If $n\left(U\right)=60,n\left(A\right)=35,n\left(B\right)=24$ and $n\left(AUB\right)\text{'}=10$ then $n(A\cap B)$ is

Given that $n\left(A\right)=35,n\left(B\right)=24,n\left(U\right)=60$ 

and $n\left({(A\cup B)}^1\right)=10$

$\begin{array}{rcl}n\left({(A\cup B)}^1\right)& =& n\left(U\right)-n(A\cup B)\\ 10& =& 60-n(A\cup B)\\ n(A\cup B)& =& 50\\ & & \end{array}$

Use the addition theorem on cardinal number of sets 

$\begin{array}{rcl}n\left(A\cap B\right)& =& n\left(A\right)+n\left(B\right)-n\left(A\cup B\right)\\ & =& 35+24-50\\ & =& 59-50\\ & =& 9\end{array}$

Therefore, $n\left(A\cap B\right)=\boxed{9}$