A cruise ship left port $A$ and traveled towards port $B$ $225$ km away. After $1.5$ hours of travel, the cruise ship was stopped for a half an hour and then it had to increase its speed by $10$ km/hour in order to reach port $B$ on time.
Find the original speed of the cruise ship.
Let the original speed be $v$ km/hour.
If it travels for $1.5$ hours, it will have travelled a distance of $1.5v$ km.
Reminaining distance is $225-1.5v$ km.
It increases its speed by $10$ km/hour, so new speed is $v+10$ km/hour.
Time taken to get to final destination is given by $\frac{distance}{speed}=\frac{225-1.5v}{v+10}$ hours.
Total time taken is $1.5 + 0.5 + \frac{225-1.5v}{v+10}$
It's not clear from your question, but I guess that being "on time" means arriving at the same time as the ship would have arrived if it had not had to stop for the half an hour. So travelling at speed $v$, the time taken would have been $\frac{225}{v}$.
That gives you the equation $1.5 + 0.5 + \frac{225-1.5v}{v+10}= \frac{225}{v}$.
Solve that and you will have the original speed.