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Re M0301
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15 Sep 2014, 23:19
Official Solution: Statement (1) by itself is insufficient. The polynomial \((x3)(x2)(x1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3. Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10. Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3. Answer: E
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Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The polynomial \((x3)(x2)(x1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3. Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10. Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.
Answer: E I incorrectly selected A.i thought that X is an integer so X1, X2 and X3 are 3 consecutive numbers and since product of 3 numbers is >0 so all 3 numbers should be either positive or any two can be ve. No any 2 can not be negative cause they are consecutive. if 2 are ve and one is +ve then it would be 2, 1 and 0 but it is given that product can not be zero. So we ultimately left with one solution of all 3 +ve no hence. and if the 3 consecutive number less than X are positive then X will be greater than 3. I got the way you solved it. I understood it before in one of your post but it did not striked in my mind when i was giving test. Can you plese tell me that what is the flaw in my thinking.
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Re: M0301
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06 Oct 2014, 06:42
him1985 wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The polynomial \((x3)(x2)(x1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3. Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10. Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.
Answer: E I incorrectly selected A.i thought that X is an integer so X1, X2 and X3 are 3 consecutive numbers and since product of 3 numbers is >0 so all 3 numbers should be either positive or any two can be ve. No any 2 can not be negative cause they are consecutive. if 2 are ve and one is +ve then it would be 2, 1 and 0 but it is given that product can not be zero. So we ultimately left with one solution of all 3 +ve no hence. and if the 3 consecutive number less than X are positive then X will be greater than 3. I got the way you solved it. I understood it before in one of your post but it did not striked in my mind when i was giving test. Can you plese tell me that what is the flaw in my thinking.You assumed (incorrectly) that x is an integer.
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Re: M0301
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07 Oct 2014, 16:20
Hello Bunuel. Can you please explain the solution in a more lay terms...i implemented similar logic as him1985. If x<3 then the product of the three terms will be negative  isn't it? and we are given that the product of these three terms is +ve...so doesn't that imply that x>3? Can you please point out the flaw in my logic? Thanks for your help.



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Re: M0301
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Re M0301
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06 Jan 2016, 06:33
I think range of X in here x>3 & 1<x<2. Am I right? I have little confusion about this. Would any like to ensure my question? thank you.



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Re: M0301
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17 Mar 2016, 10:06
In my opinion.... The solution for statement one should be: S1 therefore limits x to A) 1 > x < 2 B) x >=3Instead of S1 therefore limits x to (1,2)∪(3,∞) Unlike the official solution, x cannot be 1 or 2 since the equation would then equal zero.



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Re M0301
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18 Jul 2016, 07:19
I think this is a highquality question and I agree with explanation. Hi, i used following logic to solve this question. Pls let me know whether my logic seems ok.
Statement 1 implies x3>0, x2>0 and x1>0, and hence x>3, x>2 and x>1. So, x could be more than 3 and less than/equal to 3 (e.g 2,3). Not sufficient
Statement 2: x>1. Not sufficient
Combine: x could be 2, 3 or more than three. Not sufficient



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Re: M0301
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11 Feb 2017, 10:13
Hi BunuelOne question: In official solution,it is mentioned that  "as roots are distinct so sign will alternate between regions" . So is this concept true for all inequalities having distinct roots. So that If we find the sign for say far end of the region(as in this question x>3) then we can simply alternate the signs without checking ? Thanks.



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Re: M0301
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26 Sep 2017, 16:10
hi Bunuelif the question had said x is an integer, confirm A would have been suffient?
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26 Sep 2017, 20:15



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My solution. I always prefer to analyze from statement 2. So,
(2) Clearly insufficient. x>1 does not mean that x>3. X could be either 2 (the answer is no) or 4 (the answer is yes) (1) Clearly insufficient. Roots of the polynomial are 1, 2 and 3. Could be less or greater than 3 as well.
(2)+(1). If x>1, it means that (x3)(x2) must be both positive or negative. If x=1.5, then (x3)(x2)>0. But x is less than 3, so the answer is NO. If x=4, then (x3)(x2) also greater than 0. And x greater than 3, so the answer is YES.
Answer: E



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Re M0301
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22 May 2018, 12:44
I think this is a highquality question and I agree with explanation.



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Re M0301
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28 Dec 2018, 15:27
I think this is a highquality question and I agree with explanation. tricky. I never even thought about fractions.










