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Re: Is x + x1 = 1?
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21 Feb 2014, 10:08
VeritasPrepKarishma wrote: kanusha wrote: My answer is C its by plugging in method 1. From 1st equation , if x = 1 , then x + x1 = 1 , then its yes if x = 5 , 9 # 1 , so No so Equation 1 is No 2. From 2nd equation , of x = 1 , then its No x = 1 its yes so equation is No 3. using both , x = 0 then its Definite yes So C.. Is my approach is Right Using both, you get 0 <= x <= 1 x can lie anywhere from 0 to 1. Just trying the extreme values is not a good idea. You must try some values from the middle too i.e. x = 1/2, 1/4 etc. Though establishing something by plugging in values is always tricky. You can try to negate a universal statement by looking for one value for which the statement doesn't hold but since you cannot try every value for which a statement must hold, its treacherous to depend on number plugging in this case. You should try to use logic. Hi Karishma, Please explain with an example how to find values for X when there is modulus in the equation. For example x6 + 7x   4  x = x + 2 Thanks in Advance, Rrsnathan.



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Re: Is x + x1 = 1?
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24 Feb 2014, 02:01
rrsnathan wrote: VeritasPrepKarishma wrote: kanusha wrote: My answer is C its by plugging in method 1. From 1st equation , if x = 1 , then x + x1 = 1 , then its yes if x = 5 , 9 # 1 , so No so Equation 1 is No 2. From 2nd equation , of x = 1 , then its No x = 1 its yes so equation is No 3. using both , x = 0 then its Definite yes So C.. Is my approach is Right Using both, you get 0 <= x <= 1 x can lie anywhere from 0 to 1. Just trying the extreme values is not a good idea. You must try some values from the middle too i.e. x = 1/2, 1/4 etc. Though establishing something by plugging in values is always tricky. You can try to negate a universal statement by looking for one value for which the statement doesn't hold but since you cannot try every value for which a statement must hold, its treacherous to depend on number plugging in this case. You should try to use logic. Hi Karishma, Please explain with an example how to find values for X when there is modulus in the equation. For example x6 + 7x   4  x = x + 2 Thanks in Advance, Rrsnathan. The method stays the same though the thinking gets a little more complicated. x6 + 7x   4  x = x + 2 x6 + x  7 = x  4 + x + 2 x is a point whose sum of distance from 6 and 7 is equal to sum of distance from 4 and 2. Let's look at the various regions. We want the sum of lengths of red lines to be equal to sum of lengths of blue lines. Attachment:
Ques3.jpg [ 25.97 KiB  Viewed 1383 times ]
In Fig 1 (x lies to the left of 2), the sum of blue lines will always be greater than red lines since 6 and 7 are to the right so they will always need to cover extra distance. In Fig 2 (x lies between 2 and 4), no matter where x is the sum of length of red lines is 6 units. Can the sum of blue lines be 6? Yes, from 7 to 4, one blue line covers 3 unit and from 6 to 4, the other blue line covers 2 units. Now both together need to cover 1 more unit so x will lie 0.5 unit to the left of 4 which gives the point 3.5. In Fig 3 (x lies between 4 and 6 or between 6 and 7), the long red line will be at least 6 units and the maximum sum of blue lines can be 2 units + 3 units = 5 units. Hence sum of red lines will always be greater. In Fig 4 (x lies to the right of 7), the red lines will always cover more distance than the blue lines since their points are to the left. Hence, there is only one solution x = 3.5
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Re: Is x + x1 = 1?
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21 Aug 2014, 08:47
Hi Bunuel,
Can you pls explain , for such kind of questions how do we decide what will be the check points ?
Like in your explanation, you said there are two check points 0 & 1 . Can you tell the approach behind this ?
Thanks in advance
Kunal



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Re: Is x + x1 = 1?
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Is x + x1 = 1?
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22 Aug 2014, 22:01
Is\(x + x1 = 1 ?\)
(1) \(x \geq 0\) (2)\(x \leq 1\)
Rewriting the equation \(x + x1 = 1\) as \(x + 1x = 1\) The above equation is true for \(x(1x) \geq 0\) [ From the property, \(x + y = x+y\) holds for \(xy \geq 0\) ] Or, \(0\geq x\geq 1\), , which is exactly the common range of x we get after combining statements (1) + (2).
Hence, answer (C).



