If |3x| > |4y|, is x > y?
To check whether x > y or not, accept values of x & y for which |3x| > |4y| holds true .
To check whether x > y or not, discard values of x & y for which |3x| > |4y| does not hold true.(1) x > 0
Case-1: x > 0, y >= 0
x > y, x = 2, y = 1 then |3x| > |4y| holds true as |3*2| > |4 *1| i.e. 6 > 4. Accepted.
x > y, x = 2, y = 0 then |3x| > |4y| holds true as |3*2| > |4 *0| i.e. 6 > 0. Accepted.
x < y, x = 1, y = 2 then |3x| > |4y| does not hold true as |3*1| > |4*1| i.e. 3 > 4 is wrong. Discarded.
Hence, Answer to the question "is x > y?" is YES!
case-2: x > 0, y < 0
x > y, x = 2, y = -1 then |3x| > |4y| holds true as |3*2| > |4 * (-1)| = |6| > |-4| = 6 > 4. Accepted.
x > y, x = 2, y = -2 then |3x| > |4y| does not hold true as |3*2| > |4 *(-2)| = |6| > |-8| = 6 > 8 is wrong. Discarded.
Hence, Answer to the question "is x > y?" is YES!
In Both case-1 & 2 We get definite answer "Yes". Statement-1 Sufficient!(2) y > 0
Case-1: y > 0, x >= 0
x > y, x = 2, y = 1 then |3x| > |4y| holds true as |3*2| > |4*1| i.e. 6 > 4. Accepted.
x < y, x = 1, y = 2 then |3x| > |4y| does not hold true as |3*1| > |4*2| i.e. 3 > 8 is wrong. Discarded.
x < y, x = 0, y = 2 then |3x| > |4y| does not hold true as |3*0| > |4*2| i.e. 0 > 8 is wrong. Discarded.
Hence, Answer to the question "is x > y?" is YES!
case-2: y > 0, x < 0
x < y, x = -2, y = 1 then |3x| > |4y| holds true as |3*(-2)| > |4*1| = |-6| > |4| = 6 > 4. Accepted.
Hence, Answer to the question "is x > y?" is NO!
From Case-1 we get "Yes" and from case-2 we get "NO". Answer is not definite. Statement-2 Insufficient!Hence "A" is the final answer!