inequalities

[AJWILL]

Q) 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

[spoiler]whats the logic behind analysing statement II?? its very time consuming to c all possible ranges[/spoiler]

[theCEO]

1.The first thing to do is to list all the terms that are to be subtracted from k
5 , 1, 3 , 6

2. Add 1 to each number to creater a new list
6, 2, 4, 7

3. Subtact numbers in the 2nd list that overlap with the 1st list. Subtact numbers in the 2nd list that are greater than the highest number in the first list
2, 4

Substitue 2 and 4 for k:

for a x b x c x d < 0; 1 or 3 of the terms have to be negative
Only the number 2 satifies this equation

[Brent@GMATPrepNow]

Q) 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

This is a nice question.

Statement 1: (k - 1)(k - 3) < 0
Let'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 = D

Cheers,
Brent

[pemdas]

Q) 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

[spoiler]whats the logic behind analysing statement II?? its very time consuming to c all possible ranges[/spoiler]

I would suggest to use graphs here (to save time and shoot onto eye )
st(1) two critical points here 1 and 3. the function will increase, decrease, increase (below I will plot some lines 4 u to understand that better). We need the interval where function decreases
___+_______-________+__
_______1________3______

1<k<3 will be the right set and we get k{integer}=2 Sufficient

st(2) four critical points here 1,3,5,6 (I will draw again though felt lazy first)
_+_____-______+_____-_____+_
____1______3_____5______6___

as you see we have 2 intervals suitable for the function decrease, hence we may get 1<k<3 and 5<k<6. Considering k is integer we get only k=2 as interval 5<k<6 is devoid of any integer value. Sufficient

answer: d