Question 31

Jambmaths question: 

If p varies inversely as the square root of q, where p = 3 and q = 16, find the value of q when p = 4.

Option A: 

24

Option B: 

12

Option C: 

25

Option D: 

16

Jamb Maths Solution: 

$\begin{align}  & p\propto \frac{1}{{{q}^{2}}} \\ & p=\frac{k}{{{q}^{2}}} \\ & 9=\frac{k}{{{16}^{2}}} \\ & k=9\times {{16}^{2}}=2304 \\ & \text{when }p=4 \\ & 4=\frac{768}{{{q}^{2}}} \\ & {{q}^{2}}=\frac{2304}{4}=576 \\ & q=\sqrt{576}=24 \\\end{align}$

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