Question 14

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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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