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# Is x/3 + 3/x > 2

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is (x/3+3/x) > 2? (1) x < 3 (2) x > 1 This ds has [#permalink]

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08 Jun 2011, 03:52
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?
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08 Jun 2011, 06:33
I'm actually a little lost as to why it's C and not B. Clearly statement 1 is insufficient because since this is a yes no question. We are getting different answers with different values so insufficient. However, with regards to statement 2 all values greater than 1 which is what statement 2 is saying are greater than 2. So statement 2 sufficient. Unless im not reading something correctly.

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08 Jun 2011, 06:44
scbguy wrote:
I'm actually a little lost as to why it's C and not B. Clearly statement 1 is insufficient because since this is a yes no question. We are getting different answers with different values so insufficient. However, with regards to statement 2 all values greater than 1 which is what statement 2 is saying are greater than 2. So statement 2 sufficient. Unless im not reading something correctly.

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TRY X = 2 AND 3 AND YOU WILL GET RESULT OF YOUR QUESTION.
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08 Jun 2011, 09:18
(x^2 + 9) / 3x > 2
1) Try -1 for No and 1 for Yes.
2) Try 2 for Yes and 3 for No.
Combining 1) and 2) values between 1< x < 3 give only Yes answer.
C is the answer. No need to simplify. Baten80 is right.
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13 Jun 2011, 04:20
a+b

1<x<3 gives values where LHS > RHS always.

hence C.
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Re: Is \frac{x}{3} + \frac{3}{x} > 2 ? (1) x < 3 [#permalink]

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14 Aug 2012, 22:06

For any positive number 'a', (a+ 1/a )>=2
So , from the given options, all we need to do is ensure that x/3 is a positive quantity.

Hence B.

Please correct me if i am wrong.
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Re: Is \frac{x}{3} + \frac{3}{x} > 2 ? (1) x < 3 [#permalink]

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14 Aug 2012, 23:16

For any positive number 'a', (a+ 1/a )>=2
So , from the given options, all we need to do is ensure that x/3 is a positive quantity.

Hence B.

Please correct me if i am wrong.

OA for this question is given in the first post under the spoiler and it's C, not A.

To see that the second statement is not sufficient, consider x=3 for a NO answer (if x=3 then 3/3+3/3=2) and x=10 for an YES answer. Please refer to the solutions on the first page.

Hope it helps.
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15 Aug 2012, 02:47
shrouded1 wrote:
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

Very well illustrated why we shouldn't multiply an inequality with an expression of unknown sign.
Just a few more things:

The given inequality $$\frac{x^2-12}{x}>-1$$ is equivalent to $$\frac{x^2+x-12}{x}>0$$. The sign of the ratio of two numbers is the same as the sign of their product. So, the previous inequality is equivalent to $$x(x^2+x-12)>0$$, which in fact can be obtained from the original inequality by multiplying by $$x^2$$, which we know for sure that it is positive, because x being in the denominator, $$x\neq0.$$

When given $$A/B>0,$$ it means:
1) A and B have the same sign (either both positive or both negative)
2) Neither A nor B can be 0, A is in the numerator and the fraction is greater than 0, B is in the denominator.
It is obvious that multiplying by B the given inequality leads to $$A>0$$, incorrect.
$$A/B>0$$ is equivalent to $$AB>0,\, B\neq0.$$

When given $$A/B\geq0,$$ it means:
1) A and B have the same sign (either both positive or both negative). A can be 0.
2) B cannot be 0, being in the denominator.
It is obvious that multiplying by B the given inequality leads to $$A\geq0$$, incorrect.
$$A/B\geq0$$ is equivalent to $$AB\geq0,\, B\neq0.$$

So, when we have to compare a fraction to 0, we can just compare the product of all the factors involved to 0. We should check carefully for values of the unknown if equality to 0 is allowed and we should not forget the values for which the denominator becomes 0.
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Re: Is x/3 + 3/x > 2 [#permalink]

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16 Jan 2013, 03:48
$$\frac{x}{3}+3/x > 2$$
$$\frac{{9+x^2}}{3x}>2$$
$$\frac{{9+x^2}}{3x}-2>0$$
$$\frac{x^2-6x+9}{3x}>0$$
$$\frac{(x-3)^2}{3x}>0$$

x is not equal 3 since it will make the value 0
x is not negative since it will make the value negative

(1) $$x < 3$$
x may be negative or positive. THUS INSUFFICIENT!

