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# Is x/3 + 3/x > 2

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Is x/3 + 3/x > 2 [#permalink]

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16 Jul 2010, 00:06
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Is $$\frac{x}{3} + \frac{3}{x} > 2$$?

(1) $$x < 3$$
(2) $$x > 1$$
[Reveal] Spoiler: OA

Last edited by DenisSh on 16 Jul 2010, 03:15, edited 1 time in total.
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Re: Tricky inequality problem [#permalink]

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17 Jul 2010, 04:36
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DenisSh wrote:
Bunuel wrote:
gurpreetsingh's solution is correct.

Is $$\frac{x}{3}+\frac{3}{x}>2$$? --> is $$\frac{(x-3)^2}{x}>0$$?

How did you get $$\frac{(x-3)^2}{x}>0$$?

Step 1:
$$\frac{x}{3}+\frac{3}{x}>2$$?
Step 2 (multiply by 3x):
$$x^2+9>6x$$
Step 3 (move 6x to the left side):
$$x^2-6x+9>0$$
Step 4 (convert to the compact form):
$$(x-3)^2>0$$

This is the most common error when solve inequalities. I keep writing this over and over again:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.

So you CANNOT multiply $$\frac{x}{3}+\frac{3}{x}>2$$ by $$3x$$ since you don't know the sign of $$x$$.

What you CAN DO is: $$\frac{x}{3}+\frac{3}{x}>2$$ --> $$\frac{x}{3}+\frac{3}{x}-2>$$ --> common denominator is $$3x$$ --> $$\frac{x^2+9-6x}{3x}>0$$ --> multiply be 3 --> $$\frac{x^2+9-6x}{x}>0$$ --> $$\frac{(x-3)^2}{x}>0$$.

Hope it helps.
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Re: Tricky inequality problem [#permalink]

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16 Jul 2010, 02:28
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on rearranging the terms we get,
$$((x^2 + 9)/3x) - 2 > 0$$
$$(x^2 + 9 -6x)/3x > 0$$
$$((x - 3)^2)/3x > 0$$

$$(x-3)^2$$ is positive.
amongst options (1) and (2), we can draw a clear cut conclusion only using (2).

Therefore ans is B.
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Re: Tricky inequality problem [#permalink]

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16 Jul 2010, 03:13
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naish wrote:
$$(x-3)^2$$ is positive.
amongst options (1) and (2), we can draw a clear cut conclusion only using (2).

Therefore ans is B.

No. As I think, you should get 2 inequalities:
1) $$(x-3)^2>0$$ and 2) $$(x-3)^2<0$$ (it depends on what the sign does $$x$$ have).

After that, (2) indicates that $$x$$ has a positive sign. Thus, we should consider only 1) inequality.
There is one option when $$(x-3)^2=0$$ - if $$x=3$$.

Since (1) indicates that $$x \neq 3$$, the OA is (C).

Correct me if I'm wrong.
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Re: Tricky inequality problem [#permalink]

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16 Jul 2010, 05:06
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naish wrote:
on rearranging the terms we get,
$$((x^2 + 9)/3x) - 2 > 0$$
$$(x^2 + 9 -6x)/3x > 0$$
$$((x - 3)^2)/3x > 0$$

$$(x-3)^2$$ is positive.
amongst options (1) and (2), we can draw a clear cut conclusion only using (2).

Therefore ans is B.

Ans is C,
For all the +ve values of x the answer to the question is yes, except for x=3 for which its value is 0 thus the answer is no.

1st statement eliminates the x=3 option thus it is required.
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Re: Tricky inequality problem [#permalink]

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16 Jul 2010, 07:42
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DenisSh wrote:
Is $$\frac{x}{3} + \frac{3}{x} > 2$$?

(1) $$x < 3$$

(2) $$x > 1$$

gurpreetsingh's solution is correct.

Is $$\frac{x}{3}+\frac{3}{x}>2$$? --> is $$\frac{(x-3)^2}{x}>0$$? Now, nominator is non-negative, thus the fraction to be positive nominator must not be zero (thus it'll be positive) and denominator mut be positive --> $$x\neq{3}$$ and $$x>0$$.

Statement (1) satisfies the first requirement and statement (2) satisfies the second requirement, so taken together they are sufficient.

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Re: Tricky inequality problem [#permalink]

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19 Jul 2010, 09:18
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ulm wrote:
(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0.
The correct answer is (0;3) and (3;+inf)
It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.

Generally, unknown (or expression with unknown) in even power is NOT always positive, it's non-negative. Not knowing this is the cause of many mistakes on GMAT.

