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Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) 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.
Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) 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 X-1, X-2 and X-3 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. _________________
Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) 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 X-1, X-2 and X-3 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.
_________________
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.
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.
I think this is a high-quality 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 x-3>0, x-2>0 and x-1>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
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 ?
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.
Yes. Please check the links provided above for more on this.
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if 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.
_________________
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 (x-3)(x-2) must be both positive or negative. If x=1.5, then (x-3)(x-2)>0. But x is less than 3, so the answer is NO. If x=4, then (x-3)(x-2) also greater than 0. And x greater than 3, so the answer is YES.
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52% (00:50) correct 
