Jambmaths question:
If x varies directly as the square of P and inversely as Q, and x = 25 when P = 4 and Q = 2, find x when P = 6 and Q =3
Option A:
36.0
Option B:
24.5
Option C:
37.0
Option D:
37.5
Jamb Maths Solution:
$\begin{align} & x\propto \frac{{{P}^{2}}}{Q} \\ & x=\frac{k{{P}^{2}}}{Q}\text{ (}k=constant) \\ & \text{When }P=4,Q=2,\text{ }x=25 \\ & 25=\frac{k({{4}^{2}})}{2} \\ & k=\frac{25}{8} \\ & \text{When }P=6,\text{ }Q=3\text{ then } \\ & x=\frac{(\tfrac{25}{8})({{6}^{2}})}{3}=\frac{25\times 36}{3\times 8} \\ & x=\frac{25\times 3}{2}=\frac{75}{2}=37.5 \\\end{align}$
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