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For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Sep 2010, 06:02
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For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 12:36
Eden wrote: For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\): Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\); So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). Answer: D.
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Sep 2010, 07:17
15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 12:15
I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ?



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 12:23
Whenever you multiply an inequality by X , you should check for the ve and +ve values of X, as it inverts the sign.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 12:26
Thanks for the lightning quick reply, I was not aware of this.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 12:45
Bunuel wrote: One could also do as follows:
\(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\):
Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\);
So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\).
Answer: D. Bunuel you are awesome



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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03 Oct 2010, 18:00
Nice alternate solution.
Please note :
\(15x^2x2 > x^2  x/15  2/15 > x^2 + x/3  2x/5  2/15 > (x+1/3)(x2/5)\)
This works well, however I would suggest to rely on quadratic equation formula: \(b(+)\sqrt{D}/2a\), as you don't need to think, how to factorize an equation?
Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect rarely could be the case in GMAT) , for example when D in above formula is irrational, then you might be trying to factorize the fraction forever.)



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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12 Aug 2013, 10:48
What I have learnt, if x is in denom, take LCM after bringing all element at one side. If we have more than 2 roots, then we can use following method to know the range. Attachment:
Solution.jpg [ 16.75 KiB  Viewed 4941 times ]



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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14 Aug 2013, 22:13
hi,
can any one explain this with graphical explanation??
Thanks in advance, Rrsnathan.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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26 Nov 2013, 19:46
anshumishra wrote: 15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0. In Case 1 where x > 0, I agree. I calculated x > (1/3) and x > 2/5, and because x > 0, it must be x > 2/5. In Case 2 where x < 0, I don't understand why you say that the range must be x > 1/3 and x < 0, since the values that satisfy the inequality "15x^2  x  2 < 0" show that x < 1/3 and x < 2/5.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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02 Mar 2014, 00:22
TooLong150 wrote: anshumishra wrote: 15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0. In Case 1 where x > 0, I agree. I calculated x > (1/3) and x > 2/5, and because x > 0, it must be x > 2/5. In Case 2 where x < 0, I don't understand why you say that the range must be x > 1/3 and x < 0, since the values that satisfy the inequality "15x^2  x  2 < 0" show that x < 1/3 and x < 2/5.I have the same doubt. Shouldn't the answer be D?



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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02 Mar 2014, 04:03
siriusblack1106 wrote: TooLong150 wrote: anshumishra wrote: 15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0. In Case 1 where x > 0, I agree. I calculated x > (1/3) and x > 2/5, and because x > 0, it must be x > 2/5. In Case 2 where x < 0, I don't understand why you say that the range must be x > 1/3 and x < 0, since the values that satisfy the inequality "15x^2  x  2 < 0" show that x < 1/3 and x < 2/5.I have the same doubt. Shouldn't the answer be D? ____________ The answer IS D.
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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02 Mar 2014, 17:02
TooLong150 wrote: anshumishra wrote: 15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0. In Case 1 where x > 0, I agree. I calculated x > (1/3) and x > 2/5, and because x > 0, it must be x > 2/5. In Case 2 where x < 0, I don't understand why you say that the range must be x > 1/3 and x < 0, since the values that satisfy the inequality "15x^2  x  2 < 0" show that x < 1/3 and x < 2/5. After you determine boundary points, you still need to test the potential +ve/ve ranges per the inequality
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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07 Jun 2014, 16:21
Bunuel wrote: Eden wrote: For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\): Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\); So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). Answer: D. Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or ive.



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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08 Jun 2014, 02:35
Enael wrote: Bunuel wrote: Eden wrote: For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\): Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\); So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). Answer: D. Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or ive. I didn't multiply. \(15x\frac{2}{x}>1\) > subtract 1: \(15x\frac{2}{x}1>0\) > put in common denominator: \(\frac{15x^2x2}{x}>0\) > factorize: \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\). Hope it's clear.
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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10 Jun 2014, 07:51
anshumishra wrote: 15x  2/x > 1
Case1 : x > 0 Multiply both sides by "x" => 15x^2  2 > x => 15x^2  x  2 > 0 (solve the quadratic eqn for 15x^2  x 2 = 0 , it gives you : x = 1/3 or 2/5) Since x > 0 , so x > 2/5
Case 2 : x < 0 Multiply both sides by "x" => 15x^2  2 < x => 15x^2  x  2 < 0 Since, x < 0 so, the range of permissible value is : 1/3 < x < 0. A graphical approach supplements this solution quite well: The roots are the blue dots, the yellow line separates the two cases, and the red lines are where the inequality holds true. We assume two cases: x>0 and x<0. For the first case, x>0, we want to see where 15x^2 x2 >0. Solving for the roots using the quadratic formula, we find roots when x = 2/5 and x = 1/3. Thus, for x>0, 15x^2 x2 >0 when x>2/5. For the second case, x<0, we want to see where 15x^2 x2 <0. Again, using the roots, we know that the 15x^2 x2 < 0 when x > 1/3. We also assumed x<0 so we must include that in our solution. Thus we have x>2/5 and 1/3<x<0. Answer: D
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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15 Jul 2014, 04:31
Bunuel wrote: Eden wrote: For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\): Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\); So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). Answer: D. Hi Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2x2 ? I don't know how to solve this????



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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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15 Jul 2014, 05:26
GGMAT760 wrote: Bunuel wrote: Eden wrote: For what range of values of 'x' will the inequality 15x  2/x > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ? Only if we had x should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(x+\frac{1}{3})(x\frac{2}{5})}{x}>0\): Both denominator and nominator positive > \(x>\frac{2}{5}\); Both denominator and nominator negative > \(\frac{1}{3}<x<0\); So inequality holds true in the ranges: \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). Answer: D. Hi Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2x2 ? I don't know how to solve this???? Hope this helps.
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Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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23 Sep 2014, 01:50
PiyushK wrote: What I have learnt, if x is in denom, take LCM after bringing all element at one side. If we have more than 2 roots, then we can use following method to know the range. Attachment: Solution.jpg Can you explain why before and after the "0" you change the sign? What was the equation that got you to draw this?




Re: For what range of values of 'x' will the inequality 15x  2/x > 1?
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