Jambmaths question:
x varies directly as y2 and x = 4, when y = 6. Find the value of y when x =16.
Option A:
12
Option B:
$\tfrac{2}{3}$
Option C:
36
Option D:
48
Jamb Maths Solution:
$\begin{align} & x\propto {{y}^{2}} \\ & x=k{{y}^{2}}\text{ (}k\text{ is the constant)} \\ & \text{when }y=6,\text{ }x=4 \\ & 4=k{{(6)}^{2}} \\ & k=\frac{4}{36}=\frac{1}{9} \\ & \text{When }x=16,\text{ then }y \\ & 16=\frac{1}{9}{{y}^{2}} \\ & {{y}^{2}}=16\times 9 \\ & y=\sqrt{16\times 9}=12 \\\end{align}$
Jamb Maths Topic:
Year of Exam: