Mode critical points

For question #1

Statement 1: |x + 3| cannot be negative.
So, 4x - 3 cannot be negative
So, 4x - 3 ≥ 0
So, 4x ≥ 3
So, x ≥ 3/4 > 0

So, statement 1 is sufficient

Statement 2: This means distance between x and 3 is equal to distance between 2x and 3.
So, two obvious solution for x are 0 and 2.
So, x need not be greater than 0.
So, statement 2 is not sufficient

Answer : A

ans are A and C

vipulgoyal wrote:ans are A and C

Question 2:

Statement 1
|x+1| = 2|x-1|
x>1: x+1 = 2x-2 or x=3
-1<x<1: x+1 = 2-2x or x=1/3
x<-1: -x-1 = 2-2x or x=3 or no solution

So, From statment 1, x=3 or x=1/3
NOT SUFFICIENT

Statement 2
|x-3| > 0
This statement is valid all values of x except when x-3 = 0 or x=3
So, NOT SUFFICIENT

Combining both statements, we can eliminate x=3 in statement 1 by using statement 2.
So, answer = 1/3 (BOTH STATEMENTS USED TOGETHER)

C