UPSC Prelims 2005 Q98: Science & Technology – Quantitative Aptitude
Aryan runs at a speed of 40 metre/minute. Rahul follows him after an interval of 5 minutes
and runs at a speed of 50 metre/minute. Rahul’s dog runs at a speed of 60 metre/minute and
starts along with Rahul. The dog reaches Aryan and then comes back to Rahul, and continues to
do so till Rahul reaches Aryan. What is the total distance covered by the dog?
- (a) 600 metres
- (b) 750 metres
- (c) 980 metres
- (d) 1200 metres
Answer: (d) 1200 metres
Explanation
Dog running between Aryan and Rahul
| Quantity | Value |
|---|---|
| Aryan's head start | 5 min x 40 m/min = 200 m |
| Rahul gains on Aryan | 50 - 40 = 10 m/min |
| Time to catch Aryan | 200 / 10 = 20 min |
| Dog's distance | 60 m/min x 20 min = 1200 m |
The dog runs for exactly as long as Rahul takes to catch Aryan, which is 20 minutes, so it covers 1200 m.
Answer: (d) 1200 metres.
Why each option is right or wrong
- (a) Incorrect. 600 m is the dog's distance in 10 minutes only.
- (b) Incorrect. 750 m does not match the 20 minute time.
- (c) Incorrect. 980 m is not 60 times 20.
- (d) CORRECT. 60 x 20 = 1200 m in 20 minutes.
Key point
Distance of the dog = speed of the dog x time for Rahul to catch Aryan.
Remember
Rahul catches Aryan in 20 min, so the dog runs 1200 m.
Common mistakes
- Summing each back and forth trip instead of using total time.
- Forgetting Aryan's 5 minute head start.
- Using the dog's speed with the wrong time.
Exam tip
Back and forth problems: find total time, multiply by constant speed.
Sources
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Time and Distance)
- Arun Sharma — How to Prepare for Quantitative Aptitude for the CAT (Time, Speed and Distance)