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

Let the distance between A and B be $D$. The cyclist takes 60 minutes (from 10:00 am to 11:00 am) to travel from A to B. Let the speed of the cyclist be $v_c=\frac{D}{60}$ per minute.

The motorcycles leave A starting at 10:01 am, 10:02 am,..., 10:$n$ am. Let the travel time for each motorcycle be $T$ minutes. Since the 45th motorcycle reaches B at or before 11:00 am, and motorcycles start every minute, the 45th motorcycle starts at 10:45 am. Thus, it must reach B at or before 11:00 am:
$10:45+T\le 11:00$ which means $T\le 15$ minutes.

Since 45 motorcycles reach B by 11:00 am, and the 46th (starting at 10:46 am) does not reach B by 11:00 am, we have:
$10:46+T>11:00$ which means $T>14$ minutes. Therefore, $T=15$ minutes.

If the cyclist doubles his speed, his travel time is halved to 30 minutes. Thus, the cyclist starts at 10:00 am and reaches B at 10:30 am.

Any motorcycle that reaches B by 10:30 am must have left A at or before:
$10:30-T=10:30-15=10:15\text{am}$

Since motorcycles start at 10:01 am, 10:02 am,..., up to 10:15 am, the total number of motorcycles that reach B by 10:30 am is 15.