If |x|>3

[rsarashi]

If |x|>3, which if the following must be true?

1) x>3

2) x^2>9

3) |x-1|> 2

a) 1 only

b) 2 only

c) 1 & 2 only

d) 2 & 3 only

e) 1,2, and 3

d

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Hi rsarashi,

When a question includes an absolute value, you have to keep your mind open to the idea that there are probably going to be multiple answers (or multiple GROUPS of answers) that fit the given information. In Roman Numeral questions, it helps to pay attention to how the answer choices are written (since you can often avoid some of the 'work' by noting which answers can be eliminated at certain points in the process).

Here, we're told that |X| > 3. This means that any number that is EITHER greater than 3 OR less than -3 will fit the given inequality. We have to consider both possibilities, since the question asks which of the following MUST be true.

I: X > 3

Since X could be -4, Roman Numeral I is NOT always true.
Eliminate Answers A, C and E.

At this point, we can tell from the remaining two answer choices that Roman Numeral II MUST be true (since it occurs in both of the remaining answers). However, here is the proof that it's true....

II: X^2 > 9

Squaring any number GREATER than 3 will result in a number that is GREATER than 9
Squaring any number LESS than -3 will result in a number that is GREATER than 9
Roman Numeral II IS always true.

III: |X - 1| > 2

IF.... X > 3, then |X - 1| will be greater than |2|.
IF... X < -3, then |X - 1| will be greater than |-4| = |4|.
Roman Numeral III IS always true.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Jay@ManhattanReview]

If |x|>3, which if the following must be true?

1) x>3

2) x^2>9

3) |x-1|> 2

a) 1 only

b) 2 only

c) 1 & 2 only

d) 2 & 3 only

e) 1,2, and 3

d

Hi,

This question is already answered by many experts in an another and is in top five posts. Pl. find it here.

https://www.beatthegmat.com/if-x-3-which ... 93536.html

Hope this helps!

-Jay
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[Matt@VeritasPrep]

One way: Try some numbers

If |x| > 3, then we either have x > 3 or -3 > x

Trying x > 3, we could use x = 4. |4 - 1| > 2, so we're set.

Trying -3 > x, we could use x = -4. |-4 - 1| > 2, so we're set.

Looks good on both sides, so we'd feel relatively comfortable guessing this works.

[Matt@VeritasPrep]

Another way: Thinking about absolute values

|x - 1| > 2 is another way of saying "The distance from x to 1 is greater than two". So whatever x is, it's more than two units away from 1 on the number line.

|x| > 3 is the same as |x - 0| > 3, or "The distance from x to 0 is greater than three". But this is the same as our first statement! If you're more than two units from 1, you're also more than 3 units from 0, since the distance from 0 to 1 is itself one.

So the statements are identical, and we're set.