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CAT 2019 Slot 2 — QA Question 12

Relative SpeedEasy

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?

Answer & solution

  • A

    22

  • B

    20

  • 15

  • D

    23

Solution

Easy

First find how long a motorcycle takes A→B from the fact that 4545 of them (leaving at 10:01, 10:02, …) all arrive by 11:00. Then the faster cyclist reaches B at 10:30, so count every motorcycle that has already arrived by then.

1

Motorcycle travel time. The 4545th motorcycle is the last to arrive by 11:00. Motorcycles leave at 10:01, 10:02, …, so the 4545th leaves at 10:45 and arrives at 11:00.

Travel time=11:00−10:45=15 min\begin{aligned} &\text{Travel time}=11{:}00-10{:}45=15\ \text{min} \end{aligned}
2

New arrival time of the cyclist. The original trip 10:00→11:00 takes 6060 min; doubling the speed halves it.

New time=30 min ⇒ cyclist reaches B at 10:30\begin{aligned} &\text{New time}=30\ \text{min}\ \Rightarrow\ \text{cyclist reaches B at }10{:}30 \end{aligned}
3

Count motorcycles arriving by 10:30. A motorcycle takes 1515 min (step 1), so to arrive by 10:30 it must leave by 10:15. Departures are 10:01 through 10:15.

From 10:01 to 10:15=15 motorcycles\begin{aligned} &\text{From 10:01 to 10:15}=15\ \text{motorcycles} \end{aligned}
15 motorcycles15\ \text{motorcycles}
CAT 2019 Slot 2 QA Q12: A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A — Solution | TheCATExam