UPSC Prelims 2001 Q140: Science & Technology – Quantitative Aptitude
A worker reaches his factory 3 minutes late if his speed from his house to the factory is 5
km/hr. If he walks at a speed of 6 km/hr, then he reaches the factory 7 minutes early. The
distance of the factory from his house is
- (a) 4 km
- (b) 5 km
- (c) 6 km
- (d) 7 km
Answer: (b) 5 km
Explanation
Distance of the factory from the worker's house
| Step | Working |
|---|---|
| Given | At 5 km/hr he is 3 minutes late; at 6 km/hr he is 7 minutes early |
| Time difference | 3 + 7 = 10 minutes = 1/6 hour |
| Equation | d/5 − d/6 = 1/6 → d/30 = 1/6 |
| Distance | d = 5 km |
Given: the worker walks from his house to the factory; at 5 km/hr he reaches 3 minutes late and at 6 km/hr 7 minutes early. The two times differ by 10 minutes, so d/5 − d/6 = 1/6 hour, giving d = 5 km.
Answer: (b) 5 km.
Why each option is right or wrong
- (a) Incorrect. 4 km gives a time difference of 8 minutes.
- (b) CORRECT. 5 km gives 60 − 50 = 10 minutes difference.
- (c) Incorrect. 6 km gives a difference of 12 minutes.
- (d) Incorrect. 7 km gives a difference of 14 minutes.
Key point
Shortcut: d = (product of speeds ÷ difference of speeds) × time difference = (30 ÷ 1) × 1/6 = 5 km.
Remember
d/5 − d/6 = 10 min → d = 5 km.
Common mistakes
- Using 7 − 3 = 4 minutes instead of 3 + 7 = 10.
- Forgetting to convert minutes to hours.
- Setting up time = speed ÷ distance.
Exam tip
Late and early add up; late and late subtract.
Sources
- NCERT Class VIII — Mathematics, chapters 'Comparing Quantities' and 'Direct and Inverse Proportions'
- R. S. Aggarwal — Quantitative Aptitude, chapter 'Time and Distance'