Question

If f varies jointly as q^2 and h, and f=96 when q=4 and h=3, find q when f=48 and h=6.​

Answer 1

If f varies jointly as q^2 and h, and f=96 when q=4 and h=3 then the value of q when f = 48 and h = 6 is 2

Solution:

Given that, f varies jointly as q^2 and h

`\begin{array}{l}{\text { Then, } f \alpha q^(2) * h} \\ {\rightarrow f=k * q^(2) * h \text { where } k \text { is proportionality constant. }}\end{array}`

And f=96 when q=4 and h=3

Now substitute f, q, h values in above formula

`\begin{array}{l}{\rightarrow 96=\mathrm{k} * 4^(2) * 3} \\\\ {\rightarrow 32=\mathrm{k} * 16} \\\\ {\rightarrow \mathrm{k}=2}\end{array}`

`\text { Then, formula is } f=2 * q^(2) * h`

We have to find q when f=48 and h=6

`\begin{array}{l}{\text { So, } 48=2 * q^(2) * 6} \\\\ {\rightarrow 48=q^(2) * 12} \\\\ {\rightarrow q^(2)=4} \\\\ {\rightarrow q=2}\end{array}`

Hence, the value of q is 2