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# M03-01

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:19
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Difficulty:

55% (hard)

Question Stats:

50% (00:49) correct 50% (00:52) wrong based on 211 sessions

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Is $$x \gt 3$$?

(1) $$(x-3)(x-2)(x-1) \gt 0$$

(2) $$x \gt 1$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:19
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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.

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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.

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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06 Oct 2014, 07:42
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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.

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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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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08 Oct 2014, 01:52
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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

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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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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 >=3

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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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]

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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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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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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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.

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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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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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25 Oct 2017, 02:06
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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