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If 4/x < 1/3 , what is the possible range of values of x? 13 Dec 2015, 02:45
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4/x < 1/3, what is the possible range of values of x??
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Merging topics.

Please refer to the solutions above.
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If 4/x < 1/3 , what is the possible range of values of x? Updated on: 07 Oct 2016, 10:44
Originally posted by @p00rv@ on 07 Oct 2016, 10:29.
Last edited by Abhishek009 on 07 Oct 2016, 10:44, edited 1 time in total.
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If 4/x < -(1/3), what is the possible range of values for x?
A. -12<x<0
B. x>-12
C. x<12
D. x<-12
E. -12<x<0



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Abhishek009 MERGED TOPIC
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If 4/x < 1/3 , what is the possible range of values of x? Updated on: 08 Oct 2016, 06:05
Originally posted by yezz on 08 Oct 2016, 05:21.
Last edited by yezz on 08 Oct 2016, 06:05, edited 1 time in total.
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[quote="@p00rv@"]If 4/x < -(1/3), what is the possible range of values for x?
A. -12<x<0
B. x>-12
C. x<12
D. x<-12
E. -12<x<0


4/x<-1/3 ......multiply by -3 and reverse sign of inequality

-12/x>1 .......... reverse again 12/x < -1 , -12<x<0
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If 4/x < 1/3 , what is the possible range of values of x? 08 Oct 2016, 05:36
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@p00rv@
If 4/x < -(1/3), what is the possible range of values for x?
A. -12<x<0
B. x>-12
C. x<12
D. x<-12
E. -12<x<0



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Abhishek009 MERGED TOPIC
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@Abhishek009: The question posted by me is different from the question in this thread in the way that it contains a "negative" term and so the answers will also be different for the 2 questions albeit yes method to solve will be the same. I did check this question before posting mine, so does that mean if questions are pretty much based on similar concepts we need not post them separately? Apologies for the inconvenience if that's the case.


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If 4/x < 1/3 , what is the possible range of values of x? 08 Oct 2016, 05:47
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If 4/x < -(1/3), what is the possible range of values for x?
A. -12<x<0
B. x>-12
C. x<12
D. x<-12
E. -12<x<0


4/x<-1/3 ......multiply by -3 and reverse sign of inequality

-12/x>1 .......... reverse again 12/x < -1 , -12>x
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No. that is not the OA. if you will solve it by the method explained in this thread, you will come to know the correct answer which is A.

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If 4/x < 1/3 , what is the possible range of values of x? 08 Oct 2016, 06:05
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yezz
@p00rv@
If 4/x < -(1/3), what is the possible range of values for x?
A. -12<x<0
B. x>-12
C. x<12
D. x<-12
E. -12<x<0


4/x<-1/3 ......multiply by -3 and reverse sign of inequality

-12/x>1 .......... reverse again 12/x < -1 , -12>x

No. that is not the OA. if you will solve it by the method explained in this thread, you will come to know the correct answer which is A.

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U r right another silly mistake ... I edited


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If 4/x < 1/3 , what is the possible range of values of x? 06 Feb 2017, 12:15
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Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem.

Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplying

Now, we have 2 scenarios:

1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible.

2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either.

Can someone please point out the obvious? It's driving me crazy...

First of all, the actual question is \(\frac{4}{x}\) < \(\frac{1}{3}\) (there is no negative with 1/3)

Also, you know what you have to do but you probably do not understand why you have to do it. That is why you are facing problem in this question.

Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplying

There is a problem here. You don't cross multiply and then take cases. You take cases and then cross multiply. Why? Because you CANNOT cross multiply till you know (or assume) the sign of x. The result of the cross multiplication depends on whether x is positive or negative. So you need to take cases and then cross multiply.

Case 1: x > 0
12 < x

Case 2: x < 0
12 > x (note that the sign has flipped here because you are multiplying by a negative number)
x should be less than 12 AND less than 0 so the range in x < 0.

Hence, two cases: x > 12 or x < 0.

Also because x is negative, you cannot just multiply it by -1 to make -x = x. That is certainly not correct. -x is positive and x is negative. They are not equal if x is not 0.
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I still dont understand when x < -12, how can we say x < 0. Since if x = -7, -6 it is not valid for x < -12. ( I understand the other range x > 12)
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If 4/x < 1/3 , what is the possible range of values of x? 07 Feb 2017, 06:36
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Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem.

Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplying

Now, we have 2 scenarios:

1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible.

2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either.

Can someone please point out the obvious? It's driving me crazy...

First of all, the actual question is \(\frac{4}{x}\) < \(\frac{1}{3}\) (there is no negative with 1/3)

Also, you know what you have to do but you probably do not understand why you have to do it. That is why you are facing problem in this question.

Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplying

There is a problem here. You don't cross multiply and then take cases. You take cases and then cross multiply. Why? Because you CANNOT cross multiply till you know (or assume) the sign of x. The result of the cross multiplication depends on whether x is positive or negative. So you need to take cases and then cross multiply.

Case 1: x > 0
12 < x

Case 2: x < 0
12 > x (note that the sign has flipped here because you are multiplying by a negative number)
x should be less than 12 AND less than 0 so the range in x < 0.

Hence, two cases: x > 12 or x < 0.

Also because x is negative, you cannot just multiply it by -1 to make -x = x. That is certainly not correct. -x is positive and x is negative. They are not equal if x is not 0.


I still dont understand when x < -12, how can we say x < 0. Since if x = -7, -6 it is not valid for x < -12. ( I understand the other range x > 12)
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Because if x < -12, then for any possible x would be true to say that it's less than 0, or less than 1,000,000.
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If 4/x < 1/3 , what is the possible range of values of x? 09 Feb 2017, 14:04
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I would like to share a nice approach.
4/X -> iperbole very similae to 1/x
1/3 -> streight line
so the solutions must be something all values at the left of the intersection of the graphs( on the first quadrant) to find the point just put (4/x)=1/3 x=12 ---> x<12
If you know the graph of 1/x this explanation will be very clear.
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If 4/x < 1/3 , what is the possible range of values of x? 09 Feb 2017, 14:04
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I would like to share a nice approach.
4/X -> iperbole very similae to 1/x
1/3 -> streight line
so the solutions must be something all values at the left of the intersection of the graphs( on the first quadrant) to find the point just put (4/x)=1/3 x=12 ---> x<12
If you know the graph of 1/x this explanation will be very clear.
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If 4/x < 1/3 , what is the possible range of values of x? 16 Feb 2017, 11:31
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There's a mistake in your second result:

x < 12 & x < 0 therefore x < 0 (instead of x < 12 )
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If 4/x < 1/3 , what is the possible range of values of x? 19 Jun 2017, 15:10
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We need to consider two cases - when x is positive and when x is negative.

when x is positive - x > 12
when x is negative - x < 0

0 < x > 12
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If 4/x < 1/3 , what is the possible range of values of x? 19 Jun 2017, 15:12
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we can't change the inequality when we have -ve nd RHS +ve in reciprocal

10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

Case 1 x is +ve

x>12

Case 2 when we consider x as -ve we will have Left hand side -ve but right hand side +ve so in that case we cnt flip the inequality.
But OA is showing both x>12 nd x<12

Pls comment which condition is wrong.

thanks

4/x <1/3
12/x-1<0
(12-x)/x < 0

Means either numerator or denominators is -ve:

Case I:
If Denominator is -ve.
x<0 ------1

Numerator must be +ve
12-x > 0
-x > -12
x< 12--------------2

In equation 1 and 2, 1 is more restrictive:
x<0

Case II:
If Denominator is +ve.
x>0 ------3

Numerator must be -ve
12-x < 0
-x < -12
x > 12


In equation 3 and 4, 4 is more restrictive:
x>12

Thus;
complete Range of x:

x<0 or x>12
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Long explanation - but very well explained. Good to understand the concept part. Great Job.
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If 4/x < 1/3 , what is the possible range of values of x? 18 Apr 2020, 05:57
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I have a very basic question to do with arithmetic rather than the inequality (which I understand).

Why does cross multiplying (4/x) < (1/3) = 12 < x, but simplifying it via x < (4 / 1 / 3) = x < 12 gives a different result?

I thought usually you can rearrange by taking the numerator to the other side and the inequality wouldn't flip since you are taking over a positive 4. As a basic example if you have 4/6 = 2/3, you can take the 4 to the RHS to get 6 = 4/2/3

I'm sure there's a simple explanation but I'm having a bit of a mind block. Thanks to anyone that can assist.
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If 4/x < 1/3 , what is the possible range of values of x? 26 Oct 2021, 07:46
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