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Is x + x1 = 1?
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Is x + x1 = 1? (1) \(x \geq 0\) (2) \(x \leq 1\)
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Originally posted by abhi758 on 16 Mar 2010, 03:43.
Last edited by Bunuel on 08 Mar 2013, 03:25, edited 1 time in total.
Edited the question and added the OA.




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Re: Modulus
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16 Mar 2010, 04:06
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.
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Re: Modulus
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16 Mar 2010, 03:50
abhi758 wrote: Is l x l + lx1l= 1? (1) \(x \geq 0\) (2) \(x \leq 1\) Any explanations for the given OA.. stmnt1. let x= 0 then we have l 0 l + l01l= 1. suff x= 10> 0 then we have l 10 l + l101l not equal to 1. hence insuff stmnt2. let x = 0< 1 then we have l 0 l + l01l= 1. suff let x= 5 < 1 then we l 5 l + l5 1l not equal to 1. hence insuff taking together we have \(0 \leq x \leq 1\) if x= 0 we have l 0 l + l01l= 1 if x = 1 we have l 1 l + l11l= 1 if x = 1/2 we have l 1/2 l + l1/21l= 1 hence C



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Re: Modulus
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16 Mar 2010, 06:31
abhi758 wrote: Is l x l + lx1l= 1? (1) \(x \geq 0\) (2) \(x \leq 1\) Any explanations for the given OA.. x+x1, there can be 4 cases stmt1: x >= 0 => x is positive, if we apply this to eq 1, we cannot say that 2x1 = 1 cos x can be >= 0 but less than 1 so insuff stmt2: x<=1 => if we apply this to eq 1, we cannot say that 2x1 = 1 cos x<=1 can lead to negative numbers also. both stmt taken together 0<= x <= 1 x is a fraction between 0 to 1 so always x + 1x = 1 in our case x will be x and x  1 is actually value of 1x so it will always be 1 this is why C both the stmts are reqd.
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Re: Modulus
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16 Mar 2010, 08:21
Thanks so much all of you! the workings definately helped to understand the OA better.. Thanks Bunnel for the algebraic approach..



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Re: Modulus
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16 Mar 2010, 08:39
abhi758 wrote: Is l x l + lx1l= 1? (1) \(x \geq 0\) (2) \(x \leq 1\) Any explanations for the given OA. Either statement alone is not sufficient to answer this question. Together x can be 0 or 1, in both cases statement is true hence sufficient.



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Re: Modulus
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18 Mar 2010, 10:10
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. Can you please clarify as to how is \(xx+1=1\) derived inB when it is provided that \(0\leq{x}\leq{1}\) that means X is +ve. Thanks !!!
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Re: Modulus
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18 Mar 2010, 10:21
mustdoit wrote: 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. Can you please clarify as to how is \(xx+1=1\) derived inB when it is provided that \(0\leq{x}\leq{1}\) that means X is +ve. Thanks !!! When \(x\) is in the range \(0\leq{x}\leq{1}\) then \(x=x\) and \(x1=(x1)=x+1\), thus \(x+x1=1\) in this range becomes \(xx+1=1\). Hope it's clear.
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Re: Modulus
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07 Mar 2013, 21:30
I have got the answer as E. Can someone please clarify, what is wrong in my approach ? Statement 1: X>=0 I assume X as 0 0+01= 0+1= 1 > Equation satisfied I assume X as 3/4 3/4+(3/4)1= 3/4+1/4= 2 > Equation unsatisfied Statement 1 is not sufficient Statement 2: X<=1 I assume X as 1 1+11= 1+0= 1 > Equation satisfied I assume X as 3/4 again 3/4+(3/4)1= 3/4+1/4= 2 > Equation unsatisfied Statement 2 is not sufficient Taking together both the equation, I assume X as 0, 3/4 and 1 When X=0, the answer is 1> Equation satisfied When X=3/4, the answer is 2> Equation unsatisfied When X=1, the answer is 1> Equation satisfied Hence answer choice E is correct. Clarifications needed, please.
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Re: Modulus
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07 Mar 2013, 21:44
abhi758 wrote: Is l x l + lx1l= 1? (1) \(x \geq 0\) (2) \(x \leq 1\) Any explanations for the given OA.. You can also consider using the number line method discussed here: http://www.veritasprep.com/blog/2011/01 ... spartii/l x l + lx1l= 1 implies that sum of distance of x from 0 and from 1 should be 1. We can easily see that the distance between 0 and 1 is 1. So, x can lie on 0, on 1 or anywhere in between 0 and 1. It cannot go beyond 1 or before 0 because then the sum of the distance will become larger than 1. Both statements together tell you that 0 <= x <= 1 and hence, the equation will hold for all such values of x. Answer (C)
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Re: Modulus
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08 Mar 2013, 03:31
Backbencher wrote: I have got the answer as E. Can someone please clarify, what is wrong in my approach ?
Statement 1: X>=0
I assume X as 0 0+01= 0+1= 1 > Equation satisfied
I assume X as 3/4 3/4+(3/4)1= 3/4+1/4= 2 > Equation unsatisfied
Statement 1 is not sufficient
Statement 2: X<=1
I assume X as 1 1+11= 1+0= 1 > Equation satisfied
I assume X as 3/4 again 3/4+(3/4)1= 3/4+1/4= 2 > Equation unsatisfied
Statement 2 is not sufficient
Taking together both the equation, I assume X as 0, 3/4 and 1
When X=0, the answer is 1> Equation satisfied When X=3/4, the answer is 2> Equation unsatisfied When X=1, the answer is 1> Equation satisfied
Hence answer choice E is correct.
Clarifications needed, please. The correct answer is C, not E. If x = 3/4, then Is x + x1 = 3/4 + 3/4  1 = 3/4 + 1/4 = 1. Hope it helps.
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Re: Is x + x1 = 1?
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08 Mar 2013, 04:41
abhi758 wrote: Is x + x1 = 1?
(1) \(x \geq 0\) (2) \(x \leq 1\) So x + x1 can only equal 1 when x = x and x1 = 1x . Thus, x>=0 AND x1=<0. Which gives, 0<=x<=1. C.
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Re: Is x + x1 = 1?
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17 Jun 2013, 08:21
Is x + x1 = 1?
For #1 (x≥0) I see two possible cases:
x=1/2 x+ (x1)=1 x+ x+1=1 1=1
OR
x=2 x+x1=1 2x=2 x=1
Wouldn't that imply that for both cases of x (where we have positive and negative values inside the   signs) the equation is true?
Thanks!



