(d) 4.5 min
Let the obstruction remain for x minutes only
\(\therefore\) Part of cistern filled in x minutes + Part of cistern filled in 3 minutes = cistern filled
\(\Big[\big(\frac{7}{8}\times\frac{X}{12}\big)+\big(\frac{5}{6}\times\frac{X}{16}\big)\Big]\) + \(\Big[\frac{3}{12}+\frac{3}{16}\Big]\) = 1
\(\Rightarrow\) \(\frac{12X}{96}\) + \(\frac{7}{16}\) = 1 \(\Rightarrow\) \(\frac{12X}{96}\) = \(\frac{9}{16}\)
\(\Rightarrow\) x = \(\frac{9\times96}{16\times12}\) = 4.5 min