Question

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?

• A. 23
• B. 20
• C. 15
• D. 22

Answer 1

The Correct Option is C

The problem involves calculating how many motorcycles would reach B if the cyclist doubled his speed. Let's break down the solution:

1. The cyclist takes 1 hour to travel from A to B at his original speed. Therefore, the cyclist would take half an hour (30 minutes) if his speed is doubled. 
2. From 10:01 am onwards, motorcycles leave A for B every minute.
3. Let's calculate how many motorcycles would have left by the time the cyclist reaches B in 30 minutes, i.e., by 10:30 am.
4. Between 10:01 am and 10:30 am, the motorcycles leave each minute. Hence, a total of 30 minutes pass in this window.
5. Thus, 30 motorcycles leave from 10:01 am to 10:30 am.
6. However, the problem states that 45 motorcycles reach B by 11 am when the cyclist originally took 1 hour. Doubling the cyclist's speed means calculating motorcycles that reach by 10:30 am when the cyclist doubles his speed.
7. Since it takes the motorcycles 1 hour to reach B (from 10:01 am to 11:00 am originally), the 30 motorcycles that left between 10:01 am to 10:30 am will just be reaching B by 11 am. Halfway through, for 30 minutes, no motorcycles would reach B because they still need a full hour to reach it.
8. Therefore, only those motorcycles that left in the previous cycle, i.e., from 10:31 am up to 11:00 am minus 15 motorcycles (since 15 motorcycles would arrive between 10:30 am to 11:00 am), would have reached earlier by 11:00 am.
9. Thus, 45 - 30 = 15 motorcycles would have reached by 10:30 am when the cyclist's speed is doubled.

The correct answer is: 15