Question 16

loading...

Prev/Next links

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: 

$\begin{align}  & p\propto \frac{1}{{{q}^{2}}} \\ & p=\frac{k}{{{q}^{2}}} \\ & \text{when }q=4,p=5 \\ & \therefore 8=\frac{k}{{{4}^{2}}} \\ & k=8\times 16=128 \\ & \text{when }p=32,q=? \\ & 32=\frac{128}{{{q}^{2}}} \\ & {{q}^{2}}=\frac{128}{32}=4 \\ & q=\sqrt{4}=\pm 2 \\\end{align}$

Jamb Maths Topic: 
Year of Exam: