First thing about DS questions, I love solving them by trying to show that the statement can give yes & no answers to the question and thereby eliminating them,

Lets take 2) first, since it is easier

Statement (2): x>1
We can easily eliminate since x can be 1.5 or 2 or 10, we can't say whether x is greater than 3 or not. So, there goes Statement 2) and answer choices (B) and (D)

Statement (1): (x-3)(x-2)(x-1) > 0
take x as 1.5 -> (1.5-3)(1.5-2)(1.5-1) -> (-ve)(-ve)(+ve) -> so the answer is positive, since two neagtives make a prositive and therefore we have proved that x can be less than 3

now, let us prove that x can be greater than 3
-> take x as 4 -> (4-3)(4-2)(4-1) -> which is obviously greater than zero.

So, statement 1 alone is also not sufficient and so there goes answer choice (A) as well


Combining both statements
X> 1 and (x-3)(x-2)(x-1) > 0

from our calculations for statement 2, we already showed that taking x=1.5 and x=4, we are able to satisfy the inequality and since these two values of x are greater than 1, we are thus unable to determine whether x>3.

Hope this helps.

i can't believe i erred on this one. i overlooked one of the most fundamental principles of DS. i considered x an integer!

thanks for pointing it out.

disbelief.

rahul.s wrote:is x > 3?

1) (x-3)(x-2)(x-1) > 0
2) x > 1

OA: E

the OE was complicated. could someone simplify it?

1. (x-3)(x-2)(x-1) > 0

For numbers greater than 3 = +ve
Numbers between 2 and 3 = -ve
numbers between 1 and 2 = +ve
Numbers less than 1 = -ve

Not sufficient

2. x>1 is not sufficient to tell whether x>3 Not sufficient

Combining
Even if we combine 1 and 2 the x could be between 1 and 2 or x is greater than 3 Not sufficient again to conclude whether x>3

Hence E

couldn't you just say that:

1) INS: the 3 possible solutions of (x-3)(x-2)(x-1)>0 are x > 3; x > 2; or x > 1. so, you cannot conclude x to be greater than 3. it could be 2 or 2.5

2) INS: x >1. x could be 2

Together: still insufficient b/c x could still be 2.

Am I looking at this logically?