In the figure given above $AB \parallel CD \parallel EF \parallel $ and $PQ \parallel RS$. $\angle RQD=25^{\circ}$ and $\angle CQP=60^{\circ}$ then $\angle QRS$ is equal to

Question

$\begin{array}{1 1} (a)\;85^{\circ} \\ (b)\;135^{\circ} \\ (c)\;145^{\circ} \\(d)\;110^{\circ} \end{array} $