So, $$(x-3)^2\geq{0}$$, because if $$x=3$$, then $$(x-3)^2=0$$ and $$\frac{(x-3)^2}{x}$$ also equals to zero (and not more than zero). We need statement (1) to exclude the possibility of $$x$$ being 3 by saying that $$x<3$$. That's why the answer to this question is C, not B.

Hope it's clear.
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10 Jun 2011, 12:38
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Baten80 wrote:
is (x/3+3/x) > 2?
(1) x < 3
(2) x > 1

This ds has been discussed thoroughly at http://gmatclub.com/forum/tricky-inequality-problem-97331.html. It inequality simplified there as [(x - 3)^2]/3 > 0.
if i do not simplify then i can i solved it as:
(1) if x = 2 then the (x/3+3/x) > 2 but if x = negative the (x/3+3/x) > 2 is not true. so Insufficient.
(2) if x = 2 then (x/3+3/x) > 2 but if x = 1 then (x/3+3/x) > 2 is true. so insufficient.

for C x =2 and (x/3+3/x) > 2.

so why i will simplify as i am getting direct answer?

I would work with the inequality as is. It is symmetrical in the given form. Symmetry makes me comfortable because it makes it easy to see patterns.
This question is a play on a standard concept that if x is a positive integer, minimum value of x + 1/x is 2. (which is actually derived from another concept: If the product of two positive integers is a constant (x*1/x = 1), their sum is least when they are equal (so x + 1/x is least when x = 1/x i.e. x = 1)
Similarly,
x/2 + 2/x has a minimum value of 2 when x = 2.
x/3 + 3/x has a minimum value of 2 when x = 3.
and so on...

In this question they don't say that x is a positive integer.
So when x is negative, (x/3 + 3/x) is negative.
When 0<x<=1, infinity > (x/3 + 3/x) >= 10/3
When 1<x<3, 10/3 > (x/3 + 3/x) > 2
When x >= 3, (x/3 + 3/x) >= 2

Notice that (x/3 + 3/x) does not take values between 0 and 2.
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Re: Tricky inequality problem [#permalink]

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17 Jul 2010, 02:09
Bunuel wrote:
gurpreetsingh's solution is correct.

Is $$\frac{x}{3}+\frac{3}{x}>2$$? --> is $$\frac{(x-3)^2}{x}>0$$?

How did you get $$\frac{(x-3)^2}{x}>0$$?

Step 1:
$$\frac{x}{3}+\frac{3}{x}>2$$?
Step 2 (multiply by 3x):
$$x^2+9>6x$$
Step 3 (move 6x to the left side):
$$x^2-6x+9>0$$
Step 4 (convert to the compact form):
$$(x-3)^2>0$$

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Re: Tricky inequality problem [#permalink]

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17 Jul 2010, 07:59
Another way :

Arithmatic mean >= geometric mean ( the numbers should be +ve )

=> $$\frac{x}{3} + \frac{3}{x} >=2$$ ; equality holds when x/3 = 3/x => $$x^2 = 9$$ => x = + 3

1st statement removes the possibility of x=3 but we do not know whether x>0 or not.

2nd statement removes the possibility of x<0 but x can be equal to 3

Thus both statements taken together states x is not equal to 3 and x is +ve

Thus C is the answer.
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Re: Tricky inequality problem [#permalink]

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18 Jul 2010, 10:55
The way I solved it.

1. x < 3 : LT3 - Insuf because with x = -1 the inequality is false and with x = 2 it is true.
2. x > 1 : GT1 - Insuf because with x = 2 the inequality is true and with x = 3 it is false

Combining the two options yeilds 1 < x < 3 which clearly shows that the original expression is > 2.

So the correct answer is C. 1 and 2 together are sufficient.
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Re: Tricky inequality problem [#permalink]

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19 Jul 2010, 03:20
(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0.
The correct answer is (0;3) and (3;+inf)
It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.
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Re: Tricky inequality problem [#permalink]

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19 Jul 2010, 09:32
I'm completely agree that x not equal 3 (as i wrote earlier;)

But we miss all positive x's that a > 3, don't we?
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Re: Tricky inequality problem [#permalink]

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19 Jul 2010, 09:48
ulm wrote:
(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0.
The correct answer is (0;3) and (3;+inf)
It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.

ulm wrote:
I'm completely agree that x not equal 3 (as i wrote earlier;)

But we miss all positive x's that a > 3, don't we?

