If 5

Question

If –5 2 ?
 
(1) (k – 1)(k – 5)(k – 2) 0

Mishari · Answer

A ? if it is correct I will explain my answer

Fig · Answer

(C) for me

We have: |k| < 5 k > 2 ? A fast way to conclude is to use a table of signs. From 1 (k – 1)(k – 5)(k – 2) < 0 The table of sign (Fig 1) shows that it is possible if: o k < 1 or o 2 < k < 5 INSUFF. From 2 k - 1 > 0 <=> k > 1 INSUFF. Both (1) and (2) We have: o k < 1 or 2 < k < 5 and o k > 1 We can so conclude that 2 < k < 5. SUFF.

ggarr · Answer

I get C too.

stmnt. 1
In order for stmnt 1 to hold true, the product of (k – 1)(k – 5)(k – 2) must be negative. In order for the prod to be negative the 3 expressions will need to be signed as follows: 
a) -ve * -ve * -ve or 
b) -ve * +ve * +ve
For a or b to occur k cannot equal 1 or 2. this leaves 3,4, and -4 thru -1.   Insufficient  

stmnt. 2
(2) k – 1 > 0 = k > 1
we're asked whether k>2. this tells us that k>1. k can equal anything greater than 1 including 2.
  Insufficient  

stmnts. 1 and 2
2 tells us that k>1. this eliminates -4 thru -1 and leaves 3 and 4.
answer can be either 3 or 4. both are greater than 2. answer C.

bkk145 · Answer

The question is asking, really, that if 2<k<5?

(1) Think of this as a X^3 graph with intersection points 1, 5, and 2. Looking at the graph, we know that the portions of the graph that are less than 0 are when k<1 and 2<k<5. We don't know if 2<k<5 or k<1. INSUFFICIENT

Together, we can eliminate k<1. So we know that the only possible solution here is 2<k<5. SUFFICIENT. 

C is the answer

andrehaui · Answer

for me it is C


1) insuff. k can be either negative or bigger than 2

2) K >1 insuff

get together you eliminate negative possibilty

andrealittrell · Answer

hey, i am just sort of confused...
for the first one we get
K<1
K<5
K<2>1

so i get that neither of those can work alone...
i tried out a few things, and i found the answer to be C... but, i am confused on how to combine the equations...
if we try 2, it wont work into the first equation because 2-2=0.. and 0 is not less than 0.. so we have to start at three...which works.. that is how i got my answer.. but
how would u go about combining the 4 statements quickly??

thanks!!

Fig · Answer

By the table of signs

... U have each line representing a basic expression for which we know the sign of key interval (the ones that make flipping the sign of at least 1 basic expression). then, u multiply the - and + as they are -1 and +1.... If k 0 k > 1.... then bingo, u have only 1 remaining interval

If 5 < k < 5, is k > 2 ? (1) (k 1)(k 5)(k 2) < 0

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