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nasir: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Tue Oct 13, 2009 6:08 pm; Location: NY; Thanked: 3 times

by nasir » Sun Oct 24, 2010 12:18 pm

If x is an integer, what is the value of x?

(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0

Please explain

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Source: Beat The GMAT — Data Sufficiency | Sun Oct 24, 2010 12:18 pm

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Sun Oct 24, 2010 12:42 pm

looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
(2) x^2+4x+3>0, (x+1)(x+3)>0 valid for x<-3, x>-1 many integers are possible insuff

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nasir: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Tue Oct 13, 2009 6:08 pm; Location: NY; Thanked: 3 times

by nasir » Sun Oct 24, 2010 7:25 pm

clock60 wrote:looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
/quote]

x-1 <0 => x<1
x-3<0 => x<3

how you made the relation 1<x<3

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sun Oct 24, 2010 9:28 pm

Consider first (1) alone.
x^2 - 4x + 3 < 0 implies (x-3)(x-1) < 0.
This means case (1) (x-3) > 0 and (x-1) < 0 which is x > 3 and x < 1.
This is not possible since no real number can be greater than 3 and less than 1. This case is hence eliminated.
Or
case (2) (x-3) < 0 and (x-1) > 0 which is x<3 and x > 1.
This is same as 1<x<3.
We accept case (2).
If x is an integer, the only possible value is x = 2.
Hence (1) alone is sufficient to answer the question.
Next consider (2) alone.
x^2 + 4x + 3 > 0.
This means (x+1)(x+3) > 0.
So we have case (1) x+1 > 0 and x+3>0 which is x> -1 and x> -3.
So combining we have x > -1.
Or
case (2) x+1 < 0 and x+3 < 0 which is x<-1 and x<-3.
On combining we have that x< -3 .
Hence from (2) we get that either x< -3 or x> -1.
There can be infinite many values of x if x is an integer.
So (2) alone is not sufficient to answer the question.

The correct answer is (A).

Rahul Lakhani
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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Oct 25, 2010 4:44 am

nasir wrote:
clock60 wrote:looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
/quote]

x-1 <0 => x<1
x-3<0 => x<3

how you made the relation 1<x<3
Guys please explain , We both Don't know this rule.........

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Mon Oct 25, 2010 6:32 am

nasir wrote:If x is an integer, what is the value of x?

(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0

Please explain

(1) x2 - 4x + 3 < 0

(x-1)(x-3)<0

Rule to remember:
Whenever (x-a)(x-b) <0 then x lies between a and b.
Whenever (x-a)(x-b)>0 then x does not lie between a and b.

From above, we can see that x is between 1 and 3.
Integer lying between 1 and 3 is only 2.
Hence x should be equal to 2.
SUFF

(2) x2 + 4x +3 > 0
(x+1)(x+3)>0

This means that x does not lie between -1 and -3.
Hence x can be any value >-1 or <-3
Hence INSUFF

pick A.

Hope this helps!!

Let me know if you have any troubles understanding this.

"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Oct 25, 2010 7:40 am

kvcpk wrote:
(1) x2 - 4x + 3 < 0

(x-1)(x-3)<0

Rule to remember:
Whenever (x-a)(x-b) <0 then x lies between a and b.

if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

and if equation is like this. (x-a)(x+b) <0 Then x will lie in between -a < x < b

kvcpk wrote: Whenever (x-a)(x-b)>0 then x does not lie between a and b.

if equation is like this. (x+a)(x+b) >0 Then x will not lie in between a and b a> x, x< b

and if equation is like this. (x-a)(x+b) >0 Then x will lie in between -a > x, x < b

Please correct me if i am wrong.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Mon Oct 25, 2010 8:14 am

goyalsau wrote:
if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

Please correct me if i am wrong.

Grrr.... Man why do you need a Concrete formula always. Just give yourself sometime and analyse it (IMHO)

(x+a)(x+b) <0

Let me take a = 2 and b = 3.

Thus, (x+2) (x+3) < 0
So x+2 < 0 and x+3 > 0 OR x+2 > 0 and x+3 < 0
So x < -2 and x > -3 OR x > -2 and x < -3

Consider x<-2 and x>-3. Which decimal will satisfy this? all in between -3 and -2 say x = -2.5

What happened here?

-3 < x < -2 but what is your a and b? a and b are 2 and 3 respectively.

In general what happened... -b < x < -a (Not what you have thought of "a < x < b")

Thus we can say the solution will be -b < x < -a OR a < -x < b.

One question to you why did I select only x < -2 and x > -3, NOT x > -2 and x < -3 ?

Please Analyse the other cases what ever you have mentioned in your earlier post

PS: And also I hope I m correct. I have never encountered this so if it is wrong Sourabh please bear with me

If the problem is Easy Respect it, if the problem is tough Attack it

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Mon Oct 25, 2010 8:35 am

goyalsau wrote:
if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

and if equation is like this. (x-a)(x+b) <0 Then x will lie in between -a < x < b

Hi Saurabh,

I think you got me wrong. You missed on the sign before the values a and b.
If (x-a)(x-b)<0
then a<x<b
[Notice that signs before a and b are negative signs]

Ok, So what if (x+a)(x+b)<0??
We can rewrite it is (x- (-a))(x- (-b))<0
So x lies between -a and -b
-a<x<-b

If (x+a)(x-b)<0 then
-a<x<b
[Notice the positive sign before a in x+a]

If (x-a)(x+b)<0 then
-b<x<a
[Notice the positive sign before b in x+b]

Helpful?? If not let me know.

When >0, then the above funda remains same. Only change is that x does not lie in that range now.

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jeetu_vishnoi: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Mon Oct 25, 2010 1:51 pm; Thanked: 1 times

by jeetu_vishnoi » Mon Oct 25, 2010 2:11 pm

Answer is A.

first equation can be written as (x-3)(x-1)<0 which means that (x-3)<0 and (x-1)>0
it shows that x<3 and x>1 and it is given that x is integer hence x=2.

first equation can be written as (x+3)(x+1)<0 which means that either {(x+3)<0 and (x+1)<0} or {(x+3)>0 and (x+1)>0}
which clearly shows that it's not possible to decide tha value of x.

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