Question

If $z$ varies jointly as $x$ and the square root of $y$, and if $z=6$ when $x=3$ and $y=16$, find $z$ when $x=7$ and $y=4$

Answer 1

It is given that $z$ varies jointly as $x$ and the square root of $y$ if $z=6$ when $x=3$ and $y=16$, therefore,

$z=kx\sqrt{y}\Rightarrow 6=k\times 3\times \sqrt{16}\Rightarrow 6=k\times 3\times 4\Rightarrow 12k=6\Rightarrow k=\frac{6}{12}=\frac{1}{2}$

Thus, the general equation is $z=\frac{1}{2}x\sqrt{y}$.

Since $x=7$ and $y=4$, therefore,

$z=\frac{1}{2}x\sqrt{y}=\frac{1}{2}\times 7\times \sqrt{4}=\frac{1}{2}\times 7\times 2=7$

Hence, $z=7$ cm when $x=7$ and $y=4$.