Is (x – 2)(x – 3) > 0?

Is (x – 2) (x – 3) > 0?
1) x – 2 < 0
2) x – 3 < 0

the official answer is A, but my explanation says: it is D.
statement 2: to be sufficient (x-2) (x-3) both must be positive or negative simultaneously.
is (x-2)*negative >0?
So, x-2=what?
or, (x-3)+1=what?
or, negative+1 (positive)=what?
or, negative (-100)+1 (positive)=-99, which is negative
so, x-2 is negative for statement 2

original question stem:
Is (x – 2) (x – 3) > 0?
negative*negative=positive, which is greater than zero.......>so, YES
what is the lack in my understanding?
Thanks...

Don't follow what you've written there... To check that the answer is not D, check 2.5 and 0 for (2). Also, based on the questions you ask I think you really have to brush fundamentals on inequalities. The links are given in my previous post. Hope it helps.

Bunuel, i am writing something against D.

statement 2: to be sufficient (x-2) (x-3) both must be positive or negative simultaneously.
is (x-2)*negative >0?
So, x-2=what?
or, (x-3)+1=what?
or, negative+1 (positive)=what?
or, negative (-1)+1 (positive)=0, which is neither positive nor negative.

original question stem:
Is (x – 2) (x – 3) > 0?
i don't know (x-2) is positive or negative*negative>0?
the answer is ''i don't know'', which indicates NOT sufficient.
is my explanation right now?

is (x-2) (x-3) > 0?

This will be so only when x<2 OR x>3.

Stmt-1: x<2 -- suff

stmt-2: x<3 , here x could fall in the range 2<X<3 where it will be negative. Also x<2 is true and x will be positive.

so positive or negative both possible. insuff

Answer is A.

Roots = 2,3.
2<3
(1) Pick 0 => (x-2)(x-3)>0 => sufficient
=> option (A) or (D)
(2) Case a: Pick a number between 2 and 3 say 2.5 => (x-2)(x-3)<0
Case b: Pick 0 => (x-2)(x-3)>0
Hence, (2) is not sufficient => correct option is (A)



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Bunuel I know what is missing in the above explanation by Asad.
Asad you are missing an important summarizing point. Please look at the below solution:

For (x – 2) (x – 3) > 0 to be true.
there are two cases.

Assume A=(x-2) and B=(x-3)
A*B>0 only when both A and B are +ve
A*B>0 only when both A and B are -ve

Case1: when both (x – 2) and (x – 3) are +ve in eq (x – 2) (x – 3) > 0
Therefore => x>2 and x>3

Now trying to plot the above values on the number line
Final equation will become x>3

OR

Case 2:when both (x – 2) and (x – 3) are -ve in eq (x – 2) (x – 3) > 0
Therefore =>x<2 and x<3

Now trying to plot the above values on the number line
Final equation will become x<2

======================

Now if we summarize we see that either x>3 or x<2.
This is what we need to prove.

(1) x-2<0
which is nothing but x<2, and this is what we need to prove.
Hence sufficient.
(2) x-3<0
which is nothing but x<3. Since nothing is given about x if it is an integer or not.
Hence x could be 2.5 or 1.5.
if x=2.5 it will not satisfy x>3 or x<2
if x=1.5 it will satisfy x>3 or x<2
Hence Not Sufficient.

Hence (A)

PS: If you liked my solution, kudos are appreciated .

I think this is a simpler problem than some of the previous explanations make it seem!

When a DS question asks you "is the product of these two things positive," that's "GMAT code." What it really means is:

Do x-2 and x-3 have the same sign? (That is, are they both positive or both negative?)

You can go further with decoding this, too. Think: when are x-2 and x-3 both positive? If x is a big number, like 100, they'll definitely both be positive. If x is a medium-sized number, like 10, they'll still both be positive. This will be the case all the way down to x = 3, where the product will now be 0 instead of being positive.

So, they're both positive in the scenario where x > 3.

When are they both negative? If x is a very small number, like -100, they'll both be negative. And if x is a moderately small number, like -10, they'll still both be negative. In fact, they'll both be negative all the way up to x = 2, where the product will now be 0 instead of being negative.

So, they're both negative in the scenario where x < 2.

In other words, the answer to this DS question is "yes" if x is outside of the range from 2 to 3. And if x is in the range of 2 to 3, inclusive, then the answer would be "no."

What we're actually interested in is: is x between 2 and 3, inclusive?

Statement 1: This translates to "x < 2". If x is less than 2, we know it's definitely not between 2 and 3. So, this statement is sufficient.

Statement 2: This translates to "x < 3". If x is less than 3, it could be between 2 and 3 (for instance, it could be 2.5). Or, it could be less than 2, and therefore it would be outside of the range. So, this statement is insufficient.

Since statement 1 is sufficient and statement 2 is insufficient, the answer is A.

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