is x/3 + 3/x >2?

[guest612]

Gmat Focus:

Is X/3 + 3/X > 2?

1. X<3
2. X>1

Answer is C. I chose E.

Collectively, 1<x<3.
I plugged in x=0.5 and 1.5 and got different answers (greater than and less than 2). How is the answer C?

[Sudhan]

Is X/3 + 3/X > 2?

1. X<3
2. X>1


BDACE Grid,

X>1; ---(2)
Sub X=2;
2/3+3/2>2
~0.6+1.5 >2 Suff

Sub X=3;
-> 3^2+9>6(3)
18>18- Insuf Suff. So for two different values (2) is not sufficient

X<3;----(1)

Sub x= -1/2

(-1/2) /3 + 3/(-1/2) >2
(-1/6) -6 >2
-37/6>2 Which is Insuff

Sub x=1,
1/3+3 >2;
0.33+3 > 2 Suff. So for two different values (1) is not sufficient

Using 1 and 2,
when x >1 and X< 3, the equation is satisfied,


Thanks

[pt]

The answer should be b, because for non-negative values (and hence for x>1) the equation is x/3 + 3/x >2 will hold true. x<3 is neither necessary nor sufficient.

[guest612]

the official answer is C.

Sudhan, can you please elaborate why It's C instead of E? I appreciate the grid but I had already eliminated A & B from my answer choices.

[Sudhan]

Is X/3 + 3/X > 2?

If you substitute for X< 1 and X>3, the condition does not satisfy.
Hence it is sufficient to answer the question. since this is YES/NO type of problem.

Thanks

[RonPurewal]

first off, note that the inequality is automatically false if x is negative, since both terms on the left hand side will be negative.

x can't be 0, because 3/x is undefined in that case.

to solve this thing for positive quantities, multiply through by the common denominator (= 3x), giving x^2 + 9 > 6x.
solve this like you'd solve any other quadratic inequality: move all the terms to one side and factor:
x^2 - 6x + 9 > 0
(x - 3)^2 > 0
this is crazy: it's true for ALL positive quantities, with the single exception x = 3.
that single exception is important, too; it's the only reason that statement (2) by itself is not sufficient to satisfy the problem. (in other words, all values greater than 1 give a 'yes' answer, with the single exception x = 3 which doesn't.)

--

taking the statements together, x is restricted to 1 < x < 3. as deduced here, those values all yield Yes answers (because the only positive number that doesn't is x = 3). so the answer is c.

--

interestingly, note that the answer to the problem becomes 'b' if we make the seemingly innocuous change of replacing '>' with '>'. if you don't understand why this is the case, feel free to post.

[vgirotra]

x^2 - 6x + 9 > 0
(x - 3)^2 > 0
this is crazy: it's true for ALL positive quantities, with the single exception x = 3.
that single exception is important, too; it's the only reason that statement (2) by itself is not sufficient to satisfy the problem. (in other words, all values greater than 1 give a 'yes' answer, with the single exception x = 3 which doesn't.)

--

--

interestingly, note that the answer to the problem becomes 'b' if we make the seemingly innocuous change of replacing '>' with '>'. if you don't understand why this is the case, feel free to post.

Hi Ron

If x=3 is the only exception, why isn't statement 1 sufficient as it doesn't include 3?

Is it because it includes 0?



Vivek

[Ben Ku]

Actually in the first step, Ron deduced that x is not negative and is not 0. And from the math, we find that x < 3. So we can actually rephrase the question as:

Is 0 < x < 3?

(1) alone is insufficient because x can be negative (which is not in the interval), or x can be positive (e.g. 2, which is in the interval).

(2) alone is insufficient because x can be less than 3 (which is in the interval) or greater than 3 (which is not in the interval).

(1) and (2) together restricts x to 1 < x < 3. We know that if x is in that interval, then the answer to the question must be YES.

Hope that helps!