Time, Speed & Distance PYQs for UPSC (2011-2024) | Solved Questions & PDF Download

Master Time, Speed & Distance problems for UPSC with solved Previous Year Questions (PYQs) from 2011 to 2024. This comprehensive guide provides step-by-step solutions, shortcut formulas, and a free downloadable PDF to help you crack all types of TSD problems in UPSC CSAT.

Why Time, Speed & Distance Matters in UPSC CSAT?

TSD questions are consistently asked in UPSC CSAT (Paper II) to test:

· Your ability to calculate motion-based problems

· Understanding of relative speed concepts

· Skills in solving real-world scenario-based problems

This page covers all TSD PYQs with smart calculation techniques.

Year-Wise TSD PYQs (2011-2024)

Q1. If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then what is the ratio of the distances travelled by them in one hour? [CSAT 2011]

(a) 8:5

(b) 5:8

(d) 1:2

Solution:

Given that,

Bus travels 160 km in 4 hours

train travels 320 km in 5 hours at uniform speed

Now,

Distance travelled by bus in an hour = 160/4 = 40km

Distance travelled by train in an hour = 320/5 = 64km

Thus, the ratio = 40 : 64 = 5 : 8

Hence option (b) is correct

Q2. Mr. Kumar drives to work at an average speed of 48 km/h. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. How far is his office? [CSAT 2012]

(a) 30 km

(b) 40 km

(d) 48 km

Solution:

Given that,

Mr. Kumar drives to work at an average speed of 48 km/h.

The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance.

Now,

The ratio of time = 60t : 40t = 3 : 2

So,

3t - 2t = 10 min

t = 10 min

Total time taken = 3t + 2t = 5t = 5 x 10 = 50 min

Distance = speed x time = 48 x (50/60) = 40 km

Hence option (b) is correct

Q3. A person can walk a certain distance and drive back in six hours. He can also walk both ways in 10 hours. How much time will he take to drive both ways?[CSAT 2013]

(a) Two hours

(b) Two and a half hours

(d) Four hours

Solution:

Given that,

A person can walk a certain distance and drive back in six hours.

He can also walk both ways in 10 hours.

Now,

Let walk be W and drive be D

W + D = 6 hr

2W = 10 hr or W = 5 hr

So, D = 6 - 5 = 1hr

Time taken to drive both ways = 2 x D = 2 x 1 = 2hrs

Hence option (a) is correct

Q4. A thief running at 8 km/hr is chased by a policeman who speed is 10 km/hr. If the thief is 100 m ahead of the policeman, then the time required for the policemen to catch the thief will be [CSAT 2013]

(a) 2 minutes

(b) 3 minutes

(d) 6 minutes

Solution:

Given that,

A thief running at 8 km/hr is chased by a policeman who speed is 10 km/hr.

The thief is 100 m ahead of the policeman

Now,

Speed of thief = 8 km/hr

Speed of Policemen = 10 km/hr

Relative speed = 10 - 8 = 2 km/hr = (5/9) m/sec

Distance = 100 m

Time = 100/(5/9) = 180 sec or 3 min

Hence option (b) is correct

Q5. Two cars start towards each other, from two places A and B which are at a distance of 160 km. They start at the same time 08:10 am. If the speeds of the cars are 50 km and 30 km/h respectively, they will meet each other at [CSAT 2014]

(a) 10: 10 am

(b) 10:30 am

(d) 11:20 am

Solution:

Given that,

Two cars start towards each other, from two places A and B which are at a distance of 160 km.

They start at the same time 08:10 am.

The speeds of the cars are 50 km and 30 km/h respectively.

Now,

Relative speed = 50 + 30 = 80 km/hr (Two objects in opposite direction, Relative speed = U+V km/hr )

Distance = 160 km

Time taken = 160/80 = 2hrs

08:10 am + 2 hr = 10 : 10 am

Hence option (a) is correct

Q6. 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 [CSAT 2014]

(a) 3 km

(b) 4 km

(d) 6 km

Solution:

Given that,

A worker reaches his factory 3 minutes late if his speed from his house to the factory is 5 km/hr.

He walks at a speed of 6 km/hr, then he reaches the factory 7 minutes early.

Now,

Speed 1 = 5 km/hr

Speed 2 = 6 km/hr

Distance = constant

Let time be t

5 ( t + 3) = 6 (t - 7)

t = 57 min

Distance = [5 x (57 + 3)]/60 = 300/60 = 5km

Hence option (c) is correct

Q7. Two cities A and B are 360 km apart. A car goes from A to B with a speed of 40 km/hr and returns to A with a speed of 60 km/hr. What is the average speed of the car? [CSAT 2015]

(a) 45 km/hr

(b) 48 km/hr

(d) 55 km/hr

Solution:

Given that,

Two cities A and B are 360 km apart.

