Is x/3 + 3/x > 2 ?
This topic has expert replies
Simplify
Is x/3+3/x > 2
Is (x^2+9 ) / 3x > 2
Stmt I
x<3
x= 1 NO
x= 2 YES
INSUFF
Stmt II
x>1 (i.e. x is positive)
Now again the question is Is (x^2+9 ) / 3x > 2
Multiply by 3x on both sides (doesnt affect the sign of inequality since x is positive
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
If x = 4 YES
If x= 3 NO
INSUFF
TOGETHER:
x is between 1 and 3 so we get a definite NO
C
Regards,
CR
Is x/3+3/x > 2
Is (x^2+9 ) / 3x > 2
Stmt I
x<3
x= 1 NO
x= 2 YES
INSUFF
Stmt II
x>1 (i.e. x is positive)
Now again the question is Is (x^2+9 ) / 3x > 2
Multiply by 3x on both sides (doesnt affect the sign of inequality since x is positive
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
If x = 4 YES
If x= 3 NO
INSUFF
TOGETHER:
x is between 1 and 3 so we get a definite NO
C
Regards,
CR

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Hi Cramya,
Can you explain how did you get x < 3?
Thanks,
Vikas
Can you explain how did you get x < 3?
Thanks,
Vikas
cramya wrote:Simplify
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
C
Regards,
CR

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Nice work Cramya, but can you solve it without plugging numbers. I need to see it algebraically please.cramya wrote:Simplify
Is x/3+3/x > 2
Is (x^2+9 ) / 3x > 2
Stmt I
x<3
x= 1 NO
x= 2 YES
INSUFF
Stmt II
x>1 (i.e. x is positive)
Now again the question is Is (x^2+9 ) / 3x > 2
Multiply by 3x on both sides (doesnt affect the sign of inequality since x is positive
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
If x = 4 YES
If x= 3 NO
INSUFF
TOGETHER:
x is between 1 and 3 so we get a definite NO
C
Regards,
CR
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Actually, that solution is incorrect. The second inequality should be x < 3.mehravikas wrote:Hi Cramya,
Can you explain how did you get x < 3?
Thanks,
Vikas
cramya wrote:Simplify
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
C
Regards,
CR
We have:
Is (x3) (x3) > 0?
Well, the product of two terms is positive when both are positive or both are negative.
So, to see when that inequality holds true, we solve for those two cases:
I) x3 > 0
x > 3
II) x3 < 0
x < 3
Basically, the answer will be "YES" for all values of x other than 3.
So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
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So basically you're saying that x could equal any number but not 3 in order to make the question true?Stuart Kovinsky wrote:Actually, that solution is incorrect. The second inequality should be x < 3.mehravikas wrote:Hi Cramya,
Can you explain how did you get x < 3?
Thanks,
Vikas
cramya wrote:Simplify
Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^26x+9>0
Or Is (x3) (x3) > 0 or simply Is x>3 or x < 3
C
Regards,
CR
We have:
Is (x3) (x3) > 0?
Well, the product of two terms is positive when both are positive or both are negative.
So, to see when that inequality holds true, we solve for those two cases:
I) x3 > 0
x > 3
II) x3 < 0
x < 3
Basically, the answer will be "YES" for all values of x other than 3.
So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x3)(x3) > 0 > always yes..... so why it's insufficient.
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Because, as Cramya noted, we could only perform all those manipulations once we knew that x was positive; with just statement 1, negative values of x give us a "no" answer.Abdulla wrote:So basically you're saying that x could equal any number but not 3 in order to make the question true?
I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x3)(x3) > 0 > always yes..... so why it's insufficient.
We need statement (2) to affirm that x is positive.
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Thanks Stuart.Stuart Kovinsky wrote:Because, as Cramya noted, we could only perform all those manipulations once we knew that x was positive; with just statement 1, negative values of x give us a "no" answer.Abdulla wrote:So basically you're saying that x could equal any number but not 3 in order to make the question true?
I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x3)(x3) > 0 > always yes..... so why it's insufficient.
We need statement (2) to affirm that x is positive.
I think plugging numbers from the beginning without any manipulation is the best and easiest approach to solve this question.
Is x/3 + 3/x > 2 ?
1) x<3
plug 2 for x : 2/3 + 3/2 > 2 > yes
plug 0 for x : 0/3 + 3/0 > 2 > no (undefined)
2) x>1
Plug 2 for x : 2/3 + 3/2 > 2 > yes
Plug 3 for x : 3/3 + 3/3 > 2 > no
Combined: Any number between 1 and 3 will give a yes answer.
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Hi Stuart,
We can say x < 3 only because we do not know whether x is +ve or ve?
Vikas
We can say x < 3 only because we do not know whether x is +ve or ve?
