Time Speed and Distance Practice Set 2 (Q11–Q20) with Answers

Introduction

Welcome to Part 2 of our Time, Speed and Distance Questions with Answers series on BrainQuro!

Time, Speed and Distance is an important topic for candidates preparing for IBPS PO, SBI PO, IBPS Clerk, RBI Assistant, SSC CGL, RRB NTPC, LIC AAO and other competitive examinations.

In Time Speed and Distance practice set 2, you will solve 10 carefully selected questions covering trains, relative speed, catching a thief, speed calculations, boats and streams, circular tracks and time-based problems.

Each question includes a shortcut method and a verified final answer to help you improve your calculation speed and problem-solving skills.

Practice Challenge: Try solving all 10 questions yourself before checking the answers.

Question 11: Two Trains Moving Towards Each Other

Question

Two trains, Train A and Train B, start simultaneously from stations X and Y towards each other. After meeting, Train A takes 9 hours to reach station Y, while Train B takes 16 hours to reach station X. If the speed of Train B is 45 km/h, find the speed of Train A.

Shortcut Method

When two trains start simultaneously and move towards each other, the ratio of their speeds is equal to the square root of the inverse ratio of their remaining travel times.

Speed of Train A : Speed of Train B = √16 : √9

= 4 : 3

Speed of Train A = 45 × 4/3

= 60 km/h

Answer

60 km/h

Question 12: Journey by Train and Car

Question

A man travels a total distance of 300 km. He covers one part of the journey by train at 60 km/h and the remaining part by car at 40 km/h. If he had travelled the first part by car and the remaining part by train, his journey would have taken 30 minutes longer. Find the actual time taken to complete the journey.

Shortcut Method

Let the distance travelled by train be x km.

Distance travelled by car = 300 − x km.

The difference in journey times is 30 minutes, or 0.5 hours.

Time difference:

x/40 − x/60 + (300 − x)/60 − (300 − x)/40 = 0.5

Simplifying:

(2x − 300)/120 = 0.5

2x − 300 = 60

2x = 360

x = 180 km

Therefore:

Train distance = 180 km

Car distance = 300 − 180 = 120 km

Actual journey time:

180/60 + 120/40

= 3 + 3

= 6 hours

Answer

6 hours

Question 13: Catching a Thief

Question

A thief steals a car at 2:00 PM and drives at 40 km/h. The theft is discovered at 3:00 PM, and the owner starts pursuing the thief in another car at 60 km/h. At what time will the owner catch the thief, and what total distance will the thief have travelled?

Shortcut Method

Step 1: Calculate the thief’s head start.

Head start = 40 × 1 = 40 km

Step 2: Calculate the relative speed.

Relative speed = 60 − 40 = 20 km/h

Step 3: Calculate the catching time.

Catching time = 40/20 = 2 hours

The owner starts at 3:00 PM.

Catching time = 3:00 PM + 2 hours = 5:00 PM

Step 4: Calculate the total distance travelled by the thief.

Total travel time = 3 hours, from 2:00 PM to 5:00 PM.

Distance = 40 × 3 = 120 km

Answer

Catching time: 5:00 PM

Total distance: 120 km

Question 14: Two Trains Starting at Different Times

Question

Two stations, P and Q, are 460 km apart. Train 1 leaves station P towards Q at 50 km/h. After 2 hours, Train 2 leaves station Q towards P at 70 km/h. How far from station P will the trains meet?

Shortcut Method

Step 1: Distance covered by Train 1 in the first 2 hours.

50 × 2 = 100 km

Remaining distance between the trains:

460 − 100 = 360 km

Step 2: Calculate their relative speed.

50 + 70 = 120 km/h

Step 3: Time taken to meet after Train 2 starts.

360/120 = 3 hours

Step 4: Total time travelled by Train 1.

2 + 3 = 5 hours

Meeting point from P:

50 × 5 = 250 km

Answer

250 km from station P

Question 15: Finding the Original Speed

Question

A person covers a certain distance at a certain speed. If he increases his speed by 10 km/h, he takes 36 minutes less to cover the same distance. If he decreases his speed by 10 km/h, he takes 45 minutes more than the original time. Find his original speed.

Shortcut Method

Let the original speed be v km/h and the distance be d km.

Time saved by increasing speed = 36 minutes = 0.6 hours.

Therefore:

Distance = 0.6 × v × (v + 10)/10

Distance = 0.06v(v + 10)

When the speed decreases by 10 km/h, the extra time is 45 minutes = 0.75 hours.

Therefore:

Distance = 0.75 × v × (v − 10)/10

Distance = 0.075v(v − 10)

Equating the two distances:

0.06v(v + 10) = 0.075v(v − 10)

Cancel v and simplify:

40(v + 10) = 50(v − 10)

40v + 400 = 50v − 500

10v = 900

v = 90 km/h

Answer

90 km/h

Question 16: Finding the Speed of Sound

Question

Two guns are fired from the same place at an interval of 12 minutes. A person approaching the place in a train hears the second gunshot 11 minutes after the first. If the speed of the train is 60 km/h, find the speed of sound.

