A bag contains $\left(2n+1\right)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$ then $n$=

Given that the bag contains $\left(2n+1\right)$ coins, in which $n$ coins are biased and these coins has both sides head, and remaining coins are fair

Hence, the probability of getting head is 

   $\begin{array}{rcl}p& =& \frac{\left(n\right)·1+\left(n+1\right)·{\displaystyle \frac{1}{2}}}{2n+1}\\ & =& \frac{2n+n+1}{2\left(2n+1\right)}\\ & =& \frac{3n+1}{4n+2}\end{array}$

This is given as $\frac{31}{42}$

hence, 

$\frac{3n+1}{4n+2}=\frac{31}{42}\Rightarrow \frac{3n+1}{2n+1}=\frac{31}{21}$

Cross multiply 

$\begin{array}{rcl}63n+21& =& 62n+31\\ n& =& 10\end{array}$

Therefore, the value of $n$ is $\boxed{10}$