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Inequality Clarification

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Inequality Clarification

by knight247 » Tue Aug 23, 2011 9:38 am
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

OA is D

I am able to solve this problem using two scenarios...One where x>0 and the other where x<0
When x>0 I get x>-1/3 and x>2/5 also x<-1/3 and x<2/5
When x<0 I get x<-1/3 and x>2/5 also x>-1/3 and x<2/5

Just not able to narrow down the options after this. Hoping to get some help with this...Detailed explanations would be appreciated. Thanks
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by GMATGuruNY » Tue Aug 23, 2011 10:48 am
knight247 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

OA is D
The easiest approach is to plug in values.

Since x is in the denominator, the expression will be undefined when x=0.
Eliminate any answer choice that includes x=0 within its range.
Eliminate B and C.

Let x = - .1
15(-.1) - 2/(-.1) > 1
-1.5 + 20 > 1
18.5 > 1.
This works.
Eliminate any remaining answer choice that does not include x= -.1 within its range.
Eliminate A and E.

The correct answer is D.
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by Whitney Garner » Tue Aug 23, 2011 10:52 am
knight247 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

OA is D

I am able to solve this problem using two scenarios...One where x>0 and the other where x<0
When x>0 I get x>-1/3 and x>2/5 also x<-1/3 and x<2/5
When x<0 I get x<-1/3 and x>2/5 also x>-1/3 and x<2/5

Just not able to narrow down the options after this. Hoping to get some help with this...Detailed explanations would be appreciated. Thanks
You might want to check your solutions for the inequality. Let's start with setting up the scenarios:

x>0 (positive x)
15x-(2/x)>1
15x^2 - 2 > x
15x^2 - x - 2 >0
(3x+1)(5x-2)>0
**cut points are x=-1/3 and x=2/5.

Let's use some test points (around the cut points) to solve for the range.
- If I plug in -1, (-2+1)(-5-2)>0 so this works
- If I plug in 0, (0+1)(0-2)<0 so this does not work
- If I plug in 1, (3+1)(5-2)>0 so this works.

Therefore, x<-1/3 or x>2/5, BUT we solved this assuming x>0, SO we must include that. The only scenario that works is x>2/5

x<0 (negative x)
15x-(2/x)>1
15x^2 - 2 < x
15x^2 - x - 2 < 0
(3x+1)(5x-2)<0
**again the cut points are x=-1/3 and x=2/5.

Let's use some test points (around the cut points) to solve for the range.
- If I plug in -1, (-2+1)(-5-2)>0 so this does not work
- If I plug in 0, (0+1)(0-2)<0 so this works
- If I plug in 1, (3+1)(5-2)>0 so this does not work.

Therefore, -1/3 < x < 2/5, BUT we solved this assuming x<0, SO we must include that. The only scenario that works is -1/3<x<0

Once we combine these, we see that the answer must be [spoiler]D: -1/3<x<0 and x>2/5[/spoiler]

:)
Whit
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by naughtyboy » Fri Jul 13, 2012 11:33 pm
Thanks Whitney for the simplified approach, now I got the concept.
This approach is much better than the solutions I found elsewhere.

But Mitch Hunt's approach has really humbled me. Value substitution
can solve this problem in seconds, while I was taking around 5 mins
to solve this through traditional methods, and still got it wrong!
I think this is what makes the difference in the real exam.
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by rattan123 » Tue Jul 17, 2012 8:42 am
simple..!!! just use wavy curve.!!!!!

15x-2/x-1>0
(15x^2-x-2)/x >0
(3x+1)(5x-2)/x >0


Using Wavy curve we get.!!(https://www.askiitians.com/forums/Differ ... method.htm)

Open it..

We wil get
-1/3≤x<0 and 0<x≤2/5.
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by tabsang » Sat Oct 20, 2012 10:40 am
Hi Whitney,

I know this a very basic question but I could definitely do with some help :(
Hmm... why is it that we've considered two cases for x (-ve and +ve)?
Isn't that clear by the equation that says that it's >1 or >0 (after re-arrangin').

