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M03 #01

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Intern
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Re: M03 #01 [#permalink] New post 17 Jan 2011, 01:35
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.
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Re: M03 #01 [#permalink] New post 24 Jan 2011, 02:55
Hi,

I tried to explain the process in this post:
m03-70436.html#p624356

Let me know what exactly is not clear if anything.

gmatpapa wrote:
nvgroshar wrote:
Just sketch the graph of y=(x-1)(x-2)(x-3) and get the answer E.


Can anyone explain how to do so?

How do we plot graphs to find the signs of the roots?

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Re: M03 #01 [#permalink] New post 26 May 2011, 08:45
IEsailor wrote:
is x > 3

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


Q: x>3?

1.(x-3)(x-2)(x-1)>0
Roots: 1,2,3
Range: x>3; 1<x<2
Not Sufficient.

2. x>1
Not Sufficient.

Combining both;
x can be any number between 1 and 2 OR it can be greater than 3.
Not Sufficient.

Ans: "E"
**********************

Explanation of my approach lies here:
inequalities-trick-91482.html
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Re: M03 #01 [#permalink] New post 26 May 2011, 08:52
IEsailor wrote:
Hi Fluke,
Can you pls explain in detail the reasoning behind the explanation of the first part.

Thnx


Did you see this:
inequalities-trick-91482.html

Please let me know if you don't understand the approach.
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Re: M03 #01 [#permalink] New post 11 Jun 2011, 17:36
1. Not sufficient

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

x>3 x is greater than 3.

x>1 and x<2 => x is not greater than 3.

2. Not sufficient

x>1

x =2 =>x is not greater than 3
x =4 x is greater than 3.

together,

both the examples in 1 applies here as well. Still not sufficient.

Answer is E.
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Re: M03 #01 [#permalink] New post 17 Jan 2012, 06:11
Ans is E.
statement 1: x>3,x>2 and x>1 => not sufficient
statement 2: x>1 => not sufficient

together also both statement not sufficient. So ans is E
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Re: M03 #01 [#permalink] New post 30 Jan 2014, 03:44
Ans is E.

Consider 1st statement;
(x-3)(x-2)(x-1)>0
This is true in 4 cases:
Case 1: (x-3), (x-2), (x-1) all are > 0, which is possible if x>3 {x-3 > 0}
Case 2: (x-3) > 0 and (x-2) and (x-1) < 0 which is not possible if x>3
Case 3: (x-3) and (x-2)< 0 and (x-1) > 0 which is possible if 1<x<2
Case 4: (x-3) and (x-1)< 0 and (x-2) > 0 which is not possible if x>2 [since x-1 can not be negative for x>2]
This gives two possible answers from Case 1 (x>3) & case 3 (1<x<2). Therefore, this statement is NOT SUFFICIENT

Consider 2nd statement;
x>1
That does not tell us if x>3 or x<3, x could be 1.5, 2, 2.5 etc...Hence, this is NOT SUFFICIENT

Combining 1st and 2nd statement does not give us exact value of x. Hence, the answer should be E
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Re: M03 #01 [#permalink] New post 20 Apr 2014, 05:48
1) Both X=4 and X=1.5 satisfy (X-3)(X-2)(X-1)>0 -> insufficient
2) Clearly insufficient
Combine 2 stats: still cannot, using the same examples as in 1)

-> Choose E
Re: M03 #01   [#permalink] 20 Apr 2014, 05:48
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