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December 14 , 2018
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Manager
Joined: 07 Aug 2018
Posts: 108
Location: United States (MA)
4/x < -1/3, what is the range of x
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49% (01:08) correct
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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8661
Location: Pune, India
Re: 4/x < -1/3, what is the range of x
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T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\)
So here, -12 < x < 0
Answer (D)
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Intern
Joined: 09 Apr 2018
Posts: 31
GPA: 4
Re: 4/x < -1/3, what is the range of x
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From the question stem we can conclude that x<0.
If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):
4>\(-(\frac{1}{3})x\)
From here we can conclude that the maximum value x can be is -12, in order to make the inequality true.
Hence -12<x<0. Please hit Kudos if you liked this answer
Intern
Joined: 09 Apr 2018
Posts: 31
GPA: 4
4/x < -1/3, what is the range of x
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From the question stem we can conclude that x<0.
If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):
4>\(-(\frac{1}{3})x\)
From here we can conclude that the maximum value x can be must be smaller than -12, in order to make the inequality true.
Hence -12<x<0. Please hit Kudos if you liked this answer
Manager
Joined: 01 Jan 2018
Posts: 121
Re: 4/x < -1/3, what is the range of x
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VeritasKarishma wrote:
T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\)
So here, -12 < x < 0
Answer (D)
Hi
VeritasKarishma ,
I have doubt in second method (excuse me if very silly and basic).
Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.
Regards,
Tamal
_________________
kudos please if it helped you.
VP
Joined: 09 Mar 2016
Posts: 1213
4/x < -1/3, what is the range of x
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VeritasKarishma wrote:
T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\)
So here, -12 < x < 0
Answer (D)
can anyone draw a wavy curvy line for ths expression ?
-12 < x < 0
\(\frac{x + 12}{3x} < 0\) in nummerator x = - 12 but what do with 3x in the denominator ?
VP
Joined: 09 Mar 2016
Posts: 1213
4/x < -1/3, what is the range of x
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T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
hello
generis is my understanding correct ?
took me a while to figure out how curvy method works
\(\frac{x+12}{3x} <0\)
To solve rational inequality:
- solve for when numerator equals 0
- find the values that make denominator equal 0
\(x+12=0\)
\(x = -12\)
\(3*0=0\)
\(x=0\)
So the critical points of boundaries are -12 and 0. Now to find the range draw a number line.
Attachment:
Wavy Curve Method.png [ 41.09 KiB | Viewed 590 times ]
- = \(x<0\)
+ = \(x>0\)
We need to find X less than ZERO, so I marked that range.
Manager
Joined: 07 Aug 2018
Posts: 108
Location: United States (MA)
4/x < -1/3, what is the range of x
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tamal99 wrote:
VeritasKarishma wrote:
T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If
x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\) So here, -12 < x < 0
Answer (D)
Hi
VeritasKarishma ,
I have doubt in second method (excuse me if very silly and basic).
Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.
Regards,
Tamal
Hey
tamal99 Eventough I am not Karishma I will try to answer your question.
In the second Case marked above we are trying to find values that are below zero (x<0), therefore x cannot be greater than 0! Maybe picking values makes it clearer:
Try to plug in numbers above 0. \(\frac{4}{x}<-\frac{1}{3}\) No positive value will make this inequality true.
Therefore the range is -12<x<0.
Hope it makes sense!
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Originally posted by
T1101 on 14 Oct 2018, 01:12.
Last edited by
T1101 on 14 Oct 2018, 01:17, edited 1 time in total.
VP
Joined: 09 Mar 2016
Posts: 1213
4/x < -1/3, what is the range of x
[#permalink ]
Show Tags
tamal99 wrote:
VeritasKarishma wrote:
T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\)
So here, -12 < x < 0
Answer (D)
Hi
VeritasKarishma ,
I have doubt in second method (excuse me if very silly and basic).
Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.
Regards,
Tamal
Hi
tamal99 i think we need to set denominator equal zero , pls check the link below
https://study.com/academy/lesson/solvin ... ities.html hopefully i answered your question correctly
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8661
Location: Pune, India
Re: 4/x < -1/3, what is the range of x
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tamal99 wrote:
VeritasKarishma wrote:
T1101 wrote:
\(\frac{4}{x}<-\frac{1}{3}\), what is the range of x?
Method 1: \(\frac{4}{x}<-\frac{1}{3}\)
\(\frac{4}{x} + \frac{1}{3} < 0\)
\(\frac{x + 12}{3x} < 0\)
Using the wavy line method discussed here:
https://www.veritasprep.com/blog/2012/0 ... e-factors/ -12 < x < 0
Answer (D)
Method 2: \(\frac{4}{x}<-\frac{1}{3}\)
x can be positive, or negative so we need to take two cases:
Case 1 : If x > 0, multiply both sides by 3x without flipping the inequality sign
\(12 < -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.
Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.
\(12 > -x\)
Multiply both sides by -1 (this will flip the inequality sign)
\(x > -12\)
So here, -12 < x < 0
Answer (D)
Hi
VeritasKarishma ,
I have doubt in second method (excuse me if very silly and basic).
Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.
Regards,
Tamal
In case 2, we have assumed that x is negative (hence have flipped the inequality sign later)
From the inequality, we got x > -12.
So x is negative but greater than -12. This gives us -12 < x < 0.
_________________
[b]KarishmaGMAT Prep Options >
Re: 4/x < -1/3, what is the range of x &nbs
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15 Oct 2018, 03:41
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49% (01:08) correct 
