is x > y?
1) sqrt(x) > y
2) cube(x) > y
Here is what i think:-
1) sqrt(x) > y
Case 1: 0<x<1
Let x = 0.36 and y = 0.50
sqrt(0.36) > 0.50
=> 0.6 > 0.5 But x=0.36 and y = 0.50 --- And 0.36 > 0.50 is false
Case 2: Let x = 16 and y=3
=> and sqrt(16) > 3 -- which is true
So Case 1 is "False" and Case 2 is "true"
Hence Statement 1 is insufficient.
Statement 2:-
cube(x) > y
Case 1:
cube(2) > 7 but 2 > 7 is "false"
Case 2:
I am NOT able to find a value of x such that 0<x<1
Probably because if cube(x) > y then x>1
Is my line of reasoning correct to decide that
Answer is (B)
Is there a way to think about this problem without substituting numnbers?
Thanks a lot.
1) sqrt(x) > y
2) cube(x) > y
Here is what i think:-
1) sqrt(x) > y
Case 1: 0<x<1
Let x = 0.36 and y = 0.50
sqrt(0.36) > 0.50
=> 0.6 > 0.5 But x=0.36 and y = 0.50 --- And 0.36 > 0.50 is false
Case 2: Let x = 16 and y=3
=> and sqrt(16) > 3 -- which is true
So Case 1 is "False" and Case 2 is "true"
Hence Statement 1 is insufficient.
Statement 2:-
cube(x) > y
Case 1:
cube(2) > 7 but 2 > 7 is "false"
Case 2:
I am NOT able to find a value of x such that 0<x<1
Probably because if cube(x) > y then x>1
Is my line of reasoning correct to decide that
Answer is (B)
Is there a way to think about this problem without substituting numnbers?
Thanks a lot.












