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

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Director
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Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink] New post 17 Nov 2007, 05:33
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A
B
C
D
E

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Is x > 3 ?

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


Could anyone please explain ?
Manager
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Re: DS : X > 3 ? [#permalink] New post 17 Nov 2007, 08:23
Amit05 wrote:
Is x > 3 ?

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


Could anyone please explain ?


I make it C I think.. though I'd like to see the OA.

Statement 1 - if x is 1, 2, 3, or 4, the statement is incorrect, thus, if x is positive, then it must be above 3. However, if x is negative, then it may perhaps work. if x is -2, then the statement is correct.

Statement 2 - on it's own is insufficient - x can be anywhere between 1 and 3...

Together though it tells us x is a positive number above 3...

I think...
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Re: DS : X > 3 ? [#permalink] New post 17 Nov 2007, 08:34
Amit05 wrote:
Is x > 3 ?

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


Could anyone please explain ?


A. nice one as x>3. if not (x-3)(x-2)(x-1) will be either 0 or -ve.
Senior Manager
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Re: DS : X > 3 ? [#permalink] New post 17 Nov 2007, 08:42
Amit05 wrote:
Is x > 3 ?

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


Could anyone please explain ?


Clearly A, just draw a number line and everything will be clear
VP
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 [#permalink] New post 17 Nov 2007, 08:42
Why are you all assuming that x is an integer ?

Is x > 3 ?

statement 1

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

This will be true when:

(x-3), (x-2) and (x-1) are positive (i.e. x=4 {1,2,3})

(x-1) is positive and (x-2), (x-3) are negative (i.e x=1.5 {0.5,-1.5,-2.5}

insufficient

statement 2

clearly insufficient

both statements

still insufficient

statement 2 doesn't add any new information

so the answer is (E)

:)
Director
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 [#permalink] New post 17 Nov 2007, 15:43
Killer, I see your point but still saying A.

I forget the rule, but if you a bunch of #s together is equal to 0, then you can set each one individually to zero and solve. Something like that. I did the following

A.

x-3>0
x-2>0
x-1>0

X > 3 suff.

Killer, Also you said it yourself. You forget this is a yes/no question....Statement 1 is true when we have integers. That is how we knows it is a positive.

B. x>1 not suff.
Manager
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 [#permalink] New post 17 Nov 2007, 16:52
jimmyjamesdonkey wrote:
Killer, I see your point but still saying A.

I forget the rule, but if you a bunch of #s together is equal to 0, then you can set each one individually to zero and solve. Something like that. I did the following

A.

x-3>0
x-2>0
x-1>0

X > 3 suff.

Killer, Also you said it yourself. You forget this is a yes/no question....Statement 1 is true when we have integers. That is how we knows it is a positive.

B. x>1 not suff.


I dont follow. E seems right... I think KS is correct no?

If X is 1.5, then we end up with -1.5 x -0.5 x 0.5 = 0.325 so answer is no

If X is 4, then we end up with 1 x 2 x 3 = 6 so answer is yes.

Thus Statement 1 is insuff.

As KS said - both numbers also fit within statement 2. 1.5 > 1 as is 6.

Thus E.
Director
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 [#permalink] New post 17 Nov 2007, 21:15
KillerSquirrel wrote:
Why are you all assuming that x is an integer ?

Is x > 3 ?

statement 1

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

This will be true when:

(x-3), (x-2) and (x-1) are positive (i.e. x=4 {1,2,3})

(x-1) is positive and (x-2), (x-3) are negative (i.e x=1.5 {0.5,-1.5,-2.5}

insufficient

statement 2

clearly insufficient

both statements

still insufficient

statement 2 doesn't add any new information

so the answer is (E)

:)


You Rock man .. Got it. I had that integer thing in my mind but was to lazy to give it a try ..
  [#permalink] 17 Nov 2007, 21:15
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