arora.sumit88 wrote:
My take on this is:
Statement 1: |x+1| = 2|x-1|
Can be broken down into two cases:
a. x + 1 = 2(x-1)
b. x + 1 = -2(x-1)
solving a:
x + 1= 2x - 2
-x = -3
x = 3
solving b:
x + 1 = -2x +2
3x =1
x = 1/3
|X|=|3|<1..No
|1/3|<1.. Yes
1. Not sufficient
2. |X-3|>0
x - 3 >0 or x-3<0
Again |X|<1 can be yes or no while testing different values.
Not sufficient.
Combining 1 and 2
We get |X|=1/3 <1 Yes. So Ans is "C"..
Hello,
For statement 1 , i got X = 1/3 or X = 3 ==> not sufficient
For Statement 2, i got X < 3 or X > 3 , Can you provide number examples here?
For X>3, i take 4 , hence |4| < 1 , NO
For X<3 i take -2, hence |2| < 1 , NO... Isnt this sufficient?
Should i also consider 0 for X< 3 , which would be |0| < 1, Yes.. hence insufficient ??
How do igo forward for statement 2?
And how does Statement (1)+ (2) give x = 1/3 ??
Regards,
Mustu