Let x and y be the speed of river and boat respectively and D be the distance. Then ATQ,2D=15×(y−x) 2D=15×y{x=0}⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(i) and 16=
D
x+y
+
D
y−x
⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(ii) Also y−x=15⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(iii) (given) Solving eqn's (1),(2) and (3) we get y2−32y+240=0 ⇒(y−20)(y−12)=0 ⇒y=12,20 From eqn (3)y≥15∴y=20 ⇒x=20−15=5km/hr