A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?

- 22
- 20
- 15
- 23

It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am

Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am

If they leave one by one every minute, the 45^{th} motorcyclist would have left by 10:45 am to reach B at 11:00 am.

Thus, time taken by one motorcyclist to reach B from A = 15 minutes.

Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am

So, the last motorcyclist should have left A by 10:15 am

Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B