If |x| > 3, then which of the following must be true?
(1) x > 3
(2) x^2 > 9
(3) |x-1| > 2
A. (1) only
B. (2) only
C. (1) and (2) only
D. (2) and (3) only
E. (1),(2) and (3)
My only doubt here is with statement 3.
|x-1| > 3 => x > 3 or x < -1
x > 3 is fine as |x| > 3 for all those x but if x < -1 then all the values dont saisfy |x| > 3 for e.g x = -2 then |x| = 2 < 3 so why should D be the OA.
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Pls explain
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prachi18oct
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If |x| > 3, then it must be true that EITHER x > 3 OR x < -3prachi18oct wrote:If |x| > 3, then which of the following must be true?
(1) x > 3
(2) x^2 > 9
(3) |x-1| > 2
A. (1) only
B. (2) only
C. (1) and (2) only
D. (2) and (3) only
E. (1),(2) and (3)
My only doubt here is with statement 3.
|x-1| > 3 => x > 3 or x < -1
x > 3 is fine as |x| > 3 for all those x but if x < -1 then all the values dont saisfy |x| > 3 for e.g x = -2 then |x| = 2 < 3 so why should D be the OA.
(1) x > 3
This need not be true, since it's also possible that x < -3.
For example, x COULD equal -5
(2) x² > 9
This means that EITHER x > 3 OR x < -3
Perfect - this matches our original conclusion that EITHER x > 3 OR x < -3
(3) |x-1| > 2
Let's solve this further.
We get two cases:
case a) x - 1 > 2, which means x > 3 PERFECT
or
case b) x - 1 < -2, which means x < -1
Must it be true that x < -1?
YES.
We already learned that EITHER x > 3 OR x < -3
If x < -3, then we can be certain that x < -1
For example, if I tell you that the temperature is less than -3 degrees Celsius, can we be certain that the temperature is less than -1 degrees? Yes.
So, statement 3 must also be true.
Answer: D
If anyone is interested, we have a free video on solving inequalities involving absolute value: https://www.gmatprepnow.com/module/gmat- ... ing?id=985
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Fri Jul 31, 2015 8:47 am, edited 1 time in total.
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MartyMurray
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You went wrong because you went in the wrong direction. We already know that |x| > 3. So that's not what you are proving.prachi18oct wrote:If |x| > 3, then which of the following must be true?
(1) x > 3
(2) x^2 > 9
(3) |x-1| > 2
A. (1) only
B. (2) only
C. (1) and (2) only
D. (2) and (3) only
E. (1),(2) and (3)
My only doubt here is with statement 3.
|x-1| > 2 => x > 3 or x < -1
x > 3 is fine as |x| > 3 for all those x but if x < -1 then all the values dont saisfy |x| > 3 for e.g x = -2 then |x| = 2 < 3 so why should D be the OA.
You are proving that |x-1| > 2.
If |x| > 3, then x > 3 or x < -3. So x - 1 > 2 or x - 1 < -4. Any number > 2 or < -4 has an absolute value > 2.
So (3) holds true at for all x such that |x| > 3, and D is correct.
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|a| = the distance between a and 0.
|a-b| = the distance between a and b.
This means that the distance between x and 0 is greater than 3.
Any value in the two red ranges below satisfies this constraint:
<----(-3).......(3)---->
I: x>3
The red range on the left illustrates that x does not have to be greater than 3.
Eliminate A, C, and E.
III: |x-1| > 2.
This statement implies that the distance between x and 1 must be greater than 2.
Every value in the red ranges above is more than 2 places away from 1.
Thus, statement III must be true.
Eliminate B.
The correct answer is D.
|a-b| = the distance between a and b.
Constraint: |x| > 3If |X|>3, which of the following must be true?
A) X>3
B) X^2>9
c) |X-1|>2
I only
II only
I and II only
II and III only
I, II, and III
This means that the distance between x and 0 is greater than 3.
Any value in the two red ranges below satisfies this constraint:
<----(-3).......(3)---->
I: x>3
The red range on the left illustrates that x does not have to be greater than 3.
Eliminate A, C, and E.
III: |x-1| > 2.
This statement implies that the distance between x and 1 must be greater than 2.
Every value in the red ranges above is more than 2 places away from 1.
Thus, statement III must be true.
Eliminate B.
The correct answer is D.
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(3) |x-1| > 2
Let's solve this further.
We get two cases:
case a) x - 1 > 2, which means x > 3 PERFECT
or
case b) x - 1 < -2, which means x < -1
Must it be true that x < -1?
YES.
Hi ,
If I solve this , then I get.
|x-1|>2
x>3 and x>-1
so what will be my next step?
Please help me in solving this.
Thanks in advance.
SJ
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Hi SJ,
You have to remember the first piece of information that you've been given: |X| > 3. Using THAT information, you know that X < -3 or X > 3.
So when you determine that |X-1| > 2 means that X > 3 or X < -1 you have a basis for comparison. Does that initial piece of information 'match up' with what Roman Numeral 3 defines? YES it does. Every potential value of X in the prompt fits the possibilities that are defined by that Roman Numeral.
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You have to remember the first piece of information that you've been given: |X| > 3. Using THAT information, you know that X < -3 or X > 3.
So when you determine that |X-1| > 2 means that X > 3 or X < -1 you have a basis for comparison. Does that initial piece of information 'match up' with what Roman Numeral 3 defines? YES it does. Every potential value of X in the prompt fits the possibilities that are defined by that Roman Numeral.
GMAT assassins aren't born, they're made,
Rich
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santhosh_katkurwar
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|x-1| > 3 => x > 3 or x < -1
Can someone explain the above inequality? How is it x > 3 or X < -3
Can someone explain the above inequality? How is it x > 3 or X < -3
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According to the prompt:santhosh_katkurwar wrote:|x-1| > 3 => x > 3 or x < -1
Can someone explain the above inequality? How is it x > 3 or X < -3
|x| > 3.
This inequality implies that x < -3 or x > 3.
Values of x such x < -3 or x > 3 include the following:
...-6, -5, -4....4, 5, 6....
Statement II: |x-1| > 2
Case 1: Signs unchanged
x-1 > 2
x > 3.
Case 2: Signs changed in the absolute value
-x+1 > 2
-1 > x
x < -1.
Thus:
x < -1 or x > 3.
Every value in the blue list above is either less than -1 or greater than 3.
Thus, Statement II must be true.
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|a - b| = the distance from a to bsanthosh_katkurwar wrote:|x-1| > 3 => x > 3 or x < -1
Can someone explain the above inequality? How is it x > 3 or X < -3
With that, we can say
|x - 1| > 3
really means
the distance from x to 1 is greater than 3
That means that x is greater than 4 (since 4 is exactly 3 units from 1) or that -2 > x (since -2 is exactly 3 units from 1).
|x| > 3 is similar: the distance from x to 0 is greater than 3. That means x > 3 or -3 > x.
