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M03 #01

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M03 #01 [#permalink] New post 19 Sep 2008, 16:06
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Is x > 3?

(1) (x - 3)(x - 2)(x - 1) > 0
(2) x > 1

[Reveal] Spoiler:
E

[Reveal] Spoiler:
The explanation starts with stating that S1 has three roots. Ok, then X >1 is the correct root of 3 possible roots, according to S2. That leaves us with X>1, not X>3. Where is X>3 derived from? That does not make sense to me.

Also, assuming we don't think about roots and we just start filling in S1 numbers (as I did during the test). If I put in 3 or anything less, the product is either negative (If X is less than 1) or zero if X is 1, 2 or 3. With that reasoning, I see the answer as Yes, X>3, only needing S1 information.


Thank you!
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Re: M03 #01 [#permalink] New post 21 Sep 2008, 23:20
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Hi dczuchta.
In the question we need to know for sure if X \gt 3. Let's see what S1 and S2 can give us. As stated in the S1, solving for X gives us
X \in (1,2) \cup (3,\infty) (X is either greater than 3 or lies in the range from 1 to 2) - insufficient.
S2 only gives us X \gt 1 (X can be 2, which is less than 3) - insufficient.
Combining the two statements adds nothing. Therefore E.

Hope this helps.
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Re: M03 #01 [#permalink] New post 30 Dec 2008, 03:49
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From the statement 1 & 2 let us take the value for x > 1 and not (3,2,1)

Case 1: Take the value for x as 3>x>2 then (-ve)(+ve)(+ve) so the end reulst will be -ve
Case 2: Take the value for 2>x>1 then (-ve)(-ve)(+ve) so the end result will be +ve

hence both the statements are not sufficient
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Re: M03 #01 [#permalink] New post 03 Jan 2009, 05:24
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E is the best

1. (x-3)(x-2)(x-1)>0

x belongs to (1,2) and x>3; clearly not suff

Ex: x=4, Ok x>3, but x=1.5, not OK, x<3

2. x>1, the same as above

1 and 2 : not suff
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Re: M03 #01 [#permalink] New post 03 Jan 2009, 10:13
bhatiagp wrote:
Can someone please explain the solution as I am unable to comprehend the answer given

Is x>3

1) (x-3)(x-2)(x-1) >0
2) x>1

My Answer was A . But the mentioned Answer is E . I had ruled any ways ruled out B, D , but cannot understand why A is incorrect. Please explain


sreehari1250 wrote:
From the statement 1 & 2 let us take the value for x > 1 and not (3,2,1)

Case 1: Take the value for x as 3>x>2 then (-ve)(+ve)(+ve) so the end reulst will be -ve
Case 2: Take the value for 2>x>1 then (-ve)(-ve)(+ve) so the end result will be +ve

hence both the statements are not sufficient


1: 3 > x > 2 is not valied for (x-3) (x-2) (x-1)> 0 cuz if 3>x>2, (x-3) (x-2) (x-1) becomes -ve, which is not the statement 1. so x has to be either (2 > x > 1) or > 3. so insuffff.

2: if x > 1, x could be in between 1 and 2 excluding (2 > x > 1) or greater than 3. so again insuff.

togather also same repeats. therefore, E.
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Re: M03 #01 [#permalink] New post 30 Jan 2009, 13:43
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kman wrote:
Is x \gt 3 ?

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

Can someone explain. I have a problem with the OA


1: (x-3)(x-2)(x-1) \gt 0
If x = 5 or 1.5, the condition is met.

2: x \gt 1
Again, if x = 5 or 1.5, the condition is met.

not suff.

E.
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Re: M03 #01 [#permalink] New post 16 Apr 2009, 01:26
GMAT TIGER wrote:
kman wrote:
Is x \gt 3 ?

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

Can someone explain. I have a problem with the OA


1: (x-3)(x-2)(x-1) \gt 0
If x = 5 or 1.5, the condition is met.

2: x \gt 1
Again, if x = 5 or 1.5, the condition is met.

not suff.

E.

this is too much of a trial error..how did you choose 1.5?i actually did the same method and chose 2.5.It being sufficient i got the answer wrong.Could someone pls explain the answer that is given officially?(given below)
Statement (1) by itself is insufficient. The polynomial has roots . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits to . can be either greater or less than 3.therefore,stat 1 is insufficient.
concept is not clear.Pls help!
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Re: M03 #01 [#permalink] New post 20 Apr 2009, 17:20
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suyashjhawar wrote:
GMAT TIGER wrote:
kman wrote:
Is x \gt 3 ?

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

Can someone explain. I have a problem with the OA


1: (x-3)(x-2)(x-1) \gt 0
If x = 5 or 1.5, the condition is met.

2: x \gt 1
Again, if x = 5 or 1.5, the condition is met.

not suff.

