A) 56.25 km/h
B) 60 km/h
C) 66.50 km/h
D) 67 km/h
Correct Answer: B
$ \therefore $ Speed in onward journey $ =\frac{125}{100}x=( \frac{5}{4}x )km/h $ Average speed $ =\frac{2\times \frac{5}{4}x\times x}{\frac{5}{4}x+x}=\frac{10x}{9}km/h $
$ \Rightarrow $ $ 1600\times \frac{9}{10x}=30 $ [since, train halts for 2 h]
$ \therefore $ $ x=\frac{1600\times 9}{30\times 10}=48km/h $
$ \therefore $ Speed in onward journey $ =\frac{5}{4}x=\frac{5}{4}\times 48=60km/h $