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

Check out this video. It discusses how to solve inequalities efficiently (using the wave).

https://youtu.be/PWsUOe77__E

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

I don't understand what you mean by this?



You do need to consider 2 cases, but you only need to spend any time on one of them. We know:

4/x < 1/3

This will clearly be true if x is negative, since then the left side is negative, and the right side is positive, and negative numbers are certainly smaller than positive ones. So whenever x < 0, the inequality is true.

Now for the second case: if x > 0, we can multiply both sides by x without needing to worry about reversing the inequality:

4 < x/3
12 < x

So either x < 0, or 12 < x.

Brilliant !

And when I know the sign of x > 0 then taking the reciprocal of left hand side Vs the right hand side will reverse the direction of the inequality

4/x < 1/3 and x > 0
or x/4 > 3 (reverse the direction)
or x > 12

4/x < 1/3 => 12/x < 1
For the fraction 12/x to be <1 ,

Scenario1 : x has to be negative.
Scenario2 : The denominator should be greater than 12

=> x < 0 or x > 12 .

Please correct me if i am wrong.

If \(\frac{4}{x} <\frac{1}{3}\) , what is the possible range of values of x?

The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
\(\frac{4}{x} - \frac{1}{3} < 0\)
\(\frac{(12 - x)}{(3x)} < 0\)

Two cases:
a) \(12 - x < 0\) &\(3x > 0\) :: x > 12 & x > 0 therefore x > 12
b) \(12 - x > 0\) & \(3x < 0\) :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12

Can someone kindly explain where I went wrong here?

When we consider the case when \(12 - x > 0\) and \(3x < 0\), we have: \(x<12\)and \(x<0\), therefore \(x<0\).

Solving inequalities:
inequalities-trick-91482.html (check this one first)
x2-4x-94661.html#p731476
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Another way to solve this type of inequality:

From \(\frac{4}{x}<\frac{1}{3}\) it follows that \(x\neq0\), therefore we can multiply both sides by \(3x^2\), which is positive.
We obtain \(12x<x^2\) or \(x^2-12x>0.\)
Now, imagine the graph of the quadratic function, \(y=12x^2-x\), which is an upward parabola. See the attached drawing.
This parabola intercepts the X-axis at \(x=0\) and \(x=12,\) the two "arms" are above the X-axis, meaning the values of \(y\) are positive when \(x<12\) or \(x>0\), and the values of \(y\) are negative (graph under the X-axis) when \(0<x<12.\)

Hence the solution \(x<0\) or \(x>12.\)

Note: Just remember the shape of the parabola, then you can easily deduce the sign of the quadratic function \(y=x^2+bx+c.\) If the quadratic equation \(x^2+bx+c=0\) has two roots \(x_1<x_2\) then \(y<0\) (negative) between the two roots and \(y>0\) (positive) outside the roots.
Or succinctly, \(y>0\) if \(x_1<x<x_2\) and \(y>0\) if \(x<x_1\) or \(x>x_2.\)

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Ah! thanks for the clarification Bunuel. I guess staying up all night took its toll. I had to reread and reread and reread and...

finally what I did was (per the first link you suggested) ::

dumping the two cases,
\(\frac{(12-x)}{(3x)} < 0\)
\(\frac{(x-12)}{(3x)} > 0\)
roots: 12 and 0
and then I could ride the waves of + - + ...... weeeeeeeeeee.... weeeeeeeeeee... and weeeeeeeeeeee

Thanks again for helping out Bunuel ..

Thanks EvaJager you guys are awesome \m/

PS: I have just started out with my prep and have a lot to learn (and ask ) in little less than 3 weeks.

One way to understand this is that Sign of \(\frac{x-a}{x-b}\) is same as the sign of (x-a)(x-b)
and for (x-a)(x-b) always remember that it will be negative between a & b and positive outside a & b

WOW !
Thanks Karishma, thats one awesome collection of inequalities stuff. A kudos alone wouldnt have cut it

Sorry to bring back this old post. I know how to solve this problems using the methods mentioned above. But I tried solving it another way and that method gave me a weird solution. Let say you wanted to solve the problem this way:

\(\frac{4}{x} <\frac{1}{3}\)

Since x cannot be 0. Lets look at the positive/negative scenarios

If x>0, then \(12 <x\)
if x<0, then \(12 >x\) but x cannot be both negative and greater than 12. So this is a contradiction.

Hence the range is \(12 <x\).

But this is incomplete. The range of x for this inquality is \(x<0\) OR \(12 <x\). So what am I doing wrong?

This is a correct approach but you made a mistake in the second case.

If x<0, then 12>x (x is less than 12) --> intersection is x<0.

So, the inequality holds true for x>12 (from the first case) and x<0 (from the second case).

Hope it helps.
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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.

Hi Karishma,

That was my mistake. There is a problem with a negative -1/3 and one without - I didn't realize this one was referencing the one with the positive 1/3.

I'll search for the -1/3 problem and explanation.

Thanks.

4/x < 1/3, what is the possible range of values of x??

Follow posting guidelines (link in my signatures). Post complete question and all options. Do post the official answer along with 3 tags, 1 each for source, difficulty and topic discussed.

As for this question,

You are asked for the range of values for 4/x < 1/3 ---> 4/x-1/3 < 0 ---> (12-x)/3x < 0 ---> (x-12)/x > 0 ---> for this to be true, the range is x<0 or x>12.
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