(2) $$x > 1$$[/quote]
x may be equal to 3 or not. THUS INSUFFICIENT.

Together (1) and (2): x is positive and not equal to 3

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Re: Is x/3 + 3/x > 2 [#permalink]

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04 Jul 2014, 22:22
In general, if a number happens to be positive, the sum of the number and its reciprocal is always >=2.
Here, statement 1 is not sufficient as it includes the negative portion of the number line.
Statement 2 appears to be sufficient, but it fails alone since it includes x=3 in its set of solutions. At x=3, LHS becomes EQUAL to 2, which is not what the question is looking for. A holistic answer is obtained by clubbing both the statements!
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Re: Is x/3 + 3/x > 2 [#permalink]

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Is x/3 + 3/x > 2 [#permalink]

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23 Dec 2015, 09:05
A simpler approach to the problem can be to use AM>=GM method (Arithmetic mean and Geometric mean)

For 2 positive numbers, the AM is always greater than either equal to the GM --> This method usually should strike when you see inequalities involving inverse fractions as is in our problem. Now simply looking at the equation and applying AM>=GM approach here:

(x/3 + 3/x)/2 >= sqrt(x/3 * 3/x) or x/3 + 3/x > 2. Now we just need to fix the numbers in a way that the basic math is not broken. With

A. x<3, x could be 0 (not defined situation in our equation) or x is -ve --> For sure this is not sufficient
B. x>1 (gives us some hope to apply AM>=GM formula) BUT x=3 will still create the issue as LHS =2 which is not required. Hence, not sufficient

USING A and B => 1<x<3, for which the AM>GM eqn will always hold.

P.S. - This approach may help us save a lot of time if we are able to think back with the problem statement. Also, sometimes in PS questions, if we are asked for minimum values (given all the variables written in the equation are >0 or positive, this approach can come handy (if the equation simplifies which is generally the case as problems are always solvable )

For e.g. - We can look at a problem like what can be the maximum value of (21-2x)(x-1)(x-5) given that 5<x<10. We can immediately solve it by the method

AM>=GM or GM<=AM
= cuberoot[(21-2x)(x-1)(x-5)]<= (21- 2x + x - 1 + x - 5)/3
= Max{(21-2x)(x-1)(x-5)} = 5^3 = 125

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Is (x - 3)^2/(3x) >0 [#permalink]

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27 Oct 2016, 00:16
Is (x - 3)^2/(3x) >0

(1) x < 3
(2) x > 1

Last edited by yezz on 27 Oct 2016, 00:36, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Is (x - 3)^2/(3x) >0 [#permalink]

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27 Oct 2016, 01:53
yezz wrote:
Is (x - 3)^2/(3x) >0

(1) x < 3
(2) x > 1

Statement 1:

If x=2, then (x - 3)^2/(3x) = 1/6.. Yes

If x=-1, then (x - 3)^2/(3x) = -16/3..No

Insufficient

Statement 2:

If x=2, then (x - 3)^2/(3x) = 1/6.. Yes

If x=3then (x - 3)^2/(3x) = 0.. No

Insufficient

Statement 1&2:

If 1<x<3, then the expression will always be positive.

C

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Is x/3 + 3/x > 2 [#permalink]

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27 Oct 2016, 03:54
Is $$\frac{x}{3} + \frac{3}{x} > 2$$?

Question: is $$\frac{x}{3}+\frac{3}{x}>2$$? Let's check when this statement holds true.

$$\frac{x}{3}+\frac{3}{x}>2$$ --> $$\frac{(x^2-6x+9)}{3x}>0$$ --> $$\frac{(x-3)^2}{3x}>0$$.

Nominator (x-3)^2 is NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that $$\frac{(x-3)^2}{3x}>0$$ holds true when x>0 and x#3.

Let's move to the statements:

(1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

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Is x/3 + 3/x > 2   [#permalink] 27 Oct 2016, 03:54

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