I.$$ {15\over \sqrt {x}}-{2\over \sqrt {x}}=6\sqrt {x}$$
II.$$ {\sqrt {y}\over 4}-{7\sqrt {y}\over \sqrt {12}}={1\over \sqrt {y}}$$
Give answer
(A) x>y
(B) x≥y
(C) x<y
(D) x≤y
(E) x=y or the relationship cannot be established
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Direction (66-70): What should come in place of the question mark(?) in the following number series?
8, 39, 155, 464, ?
1735 05da05c7b00e22578c9e9ab50In the following questions two equations numbered I and II are given. You have to solve both the equations and ____
(A) x>y
(B) x≥y
(C) x<y
(D) x≤y
(E) x=y or the relationship cannot be established
I.\(20x^2 - x - 12 = 0\)
II.\(20y^2 + 27y + 9 = 0\)