Inequality Question

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Inequality Question

by GaneshMalkar » Mon Mar 17, 2014 3:19 am
Is x/3 + 3/x > 2

i) x < 3
ii) x > 1

OA - C

Please help me in easiest way...
If you cant explain it simply you dont understand it well enough!!!
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by HerrGrau » Mon Mar 17, 2014 4:23 am
Hi Ganesh,

Good question. This is a standard but tough DS question. With DS you always want to think about special numbers: 1, 0, Negative, Positive, Integers, Non-Integers, Maximum, Minimum

If X is zero then this will be undefined. If X is 1 you can answer Yes. If X is negative you can answer No. It helps to think about this information in advance of the statements.

1. X could be 1 or negative. Insufficient.

2. X could be Infinity (YES) or 3 (NO) Insufficient. Why did I think of 3? It's easy and right at the limit of 2.

Putting them together: If you test a positive number right below 3 you will see that the expression gets bigger. The same thing happens above one. The same will happen with numbers above 3. X=3 is actually the minimum for that expression (in terms of positive numbers). You can also think of this question in terms of limits. As X approaches zero then 3/x approaches infinity. As X approaches infinity then X/3 approaches infinity.

Let me know if you have any questions!

Happy Studies,

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by GMATGuruNY » Mon Mar 17, 2014 4:40 am
GaneshMalkar wrote:Is x/3 + 3/x > 2

i) x < 3
ii) x > 1

OA - C
Determine the CRITICAL POINTS.
Critical points occur when:
The LEFT HAND SIDE = the RIGHT HAND SIDE.
The value of the expression is UNDEFINED.

To determine where x/3 + 3/x = 2, test easy values:
If x=1, then x/3 + 3/x = 1/3 + 3/1 = 10/3.
If x=2, then x/3 + 3/x = 2/3 + 3/2 = 13/6.
If x=3, then x/3 + 3/x = 3/3 + 3/3 = 2.
Thus, x=3 is a critical point.

Since division by 0 is not allowed, x/3 + 3/x is undefined when x=0.
Thus, x=0 is a critical point.

To determine where x/3 + 3/x > 2, test one value TO THE LEFT and one value TO THE RIGHT of each critical point.

x<0:
If x=-6, then x/3 + 3/x = -6/3 + 3/-6 = -2.5, which is not greater than 2.
Thus, x<0 is not a valid range.

0<x<3.
If x=1, then x/3 + 3/x = 1/3 + 3/1 = 10/3, which is greater than 2.
Thus, 0<x<3 is a valid range.

x>3:
If x=6, then x/3 + 3/x = 6/3 + 3/6 = 2.5, which is greater than 2.
Thus, x>3 is a valid range.

Result:
x/3 + 3/x > 2 if 0<x<3 or x>3.

Question stem, rephrased:
Is 0<x<3 or x>3?

Statement 1: x<3
If x=1, then 0<x<3.
If x=-6, then x is not part of a valid range.
INSUFFiCIENT.

Statement 2: x>1
If x=2, then 0<x<3.
If x=3, then x is not part of a valid range.
INSUFFICIENT.

Statements combined: 1<x<3
Thus, 0<x<3.
SUFFICIENT.

The correct answer is C.
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by GaneshMalkar » Mon Mar 17, 2014 5:37 am
Thankyou very much for the explanation...
So basically we need to calculate the CRITICAL POINT in inequalities question of this type?
If you cant explain it simply you dont understand it well enough!!!
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by HerrGrau » Mon Mar 17, 2014 5:53 am
Hi Ganesh,

Yep - that's exactly right. Understanding critical points/Max/Min/Limits is super important for DS.

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by GMATGuruNY » Mon Mar 17, 2014 5:05 pm
GaneshMalkar wrote:Thankyou very much for the explanation...
So basically we need to calculate the CRITICAL POINT in inequalities question of this type?
The CRITICAL POINT APPROACH is a great way to handle complex inequality problems.
If you enter "critical points" and "gmatguruny" into the BTG search bar, you'll find a number of problems that I've solved with this approach.
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by Matt@VeritasPrep » Mon Mar 17, 2014 11:19 pm
HerrGrau wrote:Hi Ganesh,

Yep - that's exactly right. Understanding critical points/Max/Min/Limits is super important for DS.
I couldn't disagree more with this statement. Critical points, maxima, minima, and limits are "super important" for Calculus I. Calculus is not tested on the GMAT, and the test writers go out of their way NOT to privilege you for knowing it.