I know that this may be the dumbest question ever but your help will be appreciated :) :)

Cheers,
Taz
Whitney Garner wrote:
knight247 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

OA is D

I am able to solve this problem using two scenarios...One where x>0 and the other where x<0
When x>0 I get x>-1/3 and x>2/5 also x<-1/3 and x<2/5
When x<0 I get x<-1/3 and x>2/5 also x>-1/3 and x<2/5

Just not able to narrow down the options after this. Hoping to get some help with this...Detailed explanations would be appreciated. Thanks
You might want to check your solutions for the inequality. Let's start with setting up the scenarios:

x>0 (positive x)
15x-(2/x)>1
15x^2 - 2 > x
15x^2 - x - 2 >0
(3x+1)(5x-2)>0
**cut points are x=-1/3 and x=2/5.

Let's use some test points (around the cut points) to solve for the range.
- If I plug in -1, (-2+1)(-5-2)>0 so this works
- If I plug in 0, (0+1)(0-2)<0 so this does not work
- If I plug in 1, (3+1)(5-2)>0 so this works.

Therefore, x<-1/3 or x>2/5, BUT we solved this assuming x>0, SO we must include that. The only scenario that works is x>2/5

x<0 (negative x)
15x-(2/x)>1
15x^2 - 2 < x
15x^2 - x - 2 < 0
(3x+1)(5x-2)<0
**again the cut points are x=-1/3 and x=2/5.

Let's use some test points (around the cut points) to solve for the range.
- If I plug in -1, (-2+1)(-5-2)>0 so this does not work
- If I plug in 0, (0+1)(0-2)<0 so this works
- If I plug in 1, (3+1)(5-2)>0 so this does not work.

Therefore, -1/3 < x < 2/5, BUT we solved this assuming x<0, SO we must include that. The only scenario that works is -1/3<x<0

Once we combine these, we see that the answer must be [spoiler]D: -1/3<x<0 and x>2/5[/spoiler]

:)
Whit
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by Whitney Garner » Sun Oct 21, 2012 12:14 pm
tabsang wrote:Hi Whitney,

I know this a very basic question but I could definitely do with some help :(
Hmm... why is it that we've considered two cases for x (-ve and +ve)?
Isn't that clear by the equation that says that it's >1 or >0 (after re-arrangin').

I know that this may be the dumbest question ever but your help will be appreciated :) :)

Cheers,
Taz
knight247 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

OA is D
Hi Taz!

NOT a dumb question at all!! Notice that we are trying to DECIDE if the fraction is greater than 1. In order for that to happen, it doesn't always mean that BOTH the numerator and denominator values are positive. Let's look at an easier example.

What would make the fraction y/x > 1?

Well, let's break this into 2 areas of thought... to be greater than 1 it has to be positive right? Well, what needs to be true to make the fraction y/x positive? Do BOTH y and x have to be positive??

If you said NO then you were exactly right, the values of Y and X could BOTH be negative, and the negatives would cancel!! Okay, so it is totally possible for y/x > 1 if both quantities are negative. SO for the fraciton y/x to be positive, then we just need them to have the SAME sign!!

Now we need to figure out what makes a number greater than 1 (not just greater than 0). So for a fraction to be greater than 1, the numerator NUMBER (ignoring the sign) has to be a larger number than the denominator NUMBER (again, ignoring the sign) - in other words we need to have an IMPROPER fraction! So when Y and X are positive, that is the easier one - Y actually has to be LARGER (more positive) than X. But what about when Y and X are negative? Same thing (sortof) Y has to be MORE NEGATIVE than X, which would actually mean that Y would be a smaller number (further to the left on the number line, e.g. -3 compared to -2).

SO back to our original question...

In order for (15x-2)/x > 1, we need to have the numerator and denominator (15x-2) and (x) to have the same sign (so it isn't necessary for X to be positive), and we need to have one of the other things true:

If they are BOTH positive, then (15x-2) > x ...(15x-2 is MORE positive than x)

If they are BOTH negative, then (15x-2) < x ...(15x-2 is MORE negative than x)

And if we use "cases" to evaluate the original inequality, these are the exact equations we get!!

Hope this helps!
:)
Whit
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by tabsang » Sat Oct 27, 2012 4:43 am
Thanks Whitney for the awesome explanation.
Honestly, I'm still at it, to understand this concept in its entirety but this definitely helped.
Thanks a ton.

Cheers,
Taz

P.S.: I meant to thank you earlier. Apologies for the delay in response and respect and appreciation for your prompt reply and detailed explanation.
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