since u don't if x if +ve or -ve you can not cross multiply it to reduce the question stem to (x-3)^2>0

so I tried by inserting number based on the stmt 1 and 2 but IMO it should be B as when x>1 all the values are resulting into x/3+ 3/x > 2 ....can someone pointout what I am mising.

Harbinder wrote:since u don't if x if +ve or -ve you can not cross multiply it to reduce the question stem to (x-3)^2>0

so I tried by inserting number based on the stmt 1 and 2 but IMO it should be B as when x>1 all the values are resulting into x/3+ 3/x > 2 ....can someone pointout what I am mising.

If x=3, we get:

3/3 + 3/3 = 1 + 1 = 2. Is 2>2? No.

Every other value of x > 1 gives a "yes" answer.

1. x<3 :Lets take 2 first, 2/3+3/2 < 2; now take x=1, 1/3+3/1 >2 hence not suff

2 x>1 :Lets take x=2, again 2/3+3/2 < 2; now take x=3, 6/3+3/6 > 2; hence not suff.

Now combine statements mean 1<x<3 : let x = 2 or x= 2.5, its will always be < 2

Hence correct option is C

sudeeparies wrote:1. x<3 :Lets take 2 first, 2/3+3/2 < 2; now take x=1, 1/3+3/1 >2 hence not suff

2 x>1 :Lets take x=2, again 2/3+3/2 < 2; now take x=3, 6/3+3/6 > 2; hence not suff.

Now combine statements mean 1<x<3 : let x = 2 or x= 2.5, its will always be < 2

Hence correct option is C

Math alert!

2/3 + 3/2 is actually greater than 2!

Also, if x=3, we get 3/3 + 3/3 = 2.

If x is between 1 and 3, we always get a result greater than 2 (which is why C is the correct answer).

(1) is insufficient on its own because if we choose a negative value for x we get a negative sum (and if we choose x=0 we have an undefined sum).

oops, my mistake..thanks Stuart!