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What is the value of x?
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30 May 2010, 02:52
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62% (01:49) correct 38% (01:49) wrong based on 325 sessions
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What is the value of x? (1) 6  3x = x  2 (2) 5x + 3 = 2x + 9
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Re: Absolute Value
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30 May 2010, 03:24
(1) 63x = x2 <=> 63x = x2 or 63x = (x2)
solving these equations you will have x=2 or x = 2 then (1) is sufficient. X = 2
(2) 5x+3 = 2x+9 or 5x+3 = (2x+9)
3x = 6 or 7x = 12 we will have two values for x. (2) is not sufficient
Would pick (A)



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Re: Absolute Value
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30 May 2010, 03:39
OA is "A". My query:: 6  3x = x  2 As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2 What am I missing here??
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Re: Absolute Value
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30 May 2010, 03:45
madeinafrica wrote: (1) 63x = x2 <=> 63x = x2 or 63x = (x2)
solving these equations you will have x=2 or x = 2 then (1) is sufficient. X = 2
(2) 5x+3 = 2x+9 or 5x+3 = (2x+9)
3x = 6 or 7x = 12 we will have two values for x. (2) is not sufficient
Would pick (A) I agree A is right. But do not neglect 2nd statement just because it gave 2 answers. 5x+3 = (2x+ 9) only when 5x+3 <0 so the answer obtained 7x=12 must satisfy this domain. In this case it is satisfying thus making it insufficient. Imagine a case when the 2nd solution does not satisfy the domain, in that case it will be sufficient. So do not neglect the statement just because it gave two answers.
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Re: Absolute Value
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30 May 2010, 04:13
Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps.
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Re: Absolute Value
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30 May 2010, 04:28
Thanks Bunuel!! +1
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Re: Absolute Value
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31 May 2010, 07:26
very nicely explained Bunuel... +1.. thanks



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Re: Absolute Value
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27 May 2013, 05:20
Bunuel wrote: Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps. I am not satisfied with the solution. How do we solve modulus, we take two cases, 1)if (63x)>0 then given equation comes out to be 63x=x2 which gives x=2 ,but we are considering the case when 63x>0 or x<2, hence this soln x=2 is not possible. 2) When we consider (63x)<0 then equation comes out to be 6+3x= x2 which again gives x=2 , but this time we are considering 63x<0 or x>2, hence this soln x=2 is not feasible in this case also. So according to me answer is E.



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Re: Absolute Value
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27 May 2013, 05:26
gchawla123 wrote: Bunuel wrote: Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps. I am not satisfied with the solution. How do we solve modulus, we take two cases, 1)if (63x)>0 then given equation comes out to be 63x=x2 which gives x=2 ,but we are considering the case when 63x>0 or x<2, hence this soln x=2 is not possible. 2) When we consider (63x)<0 then equation comes out to be 6+3x= x2 which again gives x=2 , but this time we are considering 63x<0 or x>2, hence this soln x=2 is not feasible in this case also. So according to me answer is E. Note that OA for this questio is A, not E. Next, to check whether x=2 is a root of 6  3x = x  2 just substitute x=2 into it: 6  3*2 = 0 = 2  2. Hope it's clear.
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Re: What is the value of x?
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29 May 2013, 21:05
Hussain15 wrote: What is the value of x?
(1) 6  3x = x  2
(2) 5x + 3 = 2x + 9 Hi Bunuel, cann't it be solved by the following approach? Statement 1: 63x=x2 X=2 Again, (63x)=x2 ie 6+3x=x2 X= 2 Similar solution from in both scenario. Hence sufficient. Statement 2: Similar process and we will get x= 12/7 and 2. Two different solutions. hence Insufficient. Ans: A Atal Pandit
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Re: What is the value of x?
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30 May 2013, 01:33
atalpanditgmat wrote: Hussain15 wrote: What is the value of x?
(1) 6  3x = x  2
(2) 5x + 3 = 2x + 9 Hi Bunuel, cann't it be solved by the following approach? Statement 1: 63x=x2 X=2 Again, (63x)=x2 ie 6+3x=x2 X= 2 Similar solution from in both scenario. Hence sufficient. Statement 2: Similar process and we will get x= 12/7 and 2. Two different solutions. hence Insufficient. Ans: A Atal Pandit Yes, we can do this way too. Typo: 5x + 3 = 2x + 9 > x=12/7 or x=2.
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Re: What is the value of x?
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04 Jun 2013, 17:15
It seems that I have solved some abs. value problems by finding what makes the equation in the abs. value sign = to zero, so for #2, why do we find what makes the equation outside the abs. value sign = to zero? In other words...
5x + 3 = 2x + 9. ==> When x=4.5 2x+9 = 2(4.5)+9 = 9+9 = 0.
Thanks!



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Re: What is the value of x?
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04 Jun 2013, 17:33
WholeLottaLove wrote: It seems that I have solved some abs. value problems by finding what makes the equation in the abs. value sign = to zero, so for #2, why do we find what makes the equation outside the abs. value sign = to zero? In other words...
5x + 3 = 2x + 9. ==> When x=4.5 2x+9 = 2(4.5)+9 = 9+9 = 0.
Thanks! We can say that \(x\geq{4.5}\) by the same logic as we concluded that \(x\geq{2}\) in (1).
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Re: What is the value of x?
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04 Jun 2013, 21:29
How did you get "for X> or = 2, 6x3 = 6x+3" ?



