Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

so we have x/3 + 3/x > 2...assume x > 0 (as when x < 0, equation fails, thus statement 1 alone is not true), then we have x^2/3 + 3 - 2x > 0 => x^2 - 6x + 9 > 0 => (x-3)^2 > 0 so for all positive x except for 3 does the equation stand. thus statement 2 alone doesn't stand either...we need both. 1 < x < 3 definitely works.

I think better way to look at this as an eqn a+1/a ... see the behaviour of this eqn....first image is for eqn (a+1/a) and the later is the actual behaviour of (x/3+3/x) value of this equation cannot go below 2 (except in fractional values less than 1)..thats the property...hence OA C

If you like my post ....consider it for Kudos

Attachments

File comment: behaviour of a+1/a

graph.JPG [ 16.6 KiB | Viewed 1186 times ]

File comment: behaviour of x/3+3/x

graph.JPG [ 14.97 KiB | Viewed 1189 times ]

_________________

Bhushan S. If you like my post....Consider it for Kudos

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (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 (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.

Bunuel can you please elaborate how did you arrive at this step: x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Just noticed that you addressed the question. I simply reduced to a common denominator x/3, 3/x and -2: x/3+3/x>2 --> x/3+3/x-2>0 --> (x^2+9-6x)/3>0 --> (x-3)^2/3>0

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 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 (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.

Answer: C.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem.....

_________________

"Where are my Kudos" ............ Good Question = kudos

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 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 (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.

Answer: C.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem.....

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

So you CANNOT multiply \frac{(x-3)^2}{x}>0 by 3x since you don't know the sign of x: if x>0, then you'll have (x-3)^2>0 but if x<0, then you'll have (x-3)^2<0 (flip the sign when multiplying by negative value).