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16 Sep 2014, 00:18



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Re M0232 [#permalink]
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15 Mar 2015, 12:46
I think this question is poor and not helpful. How can statement (2) be correct if there are two value for x? (x= 0 & 1)



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15 Mar 2015, 14:12



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Re: M0232 [#permalink]
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20 Mar 2015, 07:17
KPMM07 wrote: I think this question is poor and not helpful. How can statement (2) be correct if there are two value for x? (x= 0 & 1) Hi, Thats exactly the reason why (2) is not correct. We will have two values for x^2. when x= 0, x^2 = 0 and when x = 1 , x^2 = 1 Hence (2) is insufficient. Thanks!



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20 Mar 2015, 07:19
sytabish wrote: KPMM07 wrote: I think this question is poor and not helpful. How can statement (2) be correct if there are two value for x? (x= 0 & 1) Hi KPMM07 , Thats exactly the reason why (2) is not correct. We will have two values for x^2. when x= 0, x^2 = 0 and when x = 1 , x^2 = 1 Hence (2) is insufficient. Thanks!



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Re: M0232 [#permalink]
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27 Apr 2015, 05:37
Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From Statement (1): \(x^2 = 2x  1\) \(x^2  2x + 1 = 0\) \((x1)^2 = 0\) \(x 1 = 0\) \(x = 1\) Thus, \(x^2 = 1\). Statement (2) by itself is insufficient. From Statement (2): \(x^2  x^3 = 0\) \(x^2 (1  x) = 0\) \(x = 0\) or \(x =1\).
Answer: A Hi Bunuel I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x 1. Squaring the given equation will not give 'x^2 = 2x  1' although it will give you a unique value of x that is real. Please check. Thanks



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Re: M0232 [#permalink]
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27 Apr 2015, 05:44
Engr2012 wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From Statement (1): \(x^2 = 2x  1\) \(x^2  2x + 1 = 0\) \((x1)^2 = 0\) \(x 1 = 0\) \(x = 1\) Thus, \(x^2 = 1\). Statement (2) by itself is insufficient. From Statement (2): \(x^2  x^3 = 0\) \(x^2 (1  x) = 0\) \(x = 0\) or \(x =1\).
Answer: A Hi Bunuel I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x 1. Squaring the given equation will not give 'x^2 = 2x  1' although it will give you a unique value of x that is real. Please check. Thanks Not sure what exactly are you doing there... x=1 satisfies x^2 = 2x  1: 1^1 = 2*1  1.
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Re: M0232 [#permalink]
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27 Apr 2015, 06:16
Bunuel wrote: Engr2012 wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From Statement (1): \(x^2 = 2x  1\) \(x^2  2x + 1 = 0\) \((x1)^2 = 0\) \(x 1 = 0\) \(x = 1\) Thus, \(x^2 = 1\). Statement (2) by itself is insufficient. From Statement (2): \(x^2  x^3 = 0\) \(x^2 (1  x) = 0\) \(x = 0\) or \(x =1\).
Answer: A Hi Bunuel I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x 1. Squaring the given equation will not give 'x^2 = 2x  1' although it will give you a unique value of x that is real. Please check. Thanks Not sure what exactly are you doing there... x=1 satisfies x^2 = 2x  1: 1^1 = 2*1  1. How did you get from x = (2)^0.5 *x 1 to x^2 = 2x  1?



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Re: M0232 [#permalink]
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27 Apr 2015, 06:20



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27 Apr 2015, 06:36
Bunuel wrote: Engr2012 wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From Statement (1): \(x^2 = 2x  1\) \(x^2  2x + 1 = 0\) \((x1)^2 = 0\) \(x 1 = 0\) \(x = 1\) Thus, \(x^2 = 1\). Statement (2) by itself is insufficient. From Statement (2): \(x^2  x^3 = 0\) \(x^2 (1  x) = 0\) \(x = 0\) or \(x =1\).
Answer: A Hi Bunuel I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x 1. Squaring the given equation will not give 'x^2 = 2x  1' although it will give you a unique value of x that is real. Please check. Thanks Not sure what exactly are you doing there... x=1 satisfies x^2 = 2x  1: 1^1 = 2*1  1. Bunuel, If x^2 =1 , then x = 1 or 1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x 1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x  1".



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Re: M0232 [#permalink]
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27 Apr 2015, 06:40
Bunuel wrote: hUserName wrote: How did you get from x = (2)^0.5 *x 1 to x^2 = 2x  1?
Square \(x = \sqrt{2x  1}\) to get \(x^2 = 2x  1\). Bunuel The way question is getting displayed, it seems that square root is with 2 only and not with the entire thing (x1). Please re format the question so that the root is displayed as per your comment. I think I got my answer to the doubt that I had.



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Re: M0232 [#permalink]
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27 Apr 2015, 06:41
Engr2012 wrote: Bunuel,
If x^2 =1 , then x = 1 or 1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x 1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x  1". You are NOT correct. If you square \(x = \sqrt{2x  1}\) you get \(x^2 = 2x  1\). Next, if you substitute x=1 into \(x = \sqrt{2x  1}\) the equation will hold true.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M0232 [#permalink]
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27 Apr 2015, 06:44
Bunuel wrote: Engr2012 wrote: Bunuel,
If x^2 =1 , then x = 1 or 1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x 1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x  1". You are NOT correct. If you square \(x = \sqrt{2x  1}\) you get \(x^2 = 2x  1\). Next, if you substitute x=1 into \(x = \sqrt{2x  1}\) the equation will hold true. I think the confusion is between 2^0.5*x1 versus (2x1)^(0.5), in the 2nd case the problem and solution as stated is correct...



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Re: M0232 [#permalink]
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27 Apr 2015, 07:08
hUserName wrote: Bunuel wrote: Engr2012 wrote: Bunuel,
If x^2 =1 , then x = 1 or 1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x 1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x  1". You are NOT correct. If you square \(x = \sqrt{2x  1}\) you get \(x^2 = 2x  1\). Next, if you substitute x=1 into \(x = \sqrt{2x  1}\) the equation will hold true. I think the confusion is between 2^0.5*x1 versus (2x1)^(0.5), in the 2nd case the problem and solution as stated is correct... Yes. Exactly. I was checking for X=1 with the square root with 2 only and not the entire expression. It is not clear from the way the question is getting displayed.



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Re: M0232 [#permalink]
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28 Apr 2015, 11:14
hi Bunuel can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.



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29 Apr 2015, 02:10



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17 Oct 2015, 14:00
I think this is a highquality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient.



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Re: M0232 [#permalink]
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17 Oct 2015, 14:45
danjbon wrote: I think this is a highquality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient. You can not divide by x^2 as you don't know whether x=0 or not. Statement 2 is also satisfied by x=0. Be careful with dividing by variables when you either don't know the sign or the values. Dividing by x^2 implies that you are assuming x can't be =0



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Bunuel wrote: djonga wrote: hi Bunuel can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong. x^2 = x^3; x^2  x^3 = 0; x^2(1  x) = 0; x = 0 or x = 1.Hi Bunuel, x^2 = x^3. why can't we just divide by x^2 to get x = 1? What am I missing here? I see how both 0 and 1 are solutions but what theorem am I missing? I feel like it is basic and I may just be forgetting. EDIT: Never mind previous poster answered it.







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