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Is x + x1 = 1?
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03 May 2017, 15:04
Shouldn't this question be a DS question? This is the PS section. abhi758 wrote: Is x + x1 = 1?
(1) \(x \geq 0\) (2) \(x \leq 1\) You have two absolute values. The value inside the absolute values can be positive or negative. You can simplity to find all the values where x + x1 = 1. There are four possibilities. Positivepositive, positivenegative, negativepositive, and negativenegative. x + x  1 = 1. This will be 2x = 2 or x = 1. x  1 + 1 = 1. This will be x = 1. x + x  1 = 1. This will be 1 = 1. This makes no sense, so you can ignore it. x  x + 1 = 1. This will be 2x = 0 or x = 0. So according to the given formula, the only possible values of x that can result in 1 are when x = 1 or x = 0. Statement 1: x >= 0. If x = 0 then true, but if x >1 then false. Insufficient. Statement 2: x <= 1. If x = 1 then true, but if x<0 then false. Insufficient. Statement 1 and 2: 0<=x<=1. Either way will make the statement true. Sufficient.



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Re: Is x + x1 = 1?
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03 May 2017, 16:17
x = x and x1 = 1x . given, x>=0 & x1=<0. 0<=x<=1. answer: C



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03 May 2017, 22:22



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Is x + x1 = 1?
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08 Jun 2018, 11:54
Bunuel wrote: abhi758 wrote: Is l x l + lx1l= 1? (1) \(x \geq 0\) (2) \(x \leq 1\) Any explanations for the given OA.. Algebraic approach: Let's check in which ranges of \(x\) expression \(x+x1=1\) holds true: Two check points for \(x\): 0 and 1 (x=0 and x1=0, x=1), thus three ranges to check: A. \(x<0\) > \(xx+1=1\) > \(x=0\), not good as we are checking for \(x<0\); B. \(0\leq{x}\leq{1}\) > \(xx+1=1\) > \(1=1\), which is true. This means that in this range given equation holds true for all xes; C. \(x>1\) > \(x+x1=1\) > \(x=1\), not good as we are checking for \(x>1\). So we got that the equation \(x+x1=1\) holds true only in the range \(0\leq{x}\leq{1}\). (1) \(x \geq 0\). Not sufficient. (2) \(x \leq 1\). Not sufficient. (1)+(2) gives us the range \(0\leq{x}\leq{1}\), which is exactly the range for which given equation holds true. Sufficient. Answer: C. Hope it helps. Bunuel. Hey abhimahna, could you please explain why we use this range \(0\leq{x}\leq{1}\) and not \(0\leq{x}<{1}\) in B (and \(x>1\) and not \(x\geq1\) in C)? If we look at this \( x1 \), then won't that give us two ranges: x<1 & \(x\geq1\)? From the GC Math book example, x1=4, we get two ranges: 1) \((x1)\geq0\) => \(x\geq1\) 2) (x1)<0 => x<1.
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Re: Is x + x1 = 1?
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16 Jun 2018, 11:50
dabaobao wrote: Bunuel. Hey abhimahna, could you please explain why we use this range \(0\leq{x}\leq{1}\) and not \(0\leq{x}<{1}\) in B (and \(x>1\) and not \(x\geq1\) in C)? If we look at this \( x1 \), then won't that give us two ranges: x<1 & \(x\geq1\)? From the GC Math book example, x1=4, we get two ranges: 1) \((x1)\geq0\) => \(x\geq1\) 2) (x1)<0 => x<1. Hey dabaobao , Yes, your point is completely valid. But you need to understand that our question does contain other expressions other than x1. Hence, we have to be very careful for such cases. If you check the values, you will find that the expressions Bunuel has mentioned work fine. Check out line method to solve such questions as explained by Karishma here. I hope this helps.
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Re: Is x + x1 = 1?
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16 Jun 2018, 15:42
abhi758 wrote: Is x + x1 = 1?
(1) \(x \geq 0\) (2) \(x \leq 1\) ab = the DISTANCE between a and b. Thus: x = x0 = the distance between x and 0. x1 = the distance between x and 1. x + x1 = the SUM of these two distances. The distance between 0 and 1 is 1. On the number line, if x is at or BETWEEN the two endpoints in blue, then the sum of the two distances will be EQUAL TO 1: 0 < x > x <x1> 1 Here, x + x1 = the distance between 0 and 1 = 1. By extension, if x is BEYOND either endpoint  if x is to the left of 0 or to the right of 1  then the sum of the two distances will be GREATER THAN 1. Thus, the answer to the question stem will be YES if 0≤x≤1. Question stem, rephrased: Is 0≤x≤1? Clearly. each statement on its own is INSUFFICIENT. When the statements are combined, we know that 0≤x≤1, so the answer to the rephrased question stem is YES. SUFFICIENT.
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