Jambmaths question:
If x varies directly as square root of y and x = 81 when y =9, find x when y = $1\tfrac{7}{9}$
Option A:
$2\tfrac{1}{4}$
Option B:
36
Option C:
$20\tfrac{1}{4}$
Jamb Maths Solution:
$\begin{align} & x\propto \sqrt{y} \\ & x=k\sqrt{y}=k{{y}^{\tfrac{1}{2}}} \\ & \text{when }x=81,y=9 \\ & 81=k({{9}^{\tfrac{1}{2}}}) \\ & 81=3k \\ & k=27 \\ & \text{when }y=1\tfrac{7}{9}=\tfrac{16}{9} \\ & x=k{{y}^{\tfrac{1}{2}}}=27\cdot {{(\tfrac{16}{9})}^{\tfrac{1}{2}}} \\ & x=27\cdot (\tfrac{4}{3})=36 \\\end{align}$
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