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Re: Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink]
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
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Re: Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink]
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)

:)
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Re: Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink]
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.
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Re: Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink]
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.
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Re: Is x > 3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1 [#permalink]
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 ..



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