3 + x = 2x 3 = x {or also Option #2} Option #2 > x + 1 = 2(x1) x + 1 = 2x +2 x = 2x + 1 3x = 1 x = 1/3 Statement (1) is insufficient because x could be 3 or 1/3. So the answer is sometimes yes, sometimes no which means insufficient. Statement (2) is insufficient because we are told that x 3 ≠ 0. This means that as long as x ≠ 3, then we're ok. There are far too many options so we certainly have a sometimes yes, and sometimes no. But together, from Statement (1) we have either 3 or 1/3 and from statement (2) we have everything BUT 3. So the only overlap we have for possible values to make both statements true is 1/3. So the answer is Yes, x IS less than 1 and we have enough info to answer the question. jade3 wrote: Is x< 1? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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x < 1? 1) x+1 = 2x1 2) x3 does not equal 0 [#permalink]
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01 May 2011, 19:59
dc123 wrote: x < 1?
1) x+1 = 2x1
2) x3 does not equal 0
Therefore we are looking to see if 1<x<1????
Can someone plz help. I dont know why its C In absolute equations, one should validate the equation for all possible cases w.r.t the signs of expression embedded in the modulus. Q: X<1 1<X<1 (1) x+1 = 2x1 Case 1: (x+1)>=0 AND (x1)>=0 x+1=2(x1) x=3 x=3 Case 2: (x+1)>=0 AND (x1)<0 x+1=2(x1)) x+1=2x+2 3x=1 x=1/3 Case 3: (x+1)<0 AND (x1)>=0 (x+1)=2(x1) x1=2x2 3x=1 x=1/3 Case 4: (x+1)<0 AND (x1)<0 (x+1)=2(x1) x1=2x+2 x=3 Thus, x can be 3 or 1/3. Not Sufficient. 2. x != 3 Not Sufficient. Combining both; x=1/3, which is between 1 and 1. Sufficient. Ans: "C"
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Re: DS..... [#permalink]
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02 May 2011, 03:43
1. 3 regions to be checked x<1, 1<x< 1 and x>1 for x>1 , (x+1) = 2(x1) x= 3 for x<1, (x+1) = 2(x1) x= 1/3 which is not possible.Hence no solution in this region. for 1<xx<1, (x+1) = 2(x1) x= 1/3 No sufficient. 2. x !=3 not sufficient 1+2 gives x = 1/3 Hence C
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Re: DS..... [#permalink]
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02 May 2011, 09:22
i did it a bit different. plz tell me if u think im wrong doing so: x < 1? 1) x+1 = 2x1 2) x3 does not equal 0 the question is  x^2<1 so x^2+2x+1=4(x^22x+1) ====> 3x^210x+3=0 so x = 3 or 1/3 st. 2: x<>3 not helping me. combination x=1/3 is that approach wrong?
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Re: DS..... [#permalink]
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02 May 2011, 17:08
(1) Square both sides : x^+1+2x = 4(x^2 2x + 1) 3x^2 10x +3 = 0 3x^2  9x x + 3 = 0 3x(x3) 1(x3) = 0 (x3)(3x1) = 0 x = 3, x = 1/3 Not sufficient (2) is not sufficient as well (1) + (2) x = 1/3 Answer C
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Re: DS..... [#permalink]
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02 May 2011, 17:57
dc123 wrote: x < 1?
1) x+1 = 2x1
2) x3 does not equal 0
Therefore we are looking to see if 1<x<1????
Can someone plz help. I dont know why its C You can do these questions in less than a minute if you understand the approach I discuss here: http://www.veritasprep.com/blog/2011/01 ... edoredid/On to your question. Is x < 1? Yes, essentially this just asks you whether x lies between 1 and 1. x represents the distance from 0 which will be less than 1 if x lies within 1 < x < 1. Statement 1: x+1 = 2x1 Draw the number line. This equation says that distance from 1 is twice the distance from 1. At which point is the distance from 1, twice the distance from 1? If you split the distance (2 units) between 1 and 1 into 3 parts, 1 part away from 1 and 2 parts away from 1 will be the point 1/3. The diagram below will show you this situation. Similarly the distance between 1 and 1 is 2 so if you go 2 units further to the right of 1, you get the point 3 which will be 4 units away from 1 i.e. twice the distance from 1. Attachment:
Ques6.jpg [ 4.74 KiB  Viewed 6042 times ]
x could be 1/3 or 3. Hence x may or may not lie between 1 and 1. Not sufficient. Statement 2: x3 does not equal 0 This statement just says that x is not 3. It is not sufficient alone. Both together, we get that x must be either 3 or 1/3 and it is not 3. Then x must be 1/3 and must lie between 1 and 1. Sufficient. Answer (C)
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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16 Jan 2012, 07:48
+1 C However, I used using other methods: a) Squaring both sides of the equation because x = \((x^2)\)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x1)>0, (x1)<0. Therefore, I get four possible scenarios. Bunuel, could you explain the logic behing your method to solve this problem?
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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metallicafan wrote: +1 C
However, I used using other methods: a) Squaring both sides of the equation because x = \((x^2)\)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x1)>0, (x1)<0. Therefore, I get four possible scenarios.
Bunuel, could you explain the logic behing your method to solve this problem? It's about expanding the absolute value in different ranges. Discussed in Walker's post on absolute values: mathabsolutevaluemodulus86462.htmlIf you're more comfortable with "squaring" method (note that it's not always applicable) then you can apply it to the first statement too (and not only to the stem). Is \(x < 1\)?Is \(x < 1\) > is \(1<x<1\)? (1) \(x + 1 = 2x  1\) > square both sides: \((x+1)^2=4(x1)^2\) > \(x^2+2x+1=4x^28x+4\) > \(3x^210x+3=0\) > \(x=\frac{1}{3}\) or \(x=3\) > 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's clear.
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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24 Apr 2012, 03:04
Hi Bunuel , I also used the squaring method.... In what all situations is the squaring method not applicable ??? Can you please explain why ?
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Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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shikhar wrote: Hi Bunuel , I also used the squaring method.... In what all situations is the squaring method not applicable ??? Can you please explain why ? Sure, for example it's not always applicable for inequalities. A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \((5)^3=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). So for our question we can not square x/x< x as we don't know the sign of either of side. Hope it helps.
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Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0 [#permalink]
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28 Apr 2012, 09:42
Nice explaination Bunuel, Thanks! 1+



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Re: Is x < 1? [#permalink]
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21 May 2012, 21:35
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. How is the sign applied for different range?can someone help please 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? [#permalink]
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21 May 2012, 22:18
sdpkind 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. How is the sign applied for different range?can someone help please X<1 x1=2(x+1) 1<=x<=1 x+1=2(x+1) X>1 x+1=2(x1) Absolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\); Applying this to \(x + 1 = 2x  1\). If \(x<1\) 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: \(x1=2(x+1)\) > \(x=3\). Similarly for other ranges. For more check Absolute Value chapter for Math Book: mathabsolutevaluemodulus86462.htmlHope it helps.
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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, 19:28
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,




Re: Is x < 1? (1) x + 1 = 2x  1 (2) x  3 0
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22 May 2012, 19:28



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