Question

Suppose $y$ varies jointly as $x$ and $z$.

How do you find $y$ when $x=6$ and $z=8$, if $y=6$ when $x$ is $4$ and $z$ is $2$?

Answer 1

Step-1: To determine constant :

It is given that $y$ varies jointly as $x$ and $z$.

So, $y$ can be written as , $y$=$kxz$ $········\left(1\right)$, where $k$ is constant.

When $y$=$6$ , $x=4$ and $z=2$ , substituting in $\left(1\right)$.

$$\begin{gathered} 6=k\times 4\times 2 \\ \Rightarrow 6=8k \\ \Rightarrow k=\frac{6}{8} \\ \therefore k=\frac{3}{4} \end{gathered}$$

Step-2: To find $\mathit{y}$

Substituting $x=6,z=8$ and $k=\frac{3}{4}$ (constant remains same ) in $\left(1\right)$ to find $y$.

$$\begin{gathered} y=\frac{3}{4}\times 6\times 8 \\ \Rightarrow y=3\times 6\times 2 \\ \therefore y=36 \end{gathered}$$

Hence , the value of $\mathit{y}$ is $\mathbf{36}$.