Question

$y$ varies directly with square root of $x\text{and}y=9$ when $x=9.$ Find the value of $x$ when $y=6$. [2 Mark]

Answer 1

As per question, $y\propto \sqrt{x}\Rightarrow y=k\sqrt{x}$ .................(1)
( where $k=$ non-zero variation constant)

When $x=9,y=9,\therefore 9=k\sqrt{9}\left[\text{by}\right(1\left)\right]$

Or, $9=k.3\text{or,}k=\frac{9}{3}=3.$

$\therefore$ from (1) we get, $y=3\sqrt{x}$...............(2)

Now, when $y=6,$ from (2) we get, $6=3\sqrt{x}$ [1 Mark]

Or, $\sqrt{x}=\frac{6}{3}=2$

Or, $x=(2{)}^2=4$ [ Squaring both the sides ]

$\therefore$ The required value of $x=4$. [1 Mark]