Madhavi1990 wrote:
so answer is A was my first thought.
|x-6| = 3
so obv x = 9 or - 3
1) x dvble 9 : so I substituted
|9-6| = 3; also |27/9 - 6| as both are dvble by 9. so X i thought could be either 27 or 9 . Insuff
2) x dvble by 3.
|x-6| =3
I put x as 9 and 18 --> |18/3 - 6| and |9/3 - 6| . so x again had two values
Answer is A. can you explain what is wrongabove?
Also, based on reading the explanations above, it appears we consider x as dvble by something when it leaves behind a remainder one (that is why 9/3 = 3 and 3/3 = 1 hence insufficient; while 9/9 = 1 was seen as sufficient. But why can't we consider 27/9 here?)Would appreciate any reply to this!
...








