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?
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Source: Beat The GMAT — Data Sufficiency |
where did you get this? good to know, if true.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?
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.
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
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?
(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?
IMO Cshahdevine wrote:where did you get this? good to know, if true.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?
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
The powers of two are bloody impolite!!