E.

this is too much of a trial error..how did you choose 1.5?i actually did the same method and chose 2.5.It being sufficient i got the answer wrong.Could someone pls explain the answer that is given officially?(given below)
Statement (1) by itself is insufficient. The polynomial has roots . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits to . can be either greater or less than 3.therefore,stat 1 is insufficient.
concept is not clear.Pls help!


Correct me if am wrong here:

Statement 1 : Lets look at it like this:

Since (x-3) (x-2) (x-1) > 0

Either x-3 > 0 or x-2 >0 or x-1>0

So we have x>3 or x>2 or x>1 So :

If x>3 it implies x>2 and x>1
What if x>2 ... then it does not imply x>3 and similarly if x>1 it does not imply x>3. Hence this option is insufficient.

From Stmnt 2: x>1 Well this is again insufficient.

Combining both: We have x>1 AND x>3 or x>2 or x>1 So option 2 is just re-emphasizing a part of what is stated in option 1. We still have x>3 or x>2. Which still leaves us with no conclusive answer. Hence Choose E.

I'm not a big fan of picking numbers - but hope this helps.

-pradeep
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Re: M03 #01 [#permalink] New post 29 Aug 2009, 17:58
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I did it the pleonasm way!However could somebody explain me the concept of roots cos i didn't get the sign changing around values thing..

The polynomial has roots (x-2)(x-3)(x-1) . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. x is therefore limited to (1,2)U(3,infinity)
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Re: M03 #01 [#permalink] New post 30 Aug 2009, 23:12
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I'm no Math expert but I'll share what I know. Suppose you have the inequality (x-2)(x-3)(x-1) > 0.
1. The first thing to do is calculate the values of x when the whole expression equals 0. These values might be called "roots" as well. So the roots for this inequality are 1, 2, and 3.
2. The second thing you do is sketch a simple number line with the roots found (1, 2, and 3 in our example).
3. The third thing is calculating the sign of the whole expression for x in different ranges formed on the number line above and marking those ranges with "+" or "-".
Let's plug 0 into the expression to see what sign it will get:
(x-2) is negative, (x-3) is negative, and (x-1) is also negative. So the whole expression is negative for x from the (-\infty,1) range. You can repeat the same operation for all ranges and find out the sign but it is usually sufficient to do only for one range because the sign of the whole expression changes in the next range on the number line. The signs for the next ranges will be "+", "-", and "+". You can check if you want to.
So, we are sure that x \in (1,2) \cup (3,\infty).

Hope this helped somebody :).
tejal777 wrote:
I did it the pleonasm way!However could somebody explain me the concept of roots cos i didn't get the sign changing around values thing..

The polynomial has roots (x-2)(x-3)(x-1) . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. x is therefore limited to (1,2)U(3,infinity)

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Re: M03 #01 [#permalink] New post 08 Jan 2010, 06:05
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1. (X-3)(X-2)(X-1) > 0

You have 2 possibilities:
X-3>0
x-2>0
x-1>0
You got: x>3
or x-3<0
x-2<0
x-1>0
You got: 1<x<2
so 1 is insuficient.

2. x>1, also insuficient.

Answer: E
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Re: M03 #01 [#permalink] New post 08 Jan 2010, 06:38
Just sketch the graph of y=(x-1)(x-2)(x-3) and get the answer E.
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Re: M03 #01 [#permalink] New post 08 Jan 2010, 07:08
2. x > 1

clear it can't be B since x can be 2 or 4 so option 2 is insufficient.

1.(x-3)(x-2)(x-1) > 0

if x is 1.5 then the value of the above equation will (-1.5)(-0.5)(0.5) > 0

if x is 10 then the value of the above equation still positive (7)(8)(9) >0

so choice 1 is also insufficient

since we picked numbers 1.5 and 10 which is x > 1 so combining both the statements we won't able to get to the answer

so the answer must be E
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Re: M03 #01 [#permalink] New post 29 Aug 2010, 16:08
seekmba wrote:
Is x>3 ?
1. (x-3)(x-2)(x-1) > 0
2. x > 1


Step 1 of the Kaplan Method for DS: Analyze the Stem

We see "is", we think "yes/no" question. So, if x is always greater than 3, sufficient; if x is never greater than 3, sufficient. If sometimes x is greater than 3 and sometimes it isn't, insufficient.

Step 2 of the Kaplan Method for DS: Evaluate the Statements

(2)x > 1. Well, x could be 1.5 ("no") or x could be 5 ("yes")... insufficient, eliminate B and D.

(1) for the product of 3 numbers to be positive, there are two possibilities:

(+)(-)(-)

or

(+)(+)(+)

so, we have to examine both cases.

In the first case, we could pickx = 1.5, giving us (-1.5)(-.5)(.5) which is greater than 0. Is 1.5 > 3? NO

In the second case, we could pick x = 5, giving us (2)(3)(4)which is greater than 0. Is 5 > 3? YES

Yes and no answer, insufficient: eliminate A.