Shortcut Method

Step 1: Find the difference between the firing interval and the hearing interval.

12 − 11 = 1 minute

Step 2: Calculate the distance travelled by the train during the 11-minute interval.

Distance = 60 × 11/60 = 11 km

Step 3: The difference in sound travel times is 1 minute.

Speed of sound = 11 km ÷ (1/60 hour)

= 11 × 60

= 660 km/h

Convert into metres per second:

660 × 5/18 = 183.33 m/s approximately

Answer

183.33 m/s approximately

Question 17: Boat and Stream

Question

A boat travels 36 km upstream and 48 km downstream in 10 hours. It can also travel 48 km upstream and 36 km downstream in 11 hours. Find the speed of the current.

Shortcut Method

Let the upstream speed be u km/h and the downstream speed be d km/h.

First condition:

36/u + 48/d = 10

Second condition:

48/u + 36/d = 11

Solving these two equations gives:

Upstream speed = 6 km/h

Downstream speed = 12 km/h

Speed of the current:

(Downstream speed − Upstream speed)/2

= (12 − 6)/2

= 3 km/h

Answer

3 km/h

Question 18: Finding the One-Way Distance Rowed

Question

A man can row a certain distance downstream in 3 hours and return the same distance upstream in 5 hours. If the speed of the current is 4 km/h, find the one-way distance he rows.

Shortcut Method

Let the one-way distance be d km.

Downstream speed = d/3

Upstream speed = d/5

The difference between downstream and upstream speeds is twice the current speed.

d/3 − d/5 = 2 × 4

Taking the LCM:

(5d − 3d)/15 = 8

2d/15 = 8

2d = 120

d = 60 km

Answer

60 km one way

Question 19: Three Runners on a Circular Track

Question

Three runners, A, B and C, run on a circular track of circumference 1,200 metres at speeds of 10 m/s, 15 m/s and 20 m/s, respectively, in the same direction. After how much time will all three meet together at the starting point for the first time?

Shortcut Method

Step 1: Calculate the time taken by each runner to complete one lap.

Runner A = 1,200/10 = 120 seconds

Runner B = 1,200/15 = 80 seconds

Runner C = 1,200/20 = 60 seconds

Step 2: Find the LCM of 120, 80 and 60.

LCM = 240 seconds

Verification:

  • Runner A completes 2 laps.
  • Runner B completes 3 laps.
  • Runner C completes 4 laps.

All three reach the starting point together after 240 seconds.

Answer

240 seconds, or 4 minutes

Question 20: Bus Passengers and Speed

Question

A bus starts with 60 passengers. At the first stop, one-fourth of the passengers get down and 15 new passengers board. At the second stop, one-third of the passengers get down and 10 new passengers board. If the bus covers 180 km between the starting point and the second stop in 2 hours, excluding stoppage time, find its speed.

Shortcut Method

Step 1: Calculate the passengers after the first stop.

Initial passengers = 60

Passengers who get down = 60/4 = 15

Passengers remaining = 60 − 15 = 45

After 15 passengers board:

45 + 15 = 60 passengers

Step 2: Calculate the passengers after the second stop.

Passengers who get down = 60/3 = 20

Passengers remaining = 60 − 20 = 40

After 10 passengers board:

40 + 10 = 50 passengers

The bus leaves the second stop with 50 passengers.

Step 3: Calculate the speed.

Speed = Distance/Time

Speed = 180/2

= 90 km/h

Answer

90 km/h

Final Answer Key

QuestionFinal Answer
Q160 km/h
Q26 hours
Q35:00 PM; 120 km
Q4250 km from P
Q590 km/h
Q6183.33 m/s approximately
Q73 km/h
Q860 km one way
Q9240 seconds (4 minutes)
Q1090 km/h

Shortcut Trick of the Day

Catch-up Problems: When two vehicles move in the same direction, calculate relative speed by subtracting the slower speed from the faster speed.

Catch-up time = Initial distance gap ÷ Relative speed.

For example, if a thief has a 40 km head start and the owner travels 20 km/h faster, the owner catches the thief in 40/20 = 2 hours.

Practice Challenge Question

A train travels at 72 km/h. After covering 120 km, its speed decreases by 20%. As a result, it reaches its destination 30 minutes late. Find the remaining distance at the time of the speed reduction.

Try solving it before checking the answer in the next part of the series!

Frequently Asked Questions

1. Which topics are covered in this Speed, Time and Distance practice set?

This set covers trains, relative speed, catching a thief, speed calculations, boats and streams, circular tracks and journey-time problems.

2. Is this practice set useful for banking exams?

Yes. These question types are useful for quantitative aptitude practice for IBPS PO, SBI PO, IBPS Clerk, RBI Assistant and other competitive examinations.

3. How can I solve Speed, Time and Distance questions faster?

Learn the basic relationships between speed, time and distance, practise relative-speed shortcuts, and verify your answers regularly.

4. What is the basic formula for speed?

Speed = Distance ÷ Time.

5. Should I practise these questions without looking at the answers?

Yes. Attempt each question independently first, then compare your working with the shortcut method and final answer.

Want More Time, Speed and Distance Questions

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