I'm not sure exactly what you're asking here. You said that the answer to this question should be B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Statement (2) is: $$x>1$$ --> denominator is positive, nominator is also positive EXCEPT for one value of $$x$$: when $$x=3>1$$, then $$\frac{(x-3)^2}{x}=0$$ (not more than zero). Hence we have TWO different answers to the question "is $$\frac{(x-3)^2}{x}>0$$": one is NO for $$x=3>1$$ and another is YES for all other values of $$x>1$$. Two different answers = not sufficient.

Hope it's clear.
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Re: Tricky inequality problem [#permalink]

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21 Jul 2010, 09:41
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C
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Re: Tricky inequality problem [#permalink]

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21 Jul 2010, 10:10
Michmax3 wrote:
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C

Unfortunately your approach is not correct.

First of all if $$x=2$$--> $$\frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2$$, so you made an error in calculations ($$\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}$$). Again $$\frac{x}{3} + \frac{3}{x}>2$$ is true for ANY value of $$x$$ but 3, for which $$\frac{x}{3} + \frac{3}{x}=2$$.

Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that $$x$$ is an integer, hence $$x<3$$ and $$x>1$$ does not mean that $$x=2$$, it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.

Hope it's clear.
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Re: Tricky inequality problem [#permalink]

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21 Jul 2010, 10:30
Bunuel wrote:
Michmax3 wrote:
Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C

Unfortunately your approach is not correct.

First of all if $$x=2$$--> $$\frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2$$, so made you an error in calculations ($$\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}$$). Again $$\frac{x}{3} + \frac{3}{x}>2$$ is true for ANY value of $$x$$ but 3, for which $$\frac{x}{3} + \frac{3}{x}=2$$.

Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that $$x$$ is an integer, hence $$x<3$$ and $$x>1$$ does not mean that $$x=2$$, it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.

Hope it's clear.

Yes, then I guess I got lucky and need a lot more practice with these. Thanks for clarifying
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Re: Tricky inequality problem [#permalink]

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05 Oct 2010, 22:25
Bunuel wrote:
DenisSh wrote:
Bunuel wrote:
gurpreetsingh's solution is correct.

Is $$\frac{x}{3}+\frac{3}{x}>2$$? --> is $$\frac{(x-3)^2}{x}>0$$?

How did you get $$\frac{(x-3)^2}{x}>0$$?

Step 1:
$$\frac{x}{3}+\frac{3}{x}>2$$?
Step 2 (multiply by 3x):
$$x^2+9>6x$$
Step 3 (move 6x to the left side):
$$x^2-6x+9>0$$
Step 4 (convert to the compact form):
$$(x-3)^2>0$$

This is the most common error when solve inequalities. I keep writing this over and over again:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

So you CAN NOT multiply $$\frac{x}{3}+\frac{3}{x}>2$$ by $$3x$$ since you don't know the sign of $$x$$.

Wheat you CAN DO is: $$\frac{x}{3}+\frac{3}{x}>2$$ --> $$\frac{x}{3}+\frac{3}{x}-2>$$ --> common denominator is $$3x$$ --> $$\frac{x^2+9-6x}{3x}>0$$ --> multiply be 3 --> $$\frac{x^2+9-6x}{x}>0$$ --> $$\frac{(x-3)^2}{x}>0$$.

Hope it helps.

thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit
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Re: Tricky inequality problem [#permalink]

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06 Oct 2010, 00:23
hirendhanak wrote:
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit

Consider the inequality $$\frac{x^2-12}{x}>-1$$

Lets say I multiply both sides by 7x without considering the signs of the variable, what happens ?

$$x^2-12>-x$$
$$x^2+x-12>0$$
$$(x+4)(x-3)>0$$

Which is true whenever x>3 (both terms positive) or when x<-4 (both terms negative)

But since we haven't kept the Sign of x in mind when we multiplied in step 1, the solution is wrong.

For eg. Take x=-1 which according to us is not a solution. It is easy to see ((-1)^2-12)/(-1)=11>-1. So it should be a solution
Similarly take x=-6 which according to us is a solution, but ((-6)^2-12)/-6=-4<-1. So it should not be a solution
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Re: Tricky inequality problem [#permalink]

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06 Oct 2010, 04:09
hirendhanak wrote:
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit

Consider a simple inequality $$4>3$$ and some variable $$x$$.

Now, you can't multiply (or divide) both parts of this inequality by $$x$$ and write: $$4x>3x$$, because if $$x=1>0$$ then yes $$4*1>3*1$$ but if $$x=-1$$ then $$4*(-1)=-4<3*(-1)=-3$$. Similarly, you can not divide an inequality by $$x$$ not knowing its sign.

Hope it's clear.
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Re: Tricky inequality problem   [#permalink] 06 Oct 2010, 04:09

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