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Question Stats:
53% (01:18) correct 47% (01:14) wrong based on 290 sessions
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Is \(x \gt 3\)? (1) \((x3)(x2)(x1) \gt 0\) (2) \(x \gt 1\)
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Re M0301
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16 Sep 2014, 00:19
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, 07:42
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, 17:20
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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08 Oct 2014, 01:52
p2bhokie wrote: 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. No. Try x = 1.5. Check other solutions here: m0370436.html#p771274Below links might help:
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Re M0301
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06 Jan 2016, 07:33
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, 11:06
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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17 Mar 2016, 11:17
eldin wrote: In my opinion.... The solution for statement one should be: S1 therefore limits x to A) 1 > x < 2 B) x >=3[/color] Instead of S1 therefore limits x to [color=#ed1c24] (1,2)∪(3,∞) Unlike the official solution, x cannot be 1 or 2 since the equation would then equal zero. 1. The highlighted part does not make any sense. 2. x cannot be 3 either. 3. (1,2) means that x is between 1 and 2, not inclusive.
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Re M0301
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18 Jul 2016, 08:19
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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26 Sep 2017, 17:10
hi Bunuelif the question had said x is an integer, confirm A would have been suffient?
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Re: M0301
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26 Sep 2017, 21:15
streetking wrote: hi Bunuelif the question had said x is an integer, confirm A would have been suffient? Yes. (1) gives 1 < x < 2 or x > 3. If x were an integer then x could be 4, 5, 6, ... and in the case we would have an YES answer to the question whether x > 3.
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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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22 May 2018, 13:44
I think this is a highquality question and I agree with explanation.



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28 Dec 2018, 16:27
I think this is a highquality question and I agree with explanation. tricky. I never even thought about fractions.



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Re: M0301
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01 Apr 2019, 11:15
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 Hi BunuelS1 says (x−3)(x−2)(x−1) > 0 This implies that x > 3, x > 2 and x > 1 So, x can be any value (integer and noninteger), hence Answer is YES or NO. As mentioned in the solution above for S1, can you please elaborate on the following? "The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"i'm not exactly familiar with this ?? Would appreciate your support! Thanks



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Re: M0301
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01 Apr 2019, 11:25
JIAA 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 Hi BunuelS1 says (x−3)(x−2)(x−1) > 0 This implies that x > 3, x > 2 and x > 1 So, x can be any value (integer and noninteger), hence Answer is YES or NO. As mentioned in the solution above for S1, can you please elaborate on the following? "The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"i'm not exactly familiar with this ?? Would appreciate your support! Thanks I provided some links on the theory above. If that does not help, here is an alternative solution. Is \(x>3\)? (1) \((x3)(x2)(x1)>0\) The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+). Note that: out of 3 numbers \(x3\) is the least one and \(x1\) is the biggest one. \((+)(+)(+)\) is when even the least one is positive so when \(x3>0\) > \(x>3\); \((+)()()\) is when the biggest one is positive (\(x1>0\) > \(x>1\)) and the next one (hence the leas one too) negative (\(x2<0\) > x<2), so when \(1<x<2\); So \((x3)(x2)(x1)>0\) means that: \(x>3\) or \(1<x<2\) > \(x\) may or may not be more than 3. Not sufficient. (2) \(x>1\). Clearly insufficient. (1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(x>3\) or \(1<x<2\), so \(x\) may or may not be more than 3. Not sufficient. Answer: E.
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Re: M0301
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02 Apr 2019, 00:32
Bunuel wrote: JIAA 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 Hi BunuelS1 says (x−3)(x−2)(x−1) > 0 This implies that x > 3, x > 2 and x > 1 So, x can be any value (integer and noninteger), hence Answer is YES or NO. As mentioned in the solution above for S1, can you please elaborate on the following? "The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"i'm not exactly familiar with this ?? Would appreciate your support! Thanks I provided some links on the theory above. If that does not help, here is an alternative solution. Is \(x>3\)? (1) \((x3)(x2)(x1)>0\) The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+). Note that: out of 3 numbers \(x3\) is the least one and \(x1\) is the biggest one. \((+)(+)(+)\) is when even the least one is positive so when \(x3>0\) > \(x>3\); \((+)()()\) is when the biggest one is positive (\(x1>0\) > \(x>1\)) and the next one (hence the leas one too) negative (\(x2<0\) > x<2), so when \(1<x<2\); So \((x3)(x2)(x1)>0\) means that: \(x>3\) or \(1<x<2\) > \(x\) may or may not be more than 3. Not sufficient. (2) \(x>1\). Clearly insufficient. (1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(x>3\) or \(1<x<2\), so \(x\) may or may not be more than 3. Not sufficient. Answer: E. Thanks Bunuel for the alternate explanation! It really did help !



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Re: M0301
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15 May 2019, 16:46
I couldn't see any 'theory' relating to polynomials in the links provided unless I overlooked it via the search function in my browser. While the theory is noted via Bunuel's explanation, I think a more simplistic approach is to test noninteger values. Statement 1 X= 1.5 produces (1.53)(1.52)(1.51) = +ve number... therefore x>0 > is X>3 ? NO X=4 produces (+)(+)(+)>0 is x>3? Yes Therefore Insufficient Statement (2) is also insufficient even without considering noninteger values i.e. since "x>1" x can be 2 or 4 producing No and Yes answers. (1+2) no new information > Insufficient.
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Re: M0301
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29 May 2019, 07:42
I don't know if my way of thinking is correct. However, I managed to answer it correctly.
I gave x a small number (3), a large number (10) and a negative number. X=3 gave me 0. x=10 gave me a number higher than 3. Therefore statement 1 could be either more or less than 3.
Statement two gave me a number that was any number (infinite) that 2 or bigger. It could be a 2 or it could be a number bigger than 3. That statement was also insufficient.
Together, both are insufficient.







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