Jambmaths question:
If p varies inversely as the square of q and p = 8 when q = 4, find when p = 32
Option A:
$\pm 16$
Option B:
$\pm 8$
Option C:
$\pm 4$
Option D:
$\pm 2$
Jamb Maths Solution:
$\begin{align} & p\propto \frac{1}{{{q}^{2}}} \\ & p=\frac{k}{{{q}^{2}}} \\ & \text{when }q=4,p=5 \\ & \therefore 8=\frac{k}{{{4}^{2}}} \\ & k=8\times 16=128 \\ & \text{when }p=32,q=? \\ & 32=\frac{128}{{{q}^{2}}} \\ & {{q}^{2}}=\frac{128}{32}=4 \\ & q=\sqrt{4}=\pm 2 \\\end{align}$
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