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DenisSh
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 Updated on: 26 Nov 2018, 03:38
Originally posted by DenisSh on 16 Jul 2010, 00:06.
Last edited by Bunuel on 26 Nov 2018, 03:38, edited 2 times in total.
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Is \(\frac{x}{3} + \frac{3}{x} > 2\)?


(1) \(x < 3\)

(2) \(x > 1\)
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C
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 17 Jul 2010, 04:36
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DenisSh
Bunuel

gurpreetsingh's solution is correct.

Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)?
Answer: C.

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):
\((x-3)^2>0\)

Please explain...
Show more

This is the most common error when solving inequalities.

Never multiply or reduce an 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:

    \(\frac{x}{3}+\frac{3}{x}>2\);

    \(\frac{x}{3}+\frac{3}{x}-2>\);

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

    \(\frac{x^2+9-6x}{x}>0\);

    \(\frac{(x-3)^2}{x}>0\).

Hope it helps.
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 27 Oct 2016, 03:54
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Is \(\frac{x}{3} + \frac{3}{x} > 2\)?

    Is \(\frac{x}{3}+\frac{3}{x}>2\)?

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

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

Since the numerator, (x - 3)^2, is NEVER negative, then the above expression holds true when the denominator, 3x, is positive, or simply when x is positive. So, \(\frac{(x-3)^2}{3x}>0\) is true for all positive values of x, except when x = 3 (in this case the whole expression becomes 0, not more than 0).

Basically the question asks: is x > 0 AND x ≠ 3 ?

(1) x < 3.

So, x ≠ 3 but it's not enough, we also need x to be positive. Not sufficient.

(2) x > 1

So, x is positive but it's not enough, we also need x not to be equal to 3. Not sufficient.

(1)+(2) When combining the statements we have that 1 < x < 3. So, x is positive and not equal to 3. Sufficient.

Answer: C.
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 16 Jul 2010, 03:13
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naish

\((x-3)^2\) 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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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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 16 Jul 2010, 05:06
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naish
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.
:)
Show more

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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 16 Jul 2010, 07:42
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DenisSh
Is \(\frac{x}{3} + \frac{3}{x} > 2\)?

(1) \(x < 3\)

(2) \(x > 1\)

Please outline your approach! :)
Show more

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.

Answer: C.
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 17 Jul 2010, 02:09
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Bunuel

gurpreetsingh's solution is correct.

Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)?
Answer: C.
Show more

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):
\((x-3)^2>0\)

Please explain...
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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 17 Jul 2010, 07:59
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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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 19 Jul 2010, 03:20
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(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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 19 Jul 2010, 09:18
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(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.
Show more

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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 21 Jul 2010, 09:41
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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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 21 Jul 2010, 10:10
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Michmax3
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
Show more

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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 05 Oct 2010, 22:25
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Bunuel
DenisSh
Bunuel

gurpreetsingh's solution is correct.

Is \(\frac{x}{3}+\frac{3}{x}>2\)? --> is \(\frac{(x-3)^2}{x}>0\)?
Answer: C.

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):
\((x-3)^2>0\)

Please explain...

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.
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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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 06 Oct 2010, 00:23
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hirendhanak

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
Show more

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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 10 Jun 2011, 12:38
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is (x/3+3/x) > 2?
(1) x < 3
(2) x > 1

This ds has been discussed thoroughly at https://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?
Show more

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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Is x/3 + 3/x > 2 (1) x < 3 (2) x > 1 15 Aug 2012, 02:47
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hirendhanak

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
Show more

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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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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