number range

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mehrasa: Master | Next Rank: 500 Posts; Posts: 279; Joined: Fri Nov 05, 2010 5:43 pmFollowed by: 1 member; Thanked: 15 times

by mehrasa » Tue Sep 13, 2011 2:04 am

f x is an integer, is |x|> 1?

1. (1-2x)(1+x) < 0
2. (1-x)(1+2x) < 0

Q is saying is x>1 or x<-1?
stat 1) is insufficient bcuz it gives us x>1/2 and x<-1
stat 2) is also insufficient bcuz the range is x>1 and x<-1/2

my Q is that when we want to consider both statements together, we have to find the intersection of answers from each of statements or their union?

[spoiler]OA: C[/spoiler]

Last edited by mehrasa on Tue Sep 13, 2011 2:49 am, edited 1 time in total.

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Source: Beat The GMAT — Data Sufficiency | Tue Sep 13, 2011 2:04 am

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Tue Sep 13, 2011 2:11 am

You combine options and find the union because both the statements are considered true.

Here if x>1 and x>1/2 is deduced from the 2 statements, you combine them and say X takes values from 1/2 to infinity

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mehrasa: Master | Next Rank: 500 Posts; Posts: 279; Joined: Fri Nov 05, 2010 5:43 pmFollowed by: 1 member; Thanked: 15 times

by mehrasa » Tue Sep 13, 2011 2:31 am

so the answer is no, [x] is not greater than 1 but it is between -1/2 and 1/2
but still two stats together are sufficient

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Tue Sep 13, 2011 2:34 am

i'd reconsider, as x<-1 and x>1 are two valid sets (ranges) for two inequalities. Hence |x|>1 is true and answer [C]

shankar.ashwin wrote:You combine options and find the union because both the statements are considered true.

Here if x>1 and x>1/2 is deduced from the 2 statements, you combine them and say X takes values from 1/2 to infinity

Success doesn't come overnight!

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Tue Sep 13, 2011 3:28 am

@mehrasa

as pemdas has pointed out, you need to take the most restrictive inequalities. Frankly, I can't understand what union or intersection in such a case even means.lol.

Now from statement 1 we have
x<-1/2 and x>1

from statement 2 we have
x<-1 and x>1/2

Lets combine the extremes on the left hand side of both statements first
x<-1/2 and x<-1 out of this x<-1 is more restrictive as it satisfies both values. Draw a number line and solve this if you are having trouble understanding it. Thats what i normally do.

For the right hand side values of both statements we have x>1 and x>1/2. Out of which x>1 is more restrictive as it satisfies both values. So we only consider x>1.

Therefore we have x<-1 and x>1 and in both cases |x| is ALWAYS greater than 1 so [spoiler]C[/spoiler]

If you're having trouble understanding this concept or need an easy way around. I would recommend reading the following link

https://blog.gmatboost.com/2011/08/29/ti ... questions/

That blog is run by Greg Michnikov from beatthegmat. The method he shows there is any extremely convenient way to solve such confusing inequality problems very quickly. That method tends to be a little tedious at first but as u get used to it, u'll just breeze thru such similar problems. He has put up some other very helpful stuff as well. Check it out.

Cheers

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Sep 13, 2011 3:36 am

mehrasa wrote:f x is an integer, is |x|> 1?

1. (1-2x)(1+x) < 0
2. (1-x)(1+2x) < 0

Q is saying is x>1 or x<-1?
stat 1) is insufficient bcuz it gives us x>1/2 and x<-1
stat 2) is also insufficient bcuz the range is x>1 and x<-1/2

my Q is that when we want to consider both statements together, we have to find the intersection of answers from each of statements or their union?

[spoiler]OA: C[/spoiler]

Question rephrased: Is the distance between 0 and x greater than 1?

Statement 1: (1-2x)(1+x) < 0.
The critical points are x=-1 and x=1/2.
These are the only values of x that make (1-2x)(1+x) equal to 0.
When x is any other value, (1-2x)(1+x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.

x<-1:
Let x=-2.
(1- 2(-2))(1+ (-2)) < 0
-5 < 0.
This works.
x<-1 is part of the range.

-1<x<1/2:
Let x=0.
(1- 2*0)(1+0) < 0
1 < 0.
Doesn't work.
-1<x<1/2 is not part of the range.

x>1/2:
Let x=1.
(1- 2*1))(1+1) < 0
-2 < 0.
This works.
x>1/2 is part of the range.

Thus, x<-1 or x>1/2:

The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.

Statement 2: (1-x)(1+2x) < 0
The critical points are x=-1/2 and x=1.
These are the only values of x that make (1-x)(1+2x) equal to 0.
When x is any other value, (1-x)(1+2x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.

x<-1/2:
Let x=-1.
(1- (-1))(1+ 2(-1)) < 0
-2< 0.
This works.
x<-1/2 is part of the range.