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Re: Is x + x1 = 1?
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17 Jun 2013, 08:41
WholeLottaLove wrote: Is x + x1 = 1?
For #1 (x≥0) I see two possible cases:
x=1/2 x+ (x1)=1 x+ x+1=1 1=1
OR
x=2 x+x1=1 2x=2 x=1
Wouldn't that imply that for both cases of x (where we have positive and negative values inside the   signs) the equation is true?
Thanks! No. In the first case for an \(x=\frac{1}{2}\) (\(0\leq{x}\leq{1}\)) you get that the equation is true \(1=1\) it's like saying ALWAYS TRUE. But in the second case for an \(x>1\) you get x=1, but since x>1 that is not true Statement 1 is not sufficient
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Re: Is x + x1 = 1?
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17 Jun 2013, 09:09
Ok, this is how I look at it: x>1 x + x1 = 1 x + (x1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient 0≤x<1 x = 1/2 x + x1 = 1 x + (x1) =1 x+ x +1 = 1 0=0I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient? Thanks! Zarrolou wrote: WholeLottaLove wrote: Is x + x1 = 1?
For #1 (x≥0) I see two possible cases:
x=1/2 x+ (x1)=1 x+ x+1=1 1=1
OR
x=2 x+x1=1 2x=2 x=1
Wouldn't that imply that for both cases of x (where we have positive and negative values inside the   signs) the equation is true?
Thanks! No. In the first case for an \(x=\frac{1}{2}\) (\(0\leq{x}\leq{1}\)) you get that the equation is true \(1=1\) it's like saying ALWAYS TRUE. But in the second case for an \(x>1\) you get x=1, but since x>1 that is not true Statement 1 is not sufficient



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Re: Is x + x1 = 1?
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17 Jun 2013, 09:14
WholeLottaLove wrote: Ok, this is how I look at it:
x>1 x + x1 = 1 x + (x1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient
0≤x<1 x = 1/2 x + x1 = 1 x + (x1) =1 x+ x +1 = 1 0=0 I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?
Thanks!
0=0, and 1=1,... means that that equation is always true. You should read the results as follow: x>1 x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1) 0≤x<1 0=0. So in this case the equation in TRUE( because 0=0 always) So from statement 1 you get two different results (F/T)=> not sufficient
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Re: Is x + x1 = 1?
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17 Jun 2013, 09:33
Okay, that's what I thought. Thank you. Zarrolou wrote: WholeLottaLove wrote: Ok, this is how I look at it:
x>1 x + x1 = 1 x + (x1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient
0≤x<1 x = 1/2 x + x1 = 1 x + (x1) =1 x+ x +1 = 1 0=0 I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?
Thanks!
0=0, and 1=1,... means that that equation is always true. You should read the results as follow: x>1 x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1) 0≤x<1 0=0. So in this case the equation in TRUE( because 0=0 always) So from statement 1 you get two different results (F/T)=> not sufficient



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Re: Is x + x1 = 1?
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30 Jan 2014, 06:02
I went through the following link x34x8xhowmanysolutionsdoestheequation148996.html and understand the following below. Quote: So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.
When we are considering ranges, say, x < 8  x is less than 8 8 <= x < 3  x is greater than or equal to 8 but less than 3 3 <= x < 4  x is greater than or equal to 3 but less than 4 x >=4  x is greater than or equal to 4
We need to include the transition points (8, 3, 4) somewhere so we include them with greater than sign.
Mind you, we could have taken the ranges as x <= 8 8 < x <= 3 3 < x <= 4 x > 4 The above explanation is easily understandable but it creates a small doubt if i compare it with this question Is x + x1 = 1? (1) \(x\geq{0}\) (2) \(x\leq{1}\) Quote: 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 \(0\leq{x}\leq{1}\) Here, we have to include the transition points (0 and 1) but only once. Are these okay, for B. \(0\leq{x}<1\) > 1=1 not in the range. for C. \(x\geq{1}\) > x = 1. It is in the range. Or A. \(x\leq{0}\) > x= 0, It is in the range. B. \(0<{x}\leq{1}\) > 1=1 It is also in the range. Are my ranges correct ? if yes, then i still don't feel close to the answer. Please give your views on it. Thanks & Regards Vinni



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Re: Is x + x1 = 1?
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15 Feb 2014, 22:18
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
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Re: Is x + x1 = 1?
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16 Feb 2014, 20:40
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.
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