Question

It is given that $x$ varies directly as $y$ and inversely as the square of $z$. When $x=10$, $y=4$ and $z=14$. Then, when $y=16$ and $z=7$, $x$ equals:

• A. $180$
• B. $160$
• C. $154$
• D. $140$

Answer 1

Answer: (B)

$160$

Given, $x\alpha \frac{y}{z^2}$

$\Rightarrow x=\frac{ky}{z^2}$ where, $k$ is constant of proportionality

$\Rightarrow 10=\frac{k\left(4\right)}{{14}^2}$

$\therefore k=\frac{{10.14}^2}{4}$

$\therefore x=\frac{\frac{{10.14}^2}{4}\times 16}{7^2}=160$