Combined: x=1.5 and x=5worked for both statements, so they're both still valid choices. Accordingly, we can still get a NO and a YES answer: insufficient, choose E.
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Re: M03 #01 [#permalink] New post 29 Aug 2010, 16:20
seekmba wrote:
Is x>3 ?
1. (x-3)(x-2)(x-1) > 0
2. x > 1


1.
x>3 and x>2 and x>1

therefore we don't know if x is greater than 3 because x could be greater than 1... INSUFFICIENT

2.
x>1

if x=2 the answer to the question is no, if x=4 answer to the question is yes -> INSUFFICIENT


combined, x>1, which is the same like B, therefore E

Last edited by zisis on 29 Aug 2010, 17:04, edited 1 time in total.
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Re: M03 #01 [#permalink] New post 29 Aug 2010, 16:32
zisis wrote:
seekmba wrote:
Is x>3 ?
1. (x-3)(x-2)(x-1) > 0
2. x > 1


1.
x>3 and x>2 x>1

therefore we don't know if x is greater than 3 because x could be greater than 1... INSUFFICIENT



Not sure how you derived that inequality, but it's false (in fact, it's impossible.. there's no number greater than 1 that's also greater than two times itself).
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Re: M03 #01 [#permalink] New post 29 Aug 2010, 17:42
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seekmba wrote:
Is x>3 ?
1. (x-3)(x-2)(x-1) > 0
2. x > 1


Use the following approach with your all such inequality questions.
Attachment:
nl.jpg
nl.jpg [ 17.81 KiB | Viewed 7161 times ]


Arrange the roots of the equal in the increasing order and create separators in the form of curves as shown.
Start from the right most i.e. x>0 to be +ve and put alternate -ve , +ve signs as you move along the left side of the number line.

If the inequality says p(x) > 0 then the domain of the inequality is in +ve curve.
If the inequality says p(x) < 0 then the domain of the inequality is in -ve curve.

For the given question in the statement 1 - p(x) > 0
=> consider +ve curve i.e. x>3 and 2>x>1
This is not sufficient as 2>x>1 is also there and we can not the question whether x>3 or not.

Consider the 2nd statement. x>1 does not answer the question x>3 as x>1 could be 2 or 4.
2 will give the answer "No" to the given question whereas 4 will give "yes". Thus not sufficient.

Take both the statements together.
we have 2>x>1, x>3 and x>1

When we combine all the given three inequalities we still can not answer as x=1.5 and x = 4 will give different answer to the question.

Thus E
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Re: M03 #01 [#permalink] New post 29 Aug 2010, 18:38
Hey gurpreetsingh, thanks for the detailed explanation but this approach is difficult to absorb for my brain. I wish I could solve the questions in the manner you did....:(

I used the same approach as "zisis" and hence messed it up.

Thanks skovinsky. I should have done something like you showed.
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Re: M03 #01 [#permalink] New post 13 Jan 2011, 15:09
E.

S1 gives three roots of equation
and is true in two conditions:
either all are positive or two of them are negative. If x is positive then equations are
x > 1 or x > 2 or x > 3 => x > 3
for two negative => 1 <x<2
insufficient
s2 says x> 1 insuff

combining together s2 does not give extra information and hence e

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Re: M03 #01 [#permalink] New post 13 Jan 2011, 20:46
tejal777 wrote:
I did it the pleonasm way!However could somebody explain me the concept of roots cos i didn't get the sign changing around values thing..

The polynomial has roots (x-2)(x-3)(x-1) . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. x is therefore limited to (1,2)U(3,infinity)



Hey ,
here is my explanation. it might help you to understand.
here we go,

the question is
x>3 ?
statement 1:(x-1)(x-2)(x-3) >0

statement 2: x>1

starting with statement 1 :
FOR (x-1)(x-2)(x-3) >0
there are 3 condition for which this eq will be +ve
case 1: (x-1)>0 , (x-2) >0, (x-3) >0 ; on plotting the point on number line, we will get x>3 ( chk with no 4)
case 2: (x-1)<0, (x-2) <0, (x-3) >0 ;; on plotting the point on number line, we will get 2<x<1 ( chk with no 1.5)
case 3: (x-1)<0, (x-2) > 0, (x-3) <0; ; on plotting the point on number line, we will get 2<x<1 ( chk with no 1.5)
case 4: (x-1)>0, (x-2) < 0, (x-3) <0; ; on plotting the point on number line, we will get 3<x<2 ( chk with no 2.5)

so, from here we cant say that x>3.

statement 2: x>1;
doesn't say anything at all. (chk with x=1.5,2.5)

on combining these two statements we cant find anything as we will get the same case as said in CASE 4.

so, either of the statement cant answer this question alone or on combination.

so, final ans is E.
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Re: M03 #01   [#permalink] 13 Jan 2011, 20:46
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