Question 5

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Jambmaths question: 

x varies directly as y2 and x = 4, when y = 6. Find the value of y when x =16.

Option A: 

12

Option B: 

$\tfrac{2}{3}$

Option C: 

36

Option D: 

48

Jamb Maths Solution: 

$\begin{align}  & x\propto {{y}^{2}} \\ & x=k{{y}^{2}}\text{ (}k\text{ is the constant)} \\ & \text{when }y=6,\text{ }x=4 \\ & 4=k{{(6)}^{2}} \\ & k=\frac{4}{36}=\frac{1}{9} \\ & \text{When }x=16,\text{ then }y \\ & 16=\frac{1}{9}{{y}^{2}} \\ & {{y}^{2}}=16\times 9 \\ & y=\sqrt{16\times 9}=12 \\\end{align}$

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