Kritisood wrote:
Is x > 3?
(1) (x - 3)(x - 2)(x - 1) > 0
(2) x > 1
Thank you for sharing,
Kritisood. A not-so-rigorous approach, often my favorite type for these types of problems, can yield a definitive answer in under a minute. Looking at the question, we know nothing about
x in the way of
number versus
integer, an important distinction that will come later. Because I am kind of lazy, I eyeballed the two statements and decided to start with the easier-looking one, Statement (2):
2) x>1
This is quite obviously NOT sufficient on its own. The value of
x could be 1.1, 1.5, 20, or anything else, as long as it was greater than 1. Mark off (B) and (D). Looking at Statement (1),
1) (x-3)(x-2)(x-1)>0
we can think of only two scenarios in which the binomial product would have to make the inequality true:
1) positive * positive * positive > 0
2) positive * negative * negative > 0
As long as we could derive either three positive values or one positive and two negative values, we would have to get a positive product. We can focus on the question, the value of
x relative to 3, later. Statement (1) tells us that
x cannot equal any of 3, 2, or 1, since the product would end up being 0 itself. If we test whole numbers, we are clearly going to have to kick up to at least 4, as in
(4 - 3)(4 - 2)(4 - 1) > 0
(1)(2)(3) > 0
6 > 0
This is clearly true. But now let the question enter into the picture. What if, because the question stem did not specify that
x is an
integer, we tried a decimal instead, one that was
less than 3? To keep it simple, try 1.5:
(1.5 - 3)(1.5 - 2)(1.5 - 1) > 0
(-1.5)(-0.5)(0.5) > 0
[something positive](0.5) > 0
This is true. As long as we get a positive product between the two negative values, we need not be concerned with the exact value of 1.5 * 0.5 (even if we might be screaming 0.75 in the back of our minds). We know now that with valid inputs of either 1.5 or 4, Statement (1) is also NOT sufficient.
Even with the statements together, I have no way to bar my earlier test values of 1.5 or 4. 1.5 is greater than 1, satisfying Statement (2), but so is 4. Thus, with
x values that can be greater than or less than 3 that create valid inequalities in both cases, we can conclude that
(E) is the correct answer. It might seem like a lot when I type everything out, but this approach took no pencil or paper and nothing more than a careful interpretation of what was given and a fundamental knowledge of how negatives and positives interact.
- Andrew
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