Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 29 May 2017, 21:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Director
Joined: 09 Aug 2006
Posts: 521
Followers: 2

Kudos [?]: 96 [0], given: 0

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

### Show Tags

17 Nov 2007, 06:33
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is x > 3 ?

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

Manager
Joined: 08 Nov 2007
Posts: 99
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: DS : X > 3 ? [#permalink]

### Show Tags

17 Nov 2007, 09:23
Amit05 wrote:
Is x > 3 ?

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

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...
SVP
Joined: 29 Aug 2007
Posts: 2476
Followers: 70

Kudos [?]: 774 [0], given: 19

Re: DS : X > 3 ? [#permalink]

### Show Tags

17 Nov 2007, 09:34
Amit05 wrote:
Is x > 3 ?

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

A. nice one as x>3. if not (x-3)(x-2)(x-1) will be either 0 or -ve.
Senior Manager
Joined: 27 Aug 2007
Posts: 253
Followers: 1

Kudos [?]: 12 [0], given: 0

Re: DS : X > 3 ? [#permalink]

### Show Tags

17 Nov 2007, 09:42
Amit05 wrote:
Is x > 3 ?

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

Clearly A, just draw a number line and everything will be clear
VP
Joined: 08 Jun 2005
Posts: 1145
Followers: 7

Kudos [?]: 213 [0], given: 0

### Show Tags

17 Nov 2007, 09:42
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

Director
Joined: 01 May 2007
Posts: 793
Followers: 2

Kudos [?]: 331 [0], given: 0

### Show Tags

17 Nov 2007, 16:43
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
Joined: 08 Nov 2007
Posts: 99
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

17 Nov 2007, 17:52
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
Joined: 09 Aug 2006
Posts: 521
Followers: 2

Kudos [?]: 96 [0], given: 0

### Show Tags

17 Nov 2007, 22:15
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

You Rock man .. Got it. I had that integer thing in my mind but was to lazy to give it a try ..
17 Nov 2007, 22:15
Display posts from previous: Sort by