Vikas
Stuart Kovinsky wrote: We have:
Is (x3) (x3) > 0?
Well, the product of two terms is positive when both are positive or both are negative.
So, to see when that inequality holds true, we solve for those two cases:
I) x3 > 0
x > 3
II) x3 < 0
x < 3
Basically, the answer will be "YES" for all values of x other than 3.
So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
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We can say that x < 3 (is one possibility) as soon as we get to the inequality:mehravikas wrote:Hi Stuart,
We can say x < 3 only because we do not know whether x is +ve or ve?
Vikas
Stuart Kovinsky wrote: We have:
Is (x3) (x3) > 0?
Well, the product of two terms is positive when both are positive or both are negative.
So, to see when that inequality holds true, we solve for those two cases:
I) x3 > 0
x > 3
II) x3 < 0
x < 3
Basically, the answer will be "YES" for all values of x other than 3.
So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
(x3) (x3) > 0
To get to that inequality in the first place, we have to know that x is positive.
Stuart Kovinsky  Kaplan GMAT Faculty  Toronto
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May be I am missing a rule completely but if we break down the inequality (x3)(x3) > 0 then
1. x  3 > 0 > x > 3
2. x  3 > 0 > x > 3
It would be the same if we have an inequality (x 1) (x + 5) > 0
therefore x > 1 or x > 5, here we can never say x > 1 and x < 5
I don't know whether I have given a good example, please don't mind if its a crap example because I am quite confused after seeing x > 3 or x < 3
1. x  3 > 0 > x > 3
2. x  3 > 0 > x > 3
It would be the same if we have an inequality (x 1) (x + 5) > 0
therefore x > 1 or x > 5, here we can never say x > 1 and x < 5
I don't know whether I have given a good example, please don't mind if its a crap example because I am quite confused after seeing x > 3 or x < 3
Stuart Kovinsky wrote:We can say that x < 3 (is one possibility) as soon as we get to the inequality:mehravikas wrote:Hi Stuart,
We can say x < 3 only because we do not know whether x is +ve or ve?
Vikas
Stuart Kovinsky wrote: We have:
Is (x3) (x3) > 0?
Well, the product of two terms is positive when both are positive or both are negative.
So, to see when that inequality holds true, we solve for those two cases:
I) x3 > 0
x > 3
II) x3 < 0
x < 3
Basically, the answer will be "YES" for all values of x other than 3.
So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
(x3) (x3) > 0
To get to that inequality in the first place, we have to know that x is positive.
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Reread the logic of my first response.
If we know that x*y > 0, there are two possibilities: both x and y are positive or both x and y are negative.
Similarly in this question, in order for (x3)(x3) to be positive, either both terms are positive or both terms are negative.
So, there are two cases:
1) Both positive, in which case (x3) and (x3) are both > 0, so:
x3 > 0
x > 3
2) Both negative, in which case (x3) and (x3) are both < 0, so:
x  3 < 0
x < 3
Therefore, that inequality will hold true whenever:
x>3 OR x<3
We can apply the same logic to the example you gave:
(x 1) (x + 5) > 0
means that either:
1) (x1) and (x+5) are BOTH positive; or
2) (x1) and (x+5) are BOTH negative.
Solving:
1) x1 > 0 AND x + 5 > 0
x > 1 AND x > 5
when we have two inequalities pointing in the same direction, the one that's more limiting applies, so the first case resolves to:
x > 1
2) x1 < 0 AND x + 5 < 0
x < 1 AND x < 5
again taking the more limiting case:
x < 5
So, the original inequality does in fact mean:
x < 5 OR x > 1
If we know that x*y > 0, there are two possibilities: both x and y are positive or both x and y are negative.
Similarly in this question, in order for (x3)(x3) to be positive, either both terms are positive or both terms are negative.
So, there are two cases:
1) Both positive, in which case (x3) and (x3) are both > 0, so:
x3 > 0
x > 3
2) Both negative, in which case (x3) and (x3) are both < 0, so:
x  3 < 0
x < 3
Therefore, that inequality will hold true whenever:
x>3 OR x<3
We can apply the same logic to the example you gave:
(x 1) (x + 5) > 0
means that either:
1) (x1) and (x+5) are BOTH positive; or
2) (x1) and (x+5) are BOTH negative.
Solving:
1) x1 > 0 AND x + 5 > 0
x > 1 AND x > 5
when we have two inequalities pointing in the same direction, the one that's more limiting applies, so the first case resolves to:
x > 1
2) x1 < 0 AND x + 5 < 0
x < 1 AND x < 5
again taking the more limiting case:
x < 5
So, the original inequality does in fact mean:
x < 5 OR x > 1
Stuart Kovinsky  Kaplan GMAT Faculty  Toronto
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