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M03-01

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Official Solution:


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
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M03-01 [#permalink]

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New post 06 Oct 2014, 06:19
Bunuel wrote:
Official Solution:


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.
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him1985 wrote:
Bunuel wrote:
Official Solution:


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.
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Re: M03-01 [#permalink]

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New post 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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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: m03-70436.html#p771274

Below links might help:

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New post 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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New post 17 Mar 2016, 11:06
In my opinion.... :peek

The solution for statement one should be:


S1 therefore limits x to A) 1 > x < 2 B) x >=3

Instead of

S1 therefore limits x to (1,2)∪(3,∞) :rocket

Unlike the official solution, x cannot be 1 or 2 since the equation would then equal zero.

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New post 17 Mar 2016, 11:17
eldin wrote:
In my opinion.... :peek

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,∞) :rocket

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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New post 18 Jul 2016, 08:19
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

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Re: M03-01 [#permalink]

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New post 11 Feb 2017, 11:13
Hi Bunuel

One 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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New post 11 Feb 2017, 12:02
pranjal123 wrote:
Hi Bunuel

One 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.


Yes. Please check the links provided above for more on this.
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Re: M03-01 [#permalink]

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New post 26 Sep 2017, 17:10
hi Bunuel

if the question had said x is an integer, confirm A would have been suffient?
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New post 26 Sep 2017, 21:15
streetking wrote:
hi Bunuel

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.
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Re: M03-01   [#permalink] 26 Sep 2017, 21:15
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