(I) x > y^2Akansha wrote:If x > y2 > z4, which of the following could be true?
x > y > z
z > y > x
x > z > y
a. I
b. I and II
c. I and III
d. II and III
e. I, II, and III
OA is E
Is hit & trial method the only way to solve this problem? I find it time consuming trying out with different numbers.
- Certainly for integer values of x and y, x is always greater than y. But what about fractional values> f y is a fraction then y^2 is < y. Now always there is a fraction x, such that y > x > y^2. Thus for fractional values x may be less than y. Therefore regarding the relation between x and y all of the options could be true.
- Same logic as above.
- This directly implies y > z^2. Now again apply same logic as above.
