If k is an integer, what is the value of k?

If k is an integer, what is the value of k?

(1) (k – 1)(k – 3) < 0 --> 1<k<3, as given that k is an integer then k=2. Sufficient.

(2) (k – 5)(k – 1)(k – 3)(k – 6) < 0 --> 5<k<6 or 1<k<3, again as given that k is an integer then k=2 (only integer IN this ranges). Sufficient.

Answer: D.

Check for more here: x2-4x-94661.html#p731476 or here: inequalities-trick-91482.html
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Hi Bunuel,

I represented these points on a graph and drew the crests and troughs to check which ranges we need to consider the point lies.
I was confused if 3<x<5 will also be considered.
Can you please share how you quickly figured out the appropriate ranges here.


Sorry if this is a silly one

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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I'm a little confused. How did k=1 & k=3 < 0 become 1<k<3? Is the <0 just there for me to set up the inequality? Thanks.

We don't have that k=1 or k=3, we have an inequality which says that (k – 1)(k – 3) < 0. This inequality means that 1 < k < 3. I think that you should brush-up fundamentals on inequalities. Please go through the following topics:

Theory on Inequalities:
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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I N T E G E R ! ! ! ! !

when the question state it clearly - i a real number
when it doesnt, i consider integers.

hope this little steam release will help me notice that in the future.

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

K is between 1& 3 only integer is 2.

(k – 5)(k – 1)(k – 3)(k – 6) < 0
so K = between 1 &3 or 5&6

Only integer 2. so D

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:

Statement 1 k cannot equal 1 or 3 since it make the equation =0 and we know its <0 and cannot be "-" or else the equation>0 so the only number that works here is sufficient
Statement 2 k cannot be 5,3,6, or 1. and cannot be negative. only works with k=2

D

st1: between 1 and 3 there is only on integer (2). Testing with k(2) < 0
st2: k=5, k=1, k=3, k=6. Between those numbers only integers are 2 and 4. ------1----(2)----3----(4)----5----6.
As this equation starts with k^4 (fourth order) and starts with +k curve would end positive, so:
-positive-1-negative-3--positive--5--negative--6--positive--.
Only negative integer possible is -2

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

Explanation:

k is given as an integer.

Considering statement 1 :
(k − 1)(k − 3) < 0

Product of 2 numbers is negative if one of the numbers is negative and other one is positive. Hence, we have 2 cases:

Case1: k − 1 is positive and k − 3 is negative
⇒ k − 1 > 0 and k − 3 < 0
⇒ k > 1 and k < 3
Or 1 < k < 3
Since k is an integer, the only value possible for k is 2.

Case 2: k − 1 is negative and k − 3 is positive
⇒ k − 1 < 0 and k − 3 > 0
⇒ k < 1 and k > 3
What above statement implies is that k must be less than 1 and at the same time k must be greater than 3, which is not possible. Hence, we get no value of k from case 2.

⇒ k = 2 is the only value possible for statement 1 to hold true.

Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.

Considering statement 2:
(k − 5)(k − 1)(k − 3)(k − 6) < 0

Applying number line concept: Mark the critical points 1, 3, 5 and 6 on the number line dividing it into 5 intervals: k > 6, 5 < k < 6, 3 < k < 5, 1 < k < 3, k < 1.



Marking the intervals alternatively as positive and negative, starting from the left gives:
(k − 5)(k − 1)(k − 3)(k − 6) < 0 (It can be verified by picking some value of k in each interval and checking the value of the expression is positive or negative.)
⇒ 1 < k < 3 or 5 < k < 6

k is given to be an integer. Between 5 and 6 no integer is there, so we get only integer possibility for k, i.e. k = 2

Since we are getting a definite answer from above statement , statement 2 also itself is sufficient to provide the answer.

Statement 1 and 2 both individually are sufficient to provide the answer.

Answer: D.

I think, I deseve a kudus :D Lol

If k is an integer, what is the value of k?

(1) (k − 1)(k − 3) < 0
So both k-1 and k-3 should have opposite sign...

a) above 3, both will be positive, so (k-1)(k-3)>0
b) below 1, both will be negative, so (k-1)(k-3)>0
c) At 1 and 3, (k-1)(k-3)=0
d) 1<k<3, (k-1)(k-3)<0

k can have only one integer value between 1 and 3 and that is 2
So Ans 2
Sufficient

(2) (k − 5)(k − 1)(k − 3)(k − 6) < 0
Either three or just one of (k − 5),(k − 1),(k − 3)and (k − 6) should be of same si

a) less than 1 all will be negative so (k − 5)(k − 1)(k − 3)(k − 6)>0
b) more than 6 all will be positive, so (k − 5)(k − 1)(k − 3)(k − 6)>0
c) At 1,3,5 and 6 (k − 5)(k − 1)(k − 3)(k − 6)=0
d) possible values left are 2 and 4
e) At 2, (k-1) will be positive and other three negative so (k − 5)(k − 1)(k − 3)(k − 6)<0
f) At 4, k-1 and k-3 will be positive and k-5and k-6 will be negative. -*-*+*+=+, so (k − 5)(k − 1)(k − 3)(k − 6)>0

Only possible value k=2
Sufficient

D
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Hi,
Thanks for the explanation , just to ask, is the below mentioned a right approach to find the range of K in Statement-2 OR any other easy approach is there?
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

Regards

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