MichelleSavina wrote:
If k is an integer, what is the value of k?
(1) (k – 1)(k – 3) < 0
(2) (k – 5)(k – 1)(k – 3)(k – 6) < 0
Statement 1: (k – 1)(k – 3) < 0Let's make a few observations.
First, k is an integer.
Notice that k=1 is not a solution to the given inequality, because if we replace k with 1, we get 0<0, which isn't true.
For the same reason, k=3 is not a solution to the given inequality,
What about k=2? We'll plug it in to get:
(k – 1)(k – 3) =(2 – 1)(2 – 3)
Important: We need not evaluate this. All we need to do is determine whether each piece is positive or negative So, (2 – 1)(2 – 3) = (positive)(negative) = (negative) < 0
So, k=2 is one possible value for k.
Now notice what happens when we try ANY number greater than 3 for a possible value of k. To do this, we'll use the following notation: biggerthan3
We get: (k – 1)(k – 3) = (biggerthan3 – 1)(biggerthan3 – 3) = (positive)(positive) = positive
However, the given inequality says: (k – 1)(k – 3) < 0
So, k cannot be greater than 3.
Is it possible for k to be less than 1?
Let's try it.
(k – 1)(k – 3) = (lessthan1 – 1)(lessthan1 – 3) = (negative)(negative) = positive
However, the given inequality says: (k – 1)(k – 3) < 0
So, k cannot be less than 1.
We've now ruled out every possible integer value of k EXCEPT k=2
So, k must equal 2, which means statement 1 IS SUFFICIENT
Statement 2: (k – 5)(k – 1)(k – 3)(k – 6) < 0 In other words, (k – 5)(k – 1)(k – 3)(k – 6) is negative.
We can see that k = 5, 1, 3 and 6 are NOT possible solutions.
What about k=2?
Plug it in to get: (k – 5)(k – 1)(k – 3)(k – 6) = (2 – 5)(2 – 1)(2 – 3)(2 – 6) = (negative)(positive)(negative)(negative) = negative
Great, k=2 is one possible value of k
What about k=4?
Plug it in to get: (k – 5)(k – 1)(k – 3)(k – 6) = (4 – 5)(4 – 1)(4 – 3)(4 – 6) = (negative)(positive)(positive)(negative) = positive
No, k=4 is not a possible value of k
What about k=lessthan1?
Plug it in to get: (lessthan1 – 5)(lessthan1 – 1)(lessthan1 – 3)(lessthan1 – 6) = (negative)(negative)(negative)(negative) = positive
No, k=lessthan1 is not a possible value of k
What about k=greaterthan6?
Plug it in to get: (greaterthan6 – 5)(greaterthan6 – 1)(greaterthan6 – 3)(greaterthan6 – 6) = (positive)(positive)(positive)(positive) = positive
No, k=greaterthan6 is not a possible value of k
We've now ruled out every possible integer value of k EXCEPT k=2
So, k must equal 2, which means statement 2 IS SUFFICIENT
Answer:
_________________
Test confidently with gmatprepnow.com