-1/2<x<1:
Let x=0.
(1-0)(1+ 2*0) < 0
1 < 0.
Doesn't work.
-1/2<x<1 is not part of the range.

x>1:
Let x=2.
(1- 2)(1+ 2*2) < 0
-5<0.
This works.
x>1 is part of the range.

Thus, x<-1/2 or x>1:

The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.

Statements 1 and 2 combined:
Only the values represented by the GREEN lines are included in BOTH ranges:

x<-1 or x>1.
Thus, the distance between 0 and x must be greater than 1.
Sufficient.

The correct answer is C.

Last edited by GMATGuruNY on Tue Sep 13, 2011 7:04 am, edited 1 time in total.

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Tue Sep 13, 2011 5:56 am

Here since x is an integer, x cannot take values from 1/2 to 1, so you can neglect those cases.
And you have got it wrong. Combining the 2 statements we get,

X ranges from -infinity to -1/2 (and)
X ranges from 1/2 to infinity.

Using both, we basically deduce x cannot be 0. Hence the absolute value of the next integer is 1.

Dont confuse with union or intersection here. The best way to me is to draw a number line and plot the range, whatever you mark would give you the exact range and you could greatly avoid confusion.

mehrasa wrote:so the answer is no, [x] is not greater than 1 but it is between -1/2 and 1/2
but still two stats together are sufficient

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1947: Master | Next Rank: 500 Posts; Posts: 215; Joined: Sat Jun 14, 2008 4:24 pmFollowed by: 1 member; Thanked: 13 times

by 1947 » Tue Sep 13, 2011 7:13 am

GMATGuruNY wrote:
mehrasa wrote:f x is an integer, is |x|> 1?

1. (1-2x)(1+x) < 0
2. (1-x)(1+2x) < 0

Q is saying is x>1 or x<-1?
stat 1) is insufficient bcuz it gives us x>1/2 and x<-1
stat 2) is also insufficient bcuz the range is x>1 and x<-1/2

my Q is that when we want to consider both statements together, we have to find the intersection of answers from each of statements or their union?

[spoiler]OA: C[/spoiler]
Question rephrased: Is the distance between 0 and x greater than 1?

Statement 1: (1-2x)(1+x) < 0.
The critical points are x=-1 and x=1/2.
These are the only values of x that make (1-2x)(1+x) equal to 0.
When x is any other value, (1-2x)(1+x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.

x<-1:
Let x=-2.
(1- 2(-2))(1+ (-2)) < 0
-5 < 0.
This works.
x<-1 is part of the range.

-1<x<1/2:
Let x=0.
(1- 2*0)(1+0) < 0
1 < 0.
Doesn't work.
-1<x<1/2 is not part of the range.

x>1/2:
Let x=1.
(1- 2*1))(1+1) < 0
-2 < 0.
This works.
x>1/2 is part of the range.

Thus, x<-1 or x>1/2:

The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.

Statement 2: (1-x)(1+2x) < 0
The critical points are x=-1/2 and x=1.
These are the only values of x that make (1-x)(1+2x) equal to 0.
When x is any other value, (1-x)(1+2x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.

x<-1/2:
Let x=-1.
(1- (-1))(1+ 2(-1)) < 0
-2< 0.
This works.
x<-1/2 is part of the range.

-1/2<x<1:
Let x=0.
(1-0)(1+ 2*0) < 0
1 < 0.
Doesn't work.
-1/2<x<1 is not part of the range.

x>1:
Let x=2.
(1- 2)(1+ 2*2) < 0
-5<0.
This works.
x>1 is part of the range.

Thus, x<-1/2 or x>1:

The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.

Statements 1 and 2 combined:
Only the values represented by the GREEN lines are included in BOTH ranges:

x<-1 or x>1.
Thus, the distance between 0 and x must be greater than 1.
Sufficient.

The correct answer is C.

Mitch,
in first inequality once we know the critical points are -1 and 0.5 then can we just check point between both of them and leave the other conditions >0.5 and less than -1/2
since the behavior changes around the critical points.
Is there a need to test all the values

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Sep 13, 2011 7:23 am

1947 wrote: Mitch,
in first inequality once we know the critical points are -1 and 0.5 then can we just check point between both of them and leave the other conditions >0.5 and less than -1/2
since the behavior changes around the critical points.
Is there a need to test all the values

In this case, once we've determined that the range between the critical points does not satisfy the inequality, we can conclude that the ranges to the left and right of the critical points WILL satisfy the inequality. But be careful when there are more than 2 critical points:

https://www.beatthegmat.com/inequality-c ... 89518.html

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