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

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

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New post 16 Oct 2011, 17:14
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 6 sessions

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Is x>3 ?

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

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New post 16 Oct 2011, 19:43
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I would choose E: none of them helps.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 19:55
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Ok we need a yes or no. Is x>3 ?

2) x > 1: it could be 2,3 so answer is No. it could be 4,5,6,7,8 so answer is YES. Two possible answers thus Insufficient.

Between A C and E

3) Lets say that you take the time to solve statement one:

(x - 1)(x - 2)
= x^2 - 2x - x + 2
= x^2 - 3x + 2

(x^2 - 3x + 2) (x - 3)
= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6
= x^3 - 6x^2+ 11x - 6
Sure x^3 - 6x^2+ 11x - 6 > 0 you probably are going to have more than one result so insufficient.

Between C or E

Put them together and:
You know that from st (2) x > 1

so try using 2 in

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

(2-1)(2-2)(2-3)>0
(1)(0)(-1)> 0
0>0 not true, thus we can not use the number two or 3, because it could give the same result.

so x could be 4,5,6,7,8,9,10...
Use 4 and

(4-1)(4-2)(4-3)>0
(3)(2)(1)>0
6>0 check

its (4,5,6,7,8,9,10...) x > 3 ? YES always.

Thus C
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 19:57
K... I thought about it. In this case X can be a fraction. X could be 5/4, in that case we cannot say that x>3. this is actually a GMAT club test question. I am leaning towards E since they have not specified that X is an integer. Thoughts?
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:06
parama79 wrote:
K... I thought about it. In this case X can be a fraction. X could be 5/4, in that case we cannot say that x>3. this is actually a GMAT club test question. I am leaning towards E since they have not specified that X is an integer. Thoughts?

Yes you're right. E is correct answer if the question does not specify that X has to be an integer.

A would be the correct answer if X must be an integer.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:16
KickingGmat wrote:
Ok we need a yes or no. Is x>3 ?

2) x > 1: it could be 2,3 so answer is No. it could be 4,5,6,7,8 so answer is YES. Two possible answers thus Insufficient.

Between A C and E

3) Lets say that you take the time to solve statement one:

(x - 1)(x - 2)
= x^2 - 2x - x + 2
= x^2 - 3x + 2

(x^2 - 3x + 2) (x - 3)
= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6
= x^3 - 6x^2+ 11x - 6
Sure x^3 - 6x^2+ 11x - 6 > 0 you probably are going to have more than one result so insufficient.

Between C or E

Put them together and:
You know that from st (2) x > 1

so try using 2 in

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

(2-1)(2-2)(2-3)>0
(1)(0)(-1)> 0
0>0 not true, thus we can not use the number two or 3, because it could give the same result.

so x could be 4,5,6,7,8,9,10... or 1.25,
Use 4 and

(4-1)(4-2)(4-3)>0
(3)(2)(1)>0
6>0 check

its (1.25,4,5,6,7,8,9,10...) x > 3 ? YES and NO two answers.

Thus E



Yes as some of other users said it does not specify that is an Integer

Last edited by KickingGmat on 16 Oct 2011, 20:30, edited 1 time in total.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:17
KickingGmat wrote:
KickingGmat wrote:
Ok we need a yes or no. Is x>3 ?

2) x > 1: it could be 2,3 so answer is No. it could be 4,5,6,7,8 so answer is YES. Two possible answers thus Insufficient.

Between A C and E

3) Lets say that you take the time to solve statement one:

(x - 1)(x - 2)
= x^2 - 2x - x + 2
= x^2 - 3x + 2

(x^2 - 3x + 2) (x - 3)
= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6
= x^3 - 6x^2+ 11x - 6
Sure x^3 - 6x^2+ 11x - 6 > 0 you probably are going to have more than one result so insufficient.

Between C or E

Put them together and:
You know that from st (2) x > 1

so try using 2 in

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

(2-1)(2-2)(2-3)>0
(1)(0)(-1)> 0
0>0 not true, thus we can not use the number two or 3, because it could give the same result.

so x could be 4,5,6,7,8,9,10...
Use 4 and

(4-1)(4-2)(4-3)>0
(3)(2)(1)>0
6>0 check

its (4,5,6,7,8,9,10...) x > 3 ? YES always.

Thus C



Sorry. I made a mistake in statement one:
(x-1)(x-2)(x-3)>0

Just numbers greater than 4 can be use to satisfy the statement, thus A

This question does not specify that X has to be an integer. There are fractions of where X=<3 that satisfies equation 1. Therefore statement 1 is NOT sufficient.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:27
^ What fractions satisfy equation 1? I'm leaning towards A.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:32
DexDee wrote:
^ What fractions satisfy equation 1? I'm leaning towards A.

Any fraction between 1 and 2

e.g. 5/3

You end up with 1 positive number and two negative numbers that when multiplied together gives you a positive number, which is >0.
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:34
DexDee wrote:
^ What fractions satisfy equation 1? I'm leaning towards A.


5/4 = 1.25

(1.25-1)(1.25-2)(1.25-3)>0
Positive Negative Negative

0.25(-0.75)(-1.75)>0

0.328125>0
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Re: DS - is x>3? [#permalink]

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New post 16 Oct 2011, 20:53
The OA is E. Since fractions cannot be ruled out in this case. Statement B is totally useless. Statement A gives us varying results.
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Re: DS - is x>3? [#permalink]

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New post 17 Oct 2011, 14:38
It looks like A to me!!!!!!!!!
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Re: DS - is x>3? [#permalink]

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New post 18 Oct 2011, 05:54
Already posted:

ds-is-x-48216.html

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Re: DS - is x>3?   [#permalink] 18 Oct 2011, 05:54
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