Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = (frac{1}{3}).
-
A.
81 -
B.
27 -
C.
(frac{1}{9}) -
D.
(frac{1}{27}) -
E.
(frac{1}{81})
Correct Answer: Option A
Explanation
(x propto frac{1}{y})
(x = frac{k}{y})
(y propto t^{2})
(y = ct^{2})
k and c are constants.
(x = frac{k}{ct^{2}})
Let (frac{k}{c} = d) (a constant)
(x = frac{d}{t^{2}})
(1 = frac{d}{3^{2}} implies d = 9)
(therefore x = frac{9}{t^{2}})
(x = 9 div (frac{1}{3})^{2} )
= ( 9 div frac{1}{9} = 9 times 9 = 81)