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Re: What is the value of x?
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05 Jun 2013, 00:35
Yahtzeefish wrote: How did you get "for X> or = 2, 6x3 = 6x+3" ? We have \(6  3x = x  2\). The left hand side (6  3x) is an absolute value. The absolute value is always more than or equal to 0, thus the right hand side (x  2) must also be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). When \(x\geq{2}\), \(6  3x\leq{0}\), thus \(6  3x=(63x)=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. Similar questions to practice: isx01x34x32x32x127978.htmlwhatisx1x22x3x126874.htmlwhatisthevalueofy127976.htmlHope it helps.
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Re: What is the value of x?
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05 Jun 2013, 01:40
Bunuel wrote: atalpanditgmat wrote: Hussain15 wrote: What is the value of x?
(1) 6  3x = x  2
(2) 5x + 3 = 2x + 9 Hi Bunuel, can't it be solved by the following approach? Statement 1: 63x=x2 X=2 Again, (63x)=x2 ie 6+3x=x2 X= 2 Similar solution from in both scenario. Hence sufficient. Statement 2: Similar process and we will get x= 12/7 and 2. Two different solutions. hence Insufficient. Ans: A Atal Pandit Yes, we can do this way too. Typo: 5x + 3 = 2x + 9 > x=12/7 or x=2. Hello Bunuel, I followed the same approach but looking at solution given by you, It is actually not required . I have seen some question types where in we just check for one condition (X>/0 or X<0) and we can conclude the answer (Like you just did in this one) Are there particular question type where in your approach to this question will work. For ex  Some expression in x = x some no. Again since LHS >/0, we can say xsome no >/ > x>/ some no. This can be helpful in saving time by avoiding checking expression for 2 possible case of x (<0,>0) Thanks Mridul
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Re: What is the value of x?
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05 Jun 2013, 01:43
mridulparashar1 wrote: Bunuel wrote: atalpanditgmat wrote: Yes, we can do this way too.
Typo: 5x + 3 = 2x + 9 > x=12/7 or x=2. Hello Bunuel, I followed the same approach but looking at solution given by you, It is actually not required . I have seen some question types where in we just check for one condition (X>/0 or X<0) and we can conclude the answer (Like you just did in this one) Are there particular question type where in your approach to this question will work. For ex  Some expression in x = x some no. Again since LHS >/0, we can say xsome no >/ > x>/ some no. This can be helpful in saving time by avoiding checking expression for 2 possible case of x (<0,>0) Thanks Mridul Yes, in such kind of questions this approach might be helpful.
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Re: What is the value of x?
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05 Jun 2013, 11:54
I just wanted to make sure the way I solved this problem is correct:
What is the value of x?
1. 63x=x2 2. 5x+3=2x+9
1. 63x is positive when x<2. For example, if x=1 then 63x = 63(1) = 3. 63x is negative when x>2. For example, if x=3 then 63(x) = 63(3) = 3.
For positive cases: 63x=x2 ==> x=2 For negative cases: 6+3x=x2 ==> x=2
My issue here is that for the positive case x must be less than 2 and for the negative case x must be greater than 2 but for both cases here, I got x =2. The answer is sufficient, but I thought that for the cases to be sufficient the results had to be in line with what x= (for example, for the positive case here, x=2 but before I found that X must be LESS than 2 for 63x to be positive)
Could someone explain? I see why #2 fails (because for the positive case, 2 is greater than 3/5 but for the negative case, 7/12 is NOT greater than 3/5 but I am a bit lost with #1)
Thanks!



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Re: What is the value of x?
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05 Jun 2013, 12:02
WholeLottaLove wrote: I just wanted to make sure the way I solved this problem is correct:
What is the value of x?
1. 63x=x2 2. 5x+3=2x+9
1. 63x is positive when x<2. For example, if x=1 then 63x = 63(1) = 3. 63x is negative when x>2. For example, if x=3 then 63(x) = 63(3) = 3.
For positive cases: 63x=x2 ==> x=2 For negative cases: 6+3x=x2 ==> x=2
My issue here is that for the positive case x must be less than 2 and for the negative case x must be greater than 2 but for both cases here, I got x =2. The answer is sufficient, but I thought that for the cases to be sufficient the results had to be in line with what x= (for example, for the positive case here, x=2 but before I found that X must be LESS than 2 for 63x to be positive)
Could someone explain? I see why #2 fails (because for the positive case, 2 is greater than 3/5 but for the negative case, 7/12 is NOT greater than 3/5 but I am a bit lost with #1)
Thanks! You have a point here. When you study a function, an abs value, you correctly define where is positive or negative \(x>a\) positve for example \(x<a\) negative for example As you see the value \(a\) is not contained in neither interval... How do we solve this? Just put an \(=\) sign under one of the above inequalities, so \(x\geq{}a\) and \(x<a\)  or \(x>a\) and \(x\leq{}a\). Where you put the sign, it doesn not matter, the result will not change. So just rewrite your equation into For positive cases (\(x\leq{2}\)): 63x=x2 ==> x=2 VALIDFor negative cases (\(x>2\)): 6+3x=x2 ==> x=2 and you have the result! Nothing changes, everything you did is absolutely correct, just put an \(=\) sign in one of the intervals Hope it's clear
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Re: What is the value of x?
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05 Jun 2013, 12:18
Ok...I see that. But isn't the whole point to test cases where 63x is positive and negative? if X=2 then 63x is neither positive or negative, it's just zero.




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