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Is |x + 2| < 3? (1) x < 1 (2) x > -5

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Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   Updated on: 01 Nov 2018, 22:27
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Is |x + 2| < 3?

(1) x < 1
(2) x > -5
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C

Originally posted by GMATBusters on 01 Nov 2018, 10:03.
Last edited by Bunuel on 01 Nov 2018, 22:27, edited 1 time in total.
Renamed the topic, edited the question and the tags.
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BrentGMATPrepNow
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   19 Feb 2022, 07:36
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Is |x + 2| < 3?

(1) x < 1
(2) x > -5

Target question: Is |x + 2| < 3?
This is a good candidate for rephrasing the target question.

-------------ASIDE-------------
When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive
------------------------------------
So, we can take the target question Is |x + 2| < 3?....
and rewrite it as Is -3 < x + 2 < 3?
To make things more clear, let's subtract 2 from all three parts of the inequality to get: and rewrite it as Is -5 < x < 1?
REPHRASED target question: Is -5 < x < 1?

Statement 1: x < 1
There are infinitely many values of x that satisfy statement 1. Here are two:
Case a: x = 0, in which case the answer to the REPHRASED target question is YES, it's true that -5 < x < 1
Case b: x = -10, in which case the answer to the REPHRASED target question is NO, it's not true that -5 < x < 1
Since we can’t answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x > -5
There are infinitely many values of x that satisfy statement 2. Here are two:
Case a: x = 0, in which case the answer to the REPHRASED target question is YES, it's true that -5 < x < 1
Case b: x = 10, in which case the answer to the REPHRASED target question is NO, it's not true that -5 < x < 1
Since we can’t answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x < 1
Statement 2 tells us that x > -5
We combine the two inequalities we get -5 < x < 1, which means the answer to the REPHRASED target question is YES, it's true that -5 < x < 1
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

VIDEO ON REPHRASING THE TARGET QUESTION:
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   01 Nov 2018, 10:09
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Is | x + 2 | < 3?
When x +ve
X+2<3
X<1

When x -ve
-x-2 <3
x>-5
rephrase question:
is -5<x<1?
Each statement alone not sufficient
Combination gives YES.
Answer C
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   23 Sep 2021, 06:27
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We are asked whether |x+2|<3.

We know that,
|x+2|=x+2 if x+2>0
=-(x+2) , if x+2<0.

If x+2>0,
x+2<3,
0r, x<-1

If x+2<0,
-(x+2)<3
or,-x-2<3
or,-x<5
or, x>-5.

So, for |x+2|<3,
-5<x<-1.

This can be determined only by combining both statements, hence the answer is C.
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Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   06 Feb 2023, 04:31
BrentGMATPrepNow avigutman Bunuel chetan2u

I'm trying to get some concept clarity on less than and greater signs in absolute value inequalities.

Are both statements necessary because the sign in the absolute value inequality is less than and therefore creates an "and" statement / intersection in the values of x?

If the absolute value inequality had a greater than sign (see below), would each statement on its own be sufficient?

Is |x+2| > 3?

1) x>1
2) x<-5

Appreciate your help so much - thank you!
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   06 Feb 2023, 07:26
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achloes
BrentGMATPrepNow avigutman Bunuel chetan2u

I'm trying to get some concept clarity on less than and greater signs in absolute value inequalities.

Are both statements necessary because the sign in the absolute value inequality is less than and therefore creates an "and" statement / intersection in the values of x?

If the absolute value inequality had a greater than sign (see below), would each statement on its own be sufficient?

Is |x+2| > 3?

1) x>1
2) x<-5

Appreciate your help so much - thank you!
From your question, achloes, it seems to me like you're looking to memorize some kind of rule. Doing so won't improve your concept clarity nor help your GMAT score, unfortunately.
Instead, I suggest that you practice translating absolute values onto a number line, as seen in many videos on my youtube channel and as explained in my book.
For example, |x+2| describes the distance of (x+2) from zero. So, the inequality |x+2| > 3 tells us that (x+2) is located more than 3 units away from zero. Draw on a number line where (x+2) might be. Then, if you're interested in where that puts x, you must reason that x is always going to be located precisely 2 units to the left of (x+2), and therefore you must shift your ranges on the number line 2 units to the left, placing x to the left of -5 or to the right of +1.
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   06 Feb 2023, 10:28
avigutman
achloes
BrentGMATPrepNow avigutman Bunuel chetan2u

I'm trying to get some concept clarity on less than and greater signs in absolute value inequalities.

Are both statements necessary because the sign in the absolute value inequality is less than and therefore creates an "and" statement / intersection in the values of x?

If the absolute value inequality had a greater than sign (see below), would each statement on its own be sufficient?

Is |x+2| > 3?

1) x>1
2) x<-5

Appreciate your help so much - thank you!
From your question, achloes, it seems to me like you're looking to memorize some kind of rule. Doing so won't improve your concept clarity nor help your GMAT score, unfortunately.
Instead, I suggest that you practice translating absolute values onto a number line, as seen in many videos on my youtube channel and as explained in my book.
For example, |x+2| describes the distance of (x+2) from zero. So, the inequality |x+2| > 3 tells us that (x+2) is located more than 3 units away from zero. Draw on a number line where (x+2) might be. Then, if you're interested in where that puts x, you must reason that x is always going to be located precisely 2 units to the left of (x+2), and therefore you must shift your ranges on the number line 2 units to the left, placing x to the left of -5 or to the right of +1.


Hi avigutman - thanks for your comment! Perhaps I wasn't clear enough earlier but it was more of a follow up question to @BrentGMATPrepNow's explanation:

Quote:
When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive

My question was around how these rules would work for a DS question such as the one above.

According to rule #1, it's clear to me why we would need to "lock in" the range of k. So, for the question above, it makes sense why statements 1 and 2 are required.

Rule#2 however isn't as clear. I do understand that the solution range would not intersect on the number line. But I'm not 100% sure whether this means that either one of these inequalities would be sufficient or whether we would need the full range and therefore both statements. Hence my example with the greater than sign in play :)
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Re: Is |x + 2| < 3? (1) x < 1 (2) x > -5 [#permalink] Is |x + 2| < 3? (1) x < 1 (2) x > -5   06 Feb 2023, 12:23
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achloes
A follow up question to @BrentGMATPrepNow's explanation:

Quote:
When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive

My question was around how these rules would work for a DS question such as the one above.

According to rule #1, it's clear to me why we would need to "lock in" the range of k. So, for the question above, it makes sense why statements 1 and 2 are required.

Rule#2 however isn't as clear. I do understand that the solution range would not intersect on the number line. But I'm not 100% sure whether this means that either one of these inequalities would be sufficient or whether we would need the full range and therefore both statements. Hence my example with the greater than sign in play :)
No rules, achloes. Only reasoning. Once we draw the ranges that your question is asking about, we must evaluate each statement on its own.
Your question
Quote:
Is |x+2| > 3?
gets a YES if x is to the right of 1 or to the left of -5, and it gets a NO if x is somewhere in between -5 and 1.
Quote:
1) x>1
can we definitively say whether it's a YES or a NO? Well, I believe this is a definite YES (can't get a NO because x is definitely not between -5 and 1).
Quote:
2) x<-5
can we definitively say whether it's a YES or a NO? Well, I believe this is a definite YES (can't get a NO because x is definitely not between -5 and 1).
Note: your statements contradict one another, so it's not a good DS problem. But, more importantly, as I have shown, don't look for rules to memorize - just do the reasoning every time. That's the only way to see substantial score improvements on this test (plus doing so will help you in your career beyond the GMAT, whereas memorizing rules has a much more limited utility).
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