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If |x + 1 | > 2x - 1, which of the following represents the correct ra
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29 Jun 2022, 09:04
29 Jun 2022, 09:04
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If |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?
A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2
A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra
[#permalink]
25 Aug 2022, 10:48
25 Aug 2022, 10:48
Given that |x + 1 | > 2x - 1 and we need to find the correct range of values of x
Let's solve the problem using two methods
Method 1: Substitution
We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)
A. x < 0
Lets take x = -1 (which falls in this range of x < 0) and substitute in the equation |x + 1 | > 2x - 1
=> |-1 + 1 | > 2*-1 - 1
=> | 0| > -3
=> 0 > -3 which is TRUE, but let's see if we can find a bigger range
B. x < 2
Lets take x = 1.9 (which falls in this range of x < 2) and substitute in the equation |x + 1 | > 2x - 1
=> |1.9 + 1 | > 2*1.9 - 1
=> | 2.9 | > 2.8
=> 0 > -3 which is TRUE.
Now, this range is bigger than option A, so we can ignore option A now.
C. -2 < x < 0
This is a subset of Option B so we don't need to check this one
D. -1 < x < 2
This is a subset of Option B so we don't need to check this one
E. 0 < x < 2
This is a subset of Option B so we don't need to check this one
So, Answer will be B
Method 2: Algebra
Now, we know that |A| > B can be opened as (Watch this video to know about the Basics of Absolute Value)
A > B for A ≥ 0 and
-A > B for A < 0
=> |x + 1 | > 2x - 1 can be written as
Case 1: x + 1 ≥ 0 or x ≥ -1
=> x + 1 > 2x - 1
=> 2x - x < 1 + 1
=> x < 2
And the condition was x ≥ -1 and we got the answer as x < 2
=> the solution will be the range common in both of them
=> -1 ≤ x < 2
-1 to 2.PNG [ 3.5 KiB | Viewed 1110 times ]
Case 2: x + 1 < 0 or x < -1
=> -(x + 1) > 2x - 1
=> 2x + x < 1 - 1
=> 3x < 0
=> x < 0
And the condition was x < -1 and we got the answer as x < 0
=> the solution will be the range common in both of them
=> x < -1
less than -1.PNG [ 3.54 KiB | Viewed 1140 times ]
Combining both the cases we get the answer as -1 ≤ x < 2 and x < -1
=> x < 2
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
Let's solve the problem using two methods
Method 1: Substitution
We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)
A. x < 0
Lets take x = -1 (which falls in this range of x < 0) and substitute in the equation |x + 1 | > 2x - 1
=> |-1 + 1 | > 2*-1 - 1
=> | 0| > -3
=> 0 > -3 which is TRUE, but let's see if we can find a bigger range
B. x < 2
Lets take x = 1.9 (which falls in this range of x < 2) and substitute in the equation |x + 1 | > 2x - 1
=> |1.9 + 1 | > 2*1.9 - 1
=> | 2.9 | > 2.8
=> 0 > -3 which is TRUE.
Now, this range is bigger than option A, so we can ignore option A now.
C. -2 < x < 0
This is a subset of Option B so we don't need to check this one
D. -1 < x < 2
This is a subset of Option B so we don't need to check this one
E. 0 < x < 2
This is a subset of Option B so we don't need to check this one
So, Answer will be B
Method 2: Algebra
Now, we know that |A| > B can be opened as (Watch this video to know about the Basics of Absolute Value)
A > B for A ≥ 0 and
-A > B for A < 0
=> |x + 1 | > 2x - 1 can be written as
Case 1: x + 1 ≥ 0 or x ≥ -1
=> x + 1 > 2x - 1
=> 2x - x < 1 + 1
=> x < 2
And the condition was x ≥ -1 and we got the answer as x < 2
=> the solution will be the range common in both of them
=> -1 ≤ x < 2
Attachment:
-1 to 2.PNG [ 3.5 KiB | Viewed 1110 times ]
Case 2: x + 1 < 0 or x < -1
=> -(x + 1) > 2x - 1
=> 2x + x < 1 - 1
=> 3x < 0
=> x < 0
And the condition was x < -1 and we got the answer as x < 0
=> the solution will be the range common in both of them
=> x < -1
Attachment:
less than -1.PNG [ 3.54 KiB | Viewed 1140 times ]
Combining both the cases we get the answer as -1 ≤ x < 2 and x < -1
=> x < 2
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
General Discussion
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra
[#permalink]
29 Jun 2022, 09:22
29 Jun 2022, 09:22
Carcass wrote:
If |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?
A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2
A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2
I find that the quickest solutions to this kind of question involve testing the answer choices
Scan the answer choices
Notice that some answer choices say that x = 1 is a solution and some say x = 1 is NOT a solution.
So, let's test x = 1
Plug it into the original inequality to get: |1 + 1 | > 2(1) - 1
Simplify to get: 2 > 1
Perfect!
So, x = 1 IS a solution to the inequality.
Since answer choices A and C do NOT include x = 1 as a solution, we can ELIMINATE them.
Now scan the remaining answer choices (B, D and E)
Some answer choices say that x = -1 is a solution and some say x = -1 is NOT a solution.
So, let's test x = -1
Plug it into the original inequality to get: |(-1) + 1 | > 2(-1) - 1
Simplify to get: 0 > -3
Perfect!
So, x = -1 IS a solution to the inequality.
Since answer choices D and E do NOT include x = -1 as a solution, we can ELIMINATE them.
We're left with B
Answer: B
Cheers,
Brent