A car goes from A to B with a speed of 40 km/hr and returns to A with a speed of 60 km/hr.

Now,

Average speed of the car = (2 x 40 x 60)/100 = 48 km/hr (Average speed when distance is constant = 2S₁ S₂ / S₁ +S₂)

Hence option (b) is correct

Q8. The figure drawn below gives the velocity graphs of two vehicles A and B. The straight line OKP represents the velocity of vehicle A at any instant, whereas the horizontal straight line CKD represents the velocity of vehicle B at any instant. In the figure, D is the point where perpendicular from P meets the 1 horizontal line CKD such that PD = LD

What is the ratio between the distances covered by vehicles A and B in the time interval OL? [CSAT 2018]

(a) 1:2

(b) 2:3

(d) 1:1

Solution:

Distance covered by A = AREA of triangle OPL

= (1/2) x OL x LP

= (1/2) x OL x (LD + PD) ......(As LD + PD = LP)

= (1/2) x OL x (LD + LD/2) (given PD = 1/2LD)

= (1/2) x OL x (3/2)(LD)

= (3/4) x OL x LD

Distance covered by B = Area of rectangle OLCD

= OL × LD

Thus ratio = [(3/4) x OL x LD]/ OL x LD = 3/4

Hence option (c) is correct

Q9. A car travels from a place X to place Y at an average speed of v km/hr, from Y to X at an average speed of 2v km/hr, again from X to Y at an average speed of 3v km/hr and again from Y to X at an average speed of 4v km/hr. Then the average speed of the car for the entire journey [CSAT 2020]

(a) Is less than v km/hr

(b) Lies between v and 2v km/hr

(d) Lies between 3v and 4v km/hr

Solution:

Given that,

A car travels from a place X to place Y at an average speed of v km/hr,

From Y to X at an average speed of 2v km/hr

From X to Y at an average speed of 3v km/hr

From Y to X at an average speed of 4v km/hr.

Now,

Let the distance be d

Total distance covered = 4d

Time taken for all trips = d/v, d/2v, d/3v and d/4v

Total time = d/v + d/2v + d/3v + d/4v = 25d/12v

Average speed = 4d/(25d/12v) = 48v/25 or 1.92v km/h

Hence option (b) is correct

Q10. A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 1.5 km/h and Y walks with a uniform speed of 1 km/h in the first hour, with a uniform speed of 1.25 km/h in the second hour and with a uniform speed of 1.5 km/h in the third hour and so on. [CSAT 2021]

Which of the following is/are correct?

1. They take 5 hours to meet.

2. They meet midway between A and B.

Select the correct answer using the code given below:

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

A person X from a place A and another person Y from a place B set out at the same time to walk towards each other.

The places are separated by a distance of 15 km.

X walks with a uniform speed of 1.5 km/h

Y walks with a uniform speed of 1 km/h in the first hour, with a uniform speed of 1.25 km/h in the second hour and with a uniform speed of 1.5 km/h in the third hour and so on.

Now,

Total distance between A and B = 15km

Speed of X = 1.5 km/h

Time = 15/1.5 = 10 hrs

Thus in 5 hr it will cover half of the distance = 7.5 km

For Y

It covers 1 km in 1 hr

1.25 km in next hour

1.5 km in next hour

1.75 km in next hour

2 km in next hour

Total distance covered = 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5 km

Total time taken = 5 hr

Therefore they meet in 5hrs

1. They take 5 hours to meet.

Hence statement 1 is correct

2. They meet midway between A and B.

Hence statement 2 is correct

Q11. A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is [CSAT 2022]

(a) 10 minutes slow

(b) 5 minutes slow

(d) 5 minutes fast

Solution:

Given that,

A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours.

After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours.

Now,

Let the speed of man be S km/hr and distance between home and village be D km

Total time taken = 16:00 - 14: 30 = 1 : 30 or 1.5hr or 3/2 hr

Total travelling time = (3/2) - (5/12) = 13/12 hr

(S/D) + (1.25S/2D) = 13/12

S/D = 2/3 hr or 40 min (time taken from home to village)

Exact time on clock = 14: 30 + 40 = 15 : 10

Thus village time is 5 min faster

Hence option (d) is correct

Q12. Two friends X and Y start running and they run together for 50 m in the same direction and reach a point. X turns right and runs 60 m, while Y turns left and runs 40 m. Then X turns left and runs 50 m and stops, while Y turns right and runs 50 m and then stops. How far are the two friends from each other now? [CSAT 2022]

(a) 100 m

(b) 90 m

(d) 50 m

Solution:

Given that,

Two friends X and Y start running and they run together for 50 m in the same direction and reach a point.

X turns right and runs 60 m, while Y turns left and runs 40 m.

