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4/x < 1/3, what is the range of x
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24 Sep 2018, 00:50
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Re: 4/x < 1/3, what is the range of x
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24 Sep 2018, 01:18
T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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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Re: 4/x < 1/3, what is the range of x
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09 Oct 2018, 22:52
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



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4/x < 1/3, what is the range of x
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09 Oct 2018, 22:53
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



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Re: 4/x < 1/3, what is the range of x
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13 Oct 2018, 02:20
VeritasKarishma wrote: T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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
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4/x < 1/3, what is the range of x
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13 Oct 2018, 06:05
VeritasKarishma wrote: T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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 ?



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4/x < 1/3, what is the range of x
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14 Oct 2018, 01:52
T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 420 times ]
 = \(x<0\) + = \(x>0\) We need to find X less than ZERO, so I marked that range.



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4/x < 1/3, what is the range of x
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Updated on: 14 Oct 2018, 02:17
tamal99 wrote: VeritasKarishma wrote: T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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, 02:12.
Last edited by T1101 on 14 Oct 2018, 02:17, edited 1 time in total.



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4/x < 1/3, what is the range of x
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14 Oct 2018, 02:15
tamal99 wrote: VeritasKarishma wrote: T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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.htmlhopefully i answered your question correctly



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Re: 4/x < 1/3, what is the range of x
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15 Oct 2018, 04:41
tamal99 wrote: VeritasKarishma wrote: T1101 wrote: \(\frac{4}{x}<\frac{1}{3}\), what is the range of x?
A. \(12<x\)
B. \(12>x>0\)
C. \(x>0\)
D. \(12<x<0\)
E. \(x<0\) 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 ... efactors/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.
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