Jambmaths question:
If y varies directly as $\sqrt{n}$and y =4 when n =4, find y when $n=1\tfrac{7}{9}$
Option A:
$\sqrt{17}$
Option B:
$\tfrac{4}{3}$
Option C:
$\tfrac{8}{3}$
Option D:
$\tfrac{2}{3}$
Jamb Maths Solution:
$\begin{align} & y\propto \sqrt{n} \\ & y=k\sqrt{n} \\ & \text{when }n=4,\text{ }y=4 \\ & 4=k\sqrt{4} \\ & 4=\pm 2k\text{ } \\ & Note:\text{For Jamb exam level, consider the positive value} \\ & 4=2k \\ & k=2 \\ & \text{When }n=1\tfrac{7}{9}=\tfrac{16}{9} \\ & y=2\sqrt{\frac{16}{9}} \\ & y=2\times \frac{4}{3}=\frac{8}{3} \\\end{align}$
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