Then X turns left and runs 50 m and stops, while Y turns right and runs 50 m and then stops.

Now,

X runs = 60 m right

Y runs = 40 m left

Total distance between them = (60 + 40) = 100 m

Thus X is 100 m apart from Y

Hence option (a) is correct

RACE

Q13. In a 500 metres race, B starts 45 meters ahead of A, but A wins the race while B is still 35 metres behind. What is the ratio of the speed of A to B assuming that both start at the same time? [CSAT 2015]

(a) 25: 21

(b) 25: 20

(d) 5:7

Solution:

Given that,

In a 500 metres race, B starts 45 meters ahead of A, but A wins the race while B is still 35 metres behind.

Now,

Distance covered by A = 500 m

Distance covered by B = 500 - 45 - 35 = 420 m

Ratio = 500/420 = 25 : 21

Hence option (a) is correct

Q14. X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many meters start will Y give to Z? [CSAT 2019]

(a) 20

(b) 25

(d) 35

Solution:

Given that,

X, Y and Z are three contestants in a race of 1000 m.

Assume that all run with different uniform speeds.

X gives Y a start of 40 m and X gives Z a start of 64 m.

Y and Z were to compete in a race of 1000 m

Now,

X completes 1000 m then Y completes 960 m

Ratio of speed = 1000 : 960 = 25 : 24 = X : Y

X completes 1000 m Z completes 936 m

Thus by unitary method

Z start - Y start = 64 - 40 = 24

So in 1000 m Y will give start to Z = (24/960) x 1000 = 25 m

Hence option (b) is correct

CIRCULAR MOTION

Q15. A and B walk around a circular park. They start at 8 am from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other after 8:00 am and before 9:30 am? [CSAT 2016]

(a) 7

(b) 6

(d) 8

Solution:

Given that,

A and B walk around a circular park.

They start at 8 am from the same point in the opposite directions.

A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively

Now,

Relative speed in an hour of A and B = 2 + 3 = 5 rounds

Distance covered in 30 min = 5 x 1/2 = 5/2 = 2.5 rounds

Total distance in 1.5 hr = 5 + 2.5 = 7.5 rounds

Thus they will meet 7 times

Hence option (a) is correct

Q16. Two persons, A and B are running on a circular track. At the start, B is ahead of A and their positions make an angle of 30° at the centre of the circle. When A reaches the point diametrically opposite to his starting point, he meets B. What is the ratio of speeds of A and B, if they are running with uniform speeds? [CSAT 2018]

(a) 6:5

(b) 4:3

(d) 4:2

Solution:

Given that,

Two persons, A and B are running on a circular track.

At the start, B is ahead of A and their positions make an angle of 30° at the centre of the circle. When A reaches the point diametrically opposite to his starting point, he meets B.

Now,

Distance travelled by A = πr

Distance travelled by B = πr - πr/6 = 5 πr/6 ( 2πr x 30/360 = πr/6 for 30 degree)

Ratio of the speed of A and B = 6 : 5

Hence option (a) is correct

Q17. X and Y run a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3:2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?

[CSAT 2022]

(a) 2

(b) 3

(d) 5

Solution:

Given that,

X and Y run a 3 km race along a circular course of length 300 m.

Their speeds are in the ratio 3:2.

They start together in the same direction

Now,

Total rounds = 3000/300 = 10 rounds

Since the ratio is 3 : 2 , the faster one meets slower one after every 3 rounds

So they will meet 3 times as there are total 10 rounds

ANSWER KEY

1. B
2. B
3. A
4. B
5. A
6. C
7. B
8. C
9. B
10. C
11. D
12. A
13. A
14. B
15. A
16. A
17. B

Download TSD PYQs PDF

✔️ Download the complete set of Time, Speed & Distance PYQs (2011-2024) with solutions in PDF format.
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FAQs on Time, Speed & Distance for UPSC

1. How many TSD questions appear in CSAT?

Typically 2-3 questions annually in CSAT Paper II.

2. What are the most common TSD problems?

· Average speed calculations (30%)

· Relative speed problems (25%)

· Boats and streams (20%)

· Train-platform problems (15%)

· Circular motion (10%)

3. Any quick solving tips?

✔️ Always draw diagrams for visualization
✔️ Convert units consistently (km/h ↔ m/s)
✔️ For average speed, use harmonic mean when distances are equal
✔️ Remember upstream/downstream speed relationships

Conclusion

Mastering TSD PYQs can secure you easy marks in CSAT. This page provides:
✅ All PYQs from 2011-2024
✅ PDF with 50+ solved problems
✅ Time-saving calculation methods

For more UPSC resources, visit iassetu.com.

Internal Links:

· UPSC CSAT Quantitative Aptitude Guide

· Shortcut Formulas Cheatsheet

· Relative Motion Concepts