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Re: Is [m]x>3[/m] ? [#permalink]
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: Is [m]x>3[/m] ? [#permalink]
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: Is [m]x>3[/m] ? [#permalink]
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.

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

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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Answer will be E. I is insufficient because X = (1,2) U (3,infinity) and II is also not sufficient as X> 1, in this cas X can be 1.5,2,2.5 etc. combining I and II is not sufficient.