x/3

a/b + b/a is always >=2

Since A.M >= G.M

Thus (a/b+b/a)/2>=Sqrt(a/b*b/a)

Or a/b+b/a >=2

This is always true.So how is this a D.S question?

Sher1 wrote:Is x/3+3/x>2

1 x<3

2 x>1

rewrite given statement

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

is (x-3)^2>0?

the answer is yes if we know that x>3

statement 1)

x<3 so sufficient

statement 2)

x>1 ---> x could be less than 3 and greater than 3. insufficient

A is answer

you got this!

vikram_k51 wrote:a/b + b/a is always >=2

Since A.M >= G.M

Thus (a/b+b/a)/2>=Sqrt(a/b*b/a)

Or a/b+b/a >=2

This is always true.So how is this a D.S question?

where did you get this? good to know, if true.

This is a question I got in GMAT Focus and the OA is C

I also incorrectly got A. Didnt like their explanation so thought if there is a better way to tackle this.

Stmt 1 is not sueffecient for the negative values of X
If X = -1, the sum of X/3 + 3/X is negative and hence less than 2
If X = 2, the sum 2/3 + 3/2 = .66 + 1.5 = 2.1 which is greater than 2
Insufficient

Stmt 2
If X = 2, the sum 2/3 + 3/2 = .66 + 1.5 = 2.1 which is greater than 2
If X = 3, the sum 3/3 + 3/3 =2 which is not greater than 2
Insufficient

Combine together 1 < x< 3 C is sufficient

So I got this question correct. I converted the original question into:

(x^2+9)/3x > 2, then plugged in numbers to test the two conditions.

For Stmt 1, positive and negative values render the stmt insufficient using the equation above.

However, I could certainly see myself converting the original question into the following (x-3)^2 > 0. This is what I don't understand. It seems as though here any number greater than 3 satisfies the requirement, so answer would be A.

Thoughts?

shahdevine wrote:

vikram_k51 wrote:a/b + b/a is always >=2


Since A.M >= G.M

Thus (a/b+b/a)/2>=Sqrt(a/b*b/a)

Or a/b+b/a >=2

This is always true.So how is this a D.S question?

where did you get this? good to know, if true.

IMO C

yeah this is always true:
A.M>=G.M>=H.M
the equality is true when a/b=b/a
in the question
x/3+3/x>=2 , equality holds when x/3=3/x, i.e X=3 or -3

1)x<3, x can be -3 not sufficient
2) x>1, X can be 3, not sufficient

combines 1<X<3, X cannot be equal to 3 or -3. sufficient