Responding to a pm:

|x| > 3 implies that x is a point whose distance from 0 is more than 3. So x could be greater than 3 or less than -3. Before you move further, think about the values x can take: 3.00001, 3.5, 4.2, 5.7, 67, 1000, -3.45, -4, -8, -100 etc. The only values it cannot take are -3 <= x <= 3

Which of the following must be true?

I. x > 3

For every value that x can take, must x be greater than 3? No. e.g. if x takes -3.45, -4 etc, it will not be greater than 3 so this is not true.

II. X^2 > 9
This is the same as |x| > 3 so it must be true

III. |x-1|>2
This implies that the distance of x from 1 must be greater than 2. So x is either greater than 3 or less than -1. Now, recall all the values that x can take. For each value, can be say that x is either greater than 3 or less than -1? Yes.
3.00001 - x is greater than 3
3.5 : x is greater than 3
4.2 : x is greater than 3
5.7 : x is greater than 3
67 : x is greater than 3
1000 : x is greater than 3
-3.45 : x is less than -1
-4 : x is less than -1
-8 : x is less than -1
-100 : x is less than -1

For every value that x can take, x will be either greater than 3 or less than -1. Note that we are not saying that every value less than -1 must be valid for x. We are saying that every value that is valid for x (found by using |x| > 3) will be either greater than 3 or less than -1. Hence |x-1|>2 must be true for every value that x can take.
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Original statement |x| > 3, which means either x>3 or x<-3
Now Check the options
Option1- x > 3 - not always true as x can be smaller than -3
Thus option A,C & E is ruled out. Only B & D are left

Option2- X^2 > 9 - Always true for x>3 or x<-3
To check - if x = 4,5,6,7.... or -4,-5,-6,-7
x^2>9

Option3-|x-1|>2,
which means (x-1)>2 ---> x>3 (if x-1>0) - True
it also means (x-1)<-2---->x<-1 (if x-1<0)

X<-1 satisfies x<-3. Thus true

Both 2 & 3 is true
Thus Answer D

Hope it helps

Now if I take x=-2 will it satisfy the condition?

I'm ok untill B and D

But the 3 option says: |x - 1| > 2. Then x > 3 AND x< -1. Our statement says: x > 3 AND x < -3. So why is always TRUE??

Please clarify
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From|x|>3, x will always have to be >3 or <-3, so "our" x will never be between -3 and 3, making the III statement always true.

Ok ok X<- 1 is not possible because our stetement says that X MUST < -3 ...........so X < -1 is not possible. Is This ?????

Silly misunderstood. I apologize

Thanks
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lol. you're doing the same error in reasoning I was doing. You have to prove statement III as true an not the other way around (prove |x|>3 true). If you know that your x will always be more than 3 OR always less than -3 (from |x|>3) then you are restricted to these values when looking to prove statement III as true. Hope you understand.

Edit: Upppss..You got it. Cheers.

Let's solve this question differently

Here following is given
I x I >3

So, Either x>3 OR x<-3
If we plot the values in line we get following -


----(-4)----(-3)-----(-2)-----(-1)-----(0)------(1)-------(2)-------(3)------(4)
<------------I-----------------------------------------------------------I------------>
(x<-3)...............................................................................(x>3)

Now, let us take some value (-4 and 4) and plug into the options.

Option I with x=4 : \(4> 3\) Correct
Option I with x=-4 \(-4>3\) Not correct. So, options I is out.

Option II with x=4 : \(4^2> 9\) Correct
Option II with x=-4 \((-4)^2>3\) Correct. So, options II is right

Option III with x=4 : \(I 4-1I > 2\) Correct
Option III with x=-4 \(I (-4-1) I >2\) Correct. So, options III is right

Hence, correct answer is D

Good explanation thanks . generally I do not this silly but crucial error but this time I could have sworn tha D was false. Indeed, III was tricky especially if I think under time pressure. The latter changes all things, all prespective.

@Corvinis
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You are confusing the question with what is the range of X
We dont have to find the range of x, the question says that distance of x from 0 is more than 3

Satement 1 says x is always positive ... ot always true x can be negative and still the distance from zero will be more than 3

Statement 2 x^2 is more than 9 ofcourse ! is distance of x from 0 is more than 3 than its square will be more than 9 always

Statement 3 distance of x from 1 iss more than 2 .. Of course ! if its distance from 0 is more than three than its distance from 1 will be always more than 3-1 = 2
Weather it is negative or positive
as if it is negative, than distance from 0 is already more than 3, and you are taking distance from 1 so it will be 1 unit more than that from 0 that is 3 + 1 and hence will surely be more than 2

Here these problems are better solved by converting expression to their definition.

Important point is you are given the value of X and you have to see, what all is ture for all those values of X not ! which values of X satisfy the expression mentioned in questions.

I just came across this one and it is clear that this is a tough cookie.

Given that |x| > 3, so that means x > 3 or x < -3. We can also test values here, only 3,4,5,6 or -3, -4, -5 etc work.

Now, let's tackle the statements:
I. x > 3

We know that x > 3 or x < -3, but it is not ALWAYS the case that x > 3. We do have values of x that are less than 3, i.e., when x is -4 or -5. So, I is false.

II. x^2 > 9.

This means x>3 or x < -3. That is what we have above so this is golden. True.

III. |x - 1| > 2

When x > 0
x -1 > 2
x > 3

x < 0
-x + 1 > 2
-x > 1
x < -1

So , III says IF x > 3 or x < -3, THEN x > 3 OR x < -1. True. Test values to prove this x > 3 or x < -3 means x can be -4 and that hits x < -1

Please check the trick to solve this in 1 min.

inequalities-trick-91482-40.html#p1145743
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i am clearly unclear how option 3 is true, since if i take the value of x as -2. It holds true for |x-1|>2. However the value of -2 does not hold true for |x| > 3. Please clarify.

The question reads "If |x| > 3, which of the following must be true?". The value of x=-2 doesn't subscribe to this condition in the first place. You have to filter all the possible values for x, BASED on the condition given in the problem.
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if x<-3 then sure x<-1 ....

The question asks is \(x>3\) or \(x<-3\)?

III tells us that \(x>3\) or \(x<-1\). So is \(x>3\) or \(x<-3\)? YES

\(x>3\) from question => \(x>3\)from III: Correct
\(x<-3\) from question => \(x<-1\) from III: Correct as well. We are asked if x<-3 and III tells us that x<-1 so for sure it will be <-3 also.

Hope it's clear now!

If some numbers confuse you, don't fixate on them. Go ahead and take some other easier examples.
Let's keep the wording of the question same but make it simple.

If n < 6, which of the following must be true?

I.

II.

III. n < 8

Can we say that III must be true? Yes!
If n is less than 6 then obviously it is less than 8 too.
If n is less than 6, it will take values such as -20, 2, 5 etc. All of these values will be less than 8 too.

Values 6 and 7 are immaterial because n cannot take these values. You are given that n is less than 6 so you only need to worry about values that n CAN take. Those should satisfy n < 8.

Similarly, your question says that x > 3 or x < -3

Then we can say that x > 3 or x < -1. All values that will be less than -3 will be less than -1 too.

Check out my post on a similar tricky question : http://www.veritasprep.com/blog/2012/07 ... -and-sets/
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What if x = -2 that is < -1 but > than -3 so IIImust be out?no






Posted from GMAT ToolKit

|x| > 3 means x>3 or x<-3
so I is false because x<-3 is true.


II. x^2 > 9 satisfies both equations (use x=-4 or x=5 )

III. is also true for x=-4 and x=5.

so D