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If 4/x <1/3, what is the possible range of values for x? [#permalink]
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07 Apr 2011, 05:01
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
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Re: we can't change the inequality when we have LHS ve nd RHS [#permalink]
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07 Apr 2011, 05:39
GMATD11 wrote: 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/x1<0 (12x)/x < 0 Means either numerator or denominators is ve: Case I: If Denominator is ve. x<0 1 Numerator must be +ve 12x > 0 x > 12 x< 122 In equation 1 and 2, 1 is more restrictive: x<0 Case II: If Denominator is +ve. x>0 3 Numerator must be ve 12x < 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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Re: we can't change the inequality when we have LHS ve nd RHS [#permalink]
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07 Apr 2011, 13:19
GMATD11 wrote: we can't change the inequality when we have ve nd RHS +ve in reciprocal I don't understand what you mean by this? GMATD11 wrote: 10) If 4/x <1/3, what is the possible range of values for x?
We need to consider 2 cases
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.
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Re: we can't change the inequality when we have LHS ve nd RHS [#permalink]
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07 Apr 2011, 13:38
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



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Re: we can't change the inequality when we have LHS ve nd RHS [#permalink]
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29 Aug 2011, 19:13
GMATD11 wrote: 10) If 4/x <1/3, what is the possible range of values for x? 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.



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If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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25 Aug 2012, 01:47
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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?
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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harshvinayak wrote: 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? 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 \(x^212x>0.\) Now, imagine the graph of the quadratic function, \(y=12x^2x\), which is an upward parabola. See the attached drawing. This parabola intercepts the Xaxis at \(x=0\) and \(x=12,\) the two "arms" are above the Xaxis, meaning the values of \(y\) are positive when \(x<12\) or \(x>0\), and the values of \(y\) are negative (graph under the Xaxis) 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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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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25 Aug 2012, 04:06
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, \(\frac{(12x)}{(3x)} < 0\) \(\frac{(x12)}{(3x)} > 0\) roots: 12 and 0 and then I could ride the waves of +  + ...... weeeeeeeeeee.... weeeeeeeeeee... and weeeeeeeeeeee Thanks again for helping out Bunuel ..
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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25 Aug 2012, 04:15
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.
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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26 Aug 2012, 11:24
harshvinayak wrote: 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 \(\frac{xa}{xb}\) is same as the sign of (xa)(xb) and for (xa)(xb) always remember that it will be negative between a & b and positive outside a & b



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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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28 Aug 2012, 21:58
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harshvinayak wrote: 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? Check out these posts. They discuss how to solve inequalities efficiently (using the wave) and how to handle various complications that can arise in a question. http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/http://www.veritasprep.com/blog/2012/07 ... qualities/
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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28 Aug 2012, 22:16
WOW ! Thanks Karishma, thats one awesome collection of inequalities stuff. A kudos alone wouldnt have cut it
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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12 Jul 2013, 14:15
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?



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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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12 Jul 2013, 14:57
alphabeta1234 wrote: 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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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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24 Oct 2013, 18:15
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 crossmultiplying
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...



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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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24 Oct 2013, 20:38
NvrEvrGvUp wrote: 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 crossmultiplying
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 crossmultiplyingThere 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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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]
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25 Oct 2013, 11:44
VeritasPrepKarishma wrote: NvrEvrGvUp wrote: 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 crossmultiplying
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 crossmultiplyingThere 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.



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What is the possible range of value for x? [#permalink]
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4/x < 1/3, what is the possible range of values of x??



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Re: What is the possible range of value for x? [#permalink]
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laveen_g wrote: 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/x1/3 < 0 > (12x)/3x < 0 > (x12)/x > 0 > for this to be true, the range is x<0 or x>12.
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