question 4

[Fiver]

Is X^2+9/3x > 2.

St1] x<3. At this stage we cannot multiply both sides by 3x, because x could be 0 and we cannot multipy by 0 on both sides of an inequality.
Hence we need to keep the inequality as above and assume negative and positive values for X. If x is a negative value then the answer to the question is no and if x =1 then the answer to the question is yes. Insufficient.

St2] x>1. Now we know that x is non-zero. Hence we can alter the above inequality by multiplying both sides by 3x.
Is x^2 + 9>6x
Is x^2 -6x+9>0
Is (x-3)^2 > 0. This is true for any value of x other than 3 and false if x=3. Hence insufficient.
Together we know that x not equal to 3. Hence sufficient.
My answer is C. What's the OA?

[aarok]

x/3 + 3/x > 2 is equal to (x-3)^2/3x>0
(x-3)^2 is always > 0 so we need information on whether 3x>0

1) not sufficient because X could be negative
2) sufficient because if x>1 3x>0

My answer B

[clock60]

my pick is C but not sure.. my reasoning
is x/3+3/x-2>0
is (x^2-6x+9)/3x>0 or
is (x-3)^2/x>0, (x-3)^2>0 in any cases exept x=3 (if x=3 the whole expression will be 0)
1/(3x)>0 if x>0
so the problem asks is x>0 and x is not 3
(1)x<3 not sufficient ,x is not 3 but it can be less than 0
(2)x>1 not sufficient x positive but it can be equal 3
together x-positive and not 3
so C
what is oa??

[NikolayZ]

C it is.
The factorized expression is ((x-3)(x-3))/3x>0 ?
1) if x<3, then (x-3)<0 anyways, the nominator is always positive. But the denominator can be either positive and negative. so Insuff.
2) Knowing that x>1, we can't conclude whether the expression is positive.

Combining 1 and 2, 1<x<3, nominator is always positive, and denominator is too. Sufficient then.

[ryan]

Is x/3 + 3/x > 2?

1. x<3 - not sufficient

for x=2, 2/3 + 3/2 >2
for x=1.5, 1.5/3 + 3/1.5 > 2
for x=-1, <2

2. x>1, not suffcient, either >2 or <2.


Combining 1&2, The answer is C.

[MBAchase]

IMO C

[sandy_online]

Originally Posted by Mits83

IMO answer is A.

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

For this condition to occur, x can be anything BUT 3.

a) sufficient.
b) not sufficient as x could be 3.

in above subtle assumptions you have made : i.e. x>0
in multiplication by 6x in inequality you need to consider the sign of x.

[kutra]

OA is C