Let  $f\left(x\right)=\left\{x\right\}\left[x\right];g\left(x\right)=a{x}^2$ the sum of all real solutions of equation satisfying $f\left(x\right)=g\left(x\right)$ is 420 (where $a$ is (+ve) rational number), then  $a$ is equal to (where $[.]$ and $\{.\}$ represents greatest integer function and fractional part function respectively)

$\left[x\right]\left\{x\right\}=a{x}^2$

Case-I  $\left[x\right].\left\{x\right\}<0$  where  $x<0$ while  $a{x}^2\ge 0\forall x\in R$ and  $x=0$ is one solution but that does not affect the sum of the solution
Thus we need to look at (+ve) solutions for  $n\le x<n+1$
$a{x}^2-nx+n^2=0\Rightarrow x=\frac{n\pm \sqrt{n^2-4a{n}^2}}{2a}=n\left(\frac{1\pm \sqrt{1-4a}}{2a}\right)$

For (+Ve) roots we consider  $x=\frac{n}{2a}(1-\sqrt{1-4a})$
Since , sum of all solutions is 420 
$$\begin{gathered} \Rightarrow \sum \frac{n}{2a}(1-\sqrt{1-4a})=420\Rightarrow \left(\frac{1-\sqrt{1-4a}}{2a}\right)\sum n=420 \\ \Rightarrow \left(\frac{1-\sqrt{1-4a}}{2a}\right)\left(\frac{n(n+1)}{2}\right)=420\Rightarrow \left(\frac{1-\sqrt{1-4a}}{2a}\right)406=420 \end{gathered}$$

(Considering n is 28 I,e   $\frac{n(n+1)}{2}= 406$ at  $n=28$  which is nearer to 420) 

 $\Rightarrow a=\frac{29}{900}$