Leakage rate= \(2.25\times\frac{60}{5.5}=\frac{135}{5.5}=\frac{270}{11}\) tonnes/hr and pumping rate= 12 tonnes/hr
Time taken by the ship to accommodate 92 tonnes= \(\frac{92}{\frac{270}{11}-12}\)
⇒\(92\times\frac{11}{138}=\frac{22}{3}\space hr.\)
⇒Average rate of sailing = \(\frac{distance}{time}\)
⇒\(\frac{77}{\frac{22}{3}}\)\(=\frac{77\times3}{22}\)\(=10.5 \space km/hr\)
The correct option is (D): 10.5 km/h