Question

x varies directly as y and inversely as the square of z. When $y=4$ and z is 14, $x=10$. If $y=16$ and $z=7$, what is x?

• A. 180
• B. 160
• C. 280
• D. 200

Answer 1

Answer: (B)

160

$x=k.\frac{y}{z^2}⟹10=k.\frac{4}{14\times 4}$ so $k=490$.
Thus $x=490\times \frac{16}{7\times 7}=160$.