I feel like I make this point every time someone posts a question like "What is the minimum of x� + x² + 8?", but it bears repeating: there is enough for people to study already without their wasting time on superfluous math. DO NOT STUDY CALCULUS WHEN PREPARING FOR THE GMAT. You shouldn't even be studying Algebra II!

Furthermore, this question doesn't seem properly formatted given the GMAT's self-imposed constraints. When the test writers give a fraction whose variable is a denominator, they take pains to stipulate that the denominator is nonzero. So the prompt should read "If x ≠ 0, is 3/x + x/3 > 2?", which, as GMATGuruNY demonstrated, is really just a complex way of asking "Is x both positive and not equal to 3?"

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by HerrGrau » Tue Mar 18, 2014 1:51 am
Hi Matt,

Hope you are well. Yes - the question is asking is X not equal to 3 and is X positive. The question is how do you get there? I'll go on record again saying that it's important for GMAT DS to think about extremes (limits), thresholds, and special numbers to focus your work. I'm not quite sure how you interpreted my solution or my previous post but I would hope that you agree with the above idea.

Have a good day,

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by subhakam » Tue Mar 18, 2014 7:17 am
GMATGuruNY wrote:
GaneshMalkar wrote:Thankyou very much for the explanation...
So basically we need to calculate the CRITICAL POINT in inequalities question of this type?
The CRITICAL POINT APPROACH is a great way to handle complex inequality problems.
If you enter "critical points" and "gmatguruny" into the BTG search bar, you'll find a number of problems that I've solved with this approach.
Is there some way/any way to solve this algebraically?

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by GMATGuruNY » Tue Mar 18, 2014 7:43 am
subhakam wrote:
GMATGuruNY wrote:
GaneshMalkar wrote:Thankyou very much for the explanation...
So basically we need to calculate the CRITICAL POINT in inequalities question of this type?
The CRITICAL POINT APPROACH is a great way to handle complex inequality problems.
If you enter "critical points" and "gmatguruny" into the BTG search bar, you'll find a number of problems that I've solved with this approach.
Is there some way/any way to solve this algebraically?
Is x/3 + 3/x > 2?
Since the answer will be YES only if x>0, we can safely clear the fractions by multiplying each side of the inequality by 3x:

x/3 + 3/x > 2

3x (x/3 + 3/x) > 3x * 2

x² + 9 > 6x

x² - 6x + 9 > 0

(x-3)² > 0.

The only positive value that does NOT satisfy the resulting inequality is x=3.
Question stem, rephrased: Is x a positive value other than 3?
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by subhakam » Tue Mar 18, 2014 7:51 am
GMATGuruNY wrote:
subhakam wrote:
GMATGuruNY wrote:
GaneshMalkar wrote:Thankyou very much for the explanation...
So basically we need to calculate the CRITICAL POINT in inequalities question of this type?
The CRITICAL POINT APPROACH is a great way to handle complex inequality problems.
If you enter "critical points" and "gmatguruny" into the BTG search bar, you'll find a number of problems that I've solved with this approach.
Is there some way/any way to solve this algebraically?
Is x/3 + 3/x > 2?
Since the answer will be YES only if x>0, we can safely clear the fractions by multiplying each side of the inequality by 3x:

x/3 + 3/x > 2

3x (x/3 + 3/x) > 3x * 2

x² + 9 > 6x

x² - 6x + 9 > 0

(x-3)² > 0.

The only positive value that does NOT satisfy the resulting inequality is x=3.
Question stem, rephrased: Is x a positive value other than 3?
Thank you - the part i was getting confused was whether and why we assume X >0 ? In case where X<0 - the answer will always be NO (no matter what the value of X is it will always be negative)

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by GMATGuruNY » Tue Mar 18, 2014 8:12 am
subhakam wrote:Thank you - the part i was getting confused was whether and why we assume X >0 ? In case where X<0 - the answer will always be NO (no matter what the value of X is it will always be negative)
We are not assuming that x>0.
We are trying to determine under what conditions x/3 + 3/x > 2.
If x<0, then x/3 + 3/x < 2.
If x>0, then x/3 + 3/x > 2 for every value other than 3 (as shown algebraically in my post above).
Thus, the answer to the question stem is YES if x is a positive value other than 3.
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by GMATGuruNY » Tue Mar 18, 2014 8:25 am
Matt@VeritasPrep wrote: Furthermore, this question doesn't seem properly formatted given the GMAT's self-imposed constraints. When the test writers give a fraction whose variable is a denominator, they take pains to stipulate that the denominator is nonzero. So the prompt should read "If x ≠ 0, is 3/x + x/3 > 2?", which, as GMATGuruNY demonstrated, is really just a complex way of asking "Is x both positive and not equal to 3?"
This problem is the real deal: an official problem from GMAC.
I believe that it used to be in the pool of questions for GMATPrep; it might now be part of GMATFocus.
That said, it's an atypical problem: as you have mentioned, GMAT problems tend to preclude division by 0.
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by Matt@VeritasPrep » Tue Mar 18, 2014 11:33 am
Thanks for the info, Guru! This might be an old-fashioned problem - I'd be surprised if you saw it today. The notion of a value not being in the domain of an expression seems way outside the scope of the modern exam.

Someone above asked for an algebraic approach. Here's one that should work on most GMAT inequalities.

If x ≠ 0, then 3/x + x/3 = (x² + 9)/3x. In light of that, we can consider three cases:

If x > 0, our question is, "Is (x² + 9)/3x > 2?" As already demonstrated, 3x is positive (by assumption), so we can multiply both sides of the inequality by 3x, giving us "Is x² + 9 > 6x?" or "Is x² - 6x + 9 > 0?" or "Is (x-3)² > 0?" The answer to this is YES if x ≠ 3 and NO if x = 3.

If x = 0, our function is undefined, so we must have x ≠ 0.

If x < 0, our question, "Is (x² + 9)/3x > 2?", simplifies to "Is (x² + 9) < 6x?" (Now 3x is negative, by assumption, so the inequality flips.) This simplifies to "Is (x-3)² < 0?", which is impossible. So x cannot be any negative value.

Considering all this, we rephrase the question as "Is x > 0 and ≠ 3?", then read each statement.

This is how I'd take most tricky inequalities: do your best to simplify the algebra and consider the appropriate cases (which often are just negative, positive, zero). Most of your work should be done BEFORE you consider the two statements, not after: you want to know what you're looking for.

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by Matt@VeritasPrep » Tue Mar 18, 2014 11:46 am
HerrGrau wrote:I'll go on record again saying that it's important for GMAT DS to think about extremes (limits), thresholds, and special numbers to focus your work. I'm not quite sure how you interpreted my solution or my previous post but I would hope that you agree with the above idea.
Special numbers sure, limits and the rest of the apparatus of calculus absolutely not. As shown in my last post, any inequality on the GMAT should succumb to a straightforward (if initially opaque) algebraic approach. Most students studying for this exam are already drowning in material - the last thing they need is to study a demanding subject that adds little or nothing on test day. Studying calculus for GMAT inequalities would be like studying Latin and French for SAT vocabulary.

If the test doesn't expect you to know plenty of fundamental algebra (polynomials, circles not centered on the origin, logarithms, quadratic vertices, ...) and geometry (medians, inscribed and circumscribed circles, angle and perpendicular bisectors, Ceva's Theorem, Heron's Formula, secants, ...), calculus is well beyond the pale. The whole point of the exam is to test your mathematical instincts (DS) and your ingenuity with and fluency in basic concepts (PS). On some level, everything you need to know for each topic (algebra, arithmetic, geometry, statistics) would be given in the first week or two of an introductory college class on the subject.