Top 130 Most Asked Speed Distance Time Questions [ 100% FREE ]

Understanding the concepts of speed, distance, and time is not just for the curious minds; it’s essential for a variety of real-world applications and competitive exams. And when it comes to cracking these exams, practicing the right set of Speed Distance Time questions is crucial.

Why, you ask? Because Speed Distance Time questions frequently pop up in examinations, and being well-versed with them can give you a notable edge. This article is a treasure trove for those who are looking for the most asked Speed Distance Time questions.

Questions related to the time and distance formula are a staple in both school and competitive exams. This formula finds its use in diverse topics like motion in a straight line, races, and even clocks. So, what is the relation between speed, distance, and time?

Well, in simple terms, distance traveled is the product of speed and the time for which an object has been in motion. We dive deeper into these concepts, answering some of the most frequently asked time speed and distance questions. Additionally, we’ve compiled some popular speed time and distance questions, as well as speed time and distance questions with answers in a PDF format, ensuring that you’re not just practicing but also checking your answers.

For those keen on aptitude, our collection includes time speed and distance aptitude problems. We’ve also incorporated speed time and distance problems with solutions to assist you in understanding where you might have gone wrong and to learn the right approach.

To give you a sneak peek, the time and distance formula revolves around the equation:

Distance (d) = Speed (v) × Time (t). This formula and its variants are the backbone of numerous speed time and distance questions. If you wish to explore the intricacies of the time and distance formula in detail, you can find more information here.

In conclusion, whether you’re preparing for an upcoming exam or just wish to brush up your knowledge, this comprehensive article promises a deeper insight into speed, time, and distance concepts. So gear up and dive into our meticulously curated list of Speed Distance Time questions. Happy learning!

121. A train A of length 120 m can cross a platform of length double the length of train A in 5 seconds and a train B moving with a speed of 108 m/second can cross a platform of length 252 m in 4 seconds. Find the time in which they cross each other moving in the opposite direction. 

एक ट्रेन A 120 मीटर लंबाई के साथ 5 सेकंड में ट्रेन A की लंबाई के दोगुने प्लेटफ़ॉर्म को पार कर सकती है और 108 मीटर/सेकंड की रफ़्तार से चलने वाली ट्रेन B 4 सेकंड में 252 मीटर की लंबाई के प्लेटफ़ॉर्म को पार कर सकती है। वह समय ज्ञात कीजिए जिसमें वे एक-दूसरे को विपरीत में चलते हुए पार करते हैं।

Option “B” is correct.

Let the length of train B be X m

Since the length of train is 120 m

So, the length of platform is 240 m

So, the speed of train A = (120 + 240)/5 = 72 m/second

Now, for train B

(X + 252) = 108 × 4

⇒ X + 252 = 432

∴ X = 180

So,

Time taken by A and B to cross each other = (120 + 180)/(72 + 108)

Hence, required time = 5/3 seconds

122. The speed of two railway engines is in the ratio 5 : 4. If they move on parallel tracks in the same direction and if the slower engine is ahead of the faster engine by 8 km when the latter starts, then how far will the faster engine have to travel before it overtakes the slower one?

दो रेल इंजन की गति का अनुपात 5 : 4 है। यदि सामानांतर पटरियों पर समान दिशा में चलते हैं और यदि धीमा इंजन तेज़ इंजन से 8 किमी आगे है जब दोनों प्रारंभ करते हैं, तो धीमे इंजन से आगे निकालने के लिए तेज़ इंजन को कितनी दूरी तय करनी होगी?

Option “B” is correct.

Speed ratio of two engines is = 5 : 4

Let the speed of the faster engine is 5 km/hr and speed of the slower engines is 4 km/hr

The distance of the two engines = 8 km

Required time to overtake by faster engine = 8/ (5 – 4) = 8 hr

Distance covered by the faster engine in 8 hrs = 5 × 8 = 40 km

123. A and B, start from X and Y towards Y and X respectively. After passing each other, they take 2.7 hours and 1.2 hours to reach Y and X respectively. If B is moving at 48 km/hr, then the speed of A is.

A और B, क्रमशः X और Y से Y और X की ओर चलना शुरू करते हैं। एक-दूसरे को पार करने के बाद, उन्हें क्रमशः Y और X तक पहुंचने के लिए 2.7 घंटे और 1.2 घंटे का समय लगता है। यदि B, 48 किमी/घंटा की गति से आगे बढ़ रहा है, तो A की गति क्या है?

Option “A” is correct.

As we know,

Speed of A/Speed of B = √(Time taken by B)/(Time taken by A)

A/48 = √(1.2)/(2.7) = √4/9 = 2/3

⇒ A = 2/3 × 48

⇒ A = 32 km/hr

124. Two friends A and B started from the same point to meet another friend C. But friend C also moving towards A and B to meet them, it is given that relative speed of A and B is 8 m/s and the relative speed of B and C is 12 m/s. If the speed of B is more than the speed of A and C, then find the relative speed (in m/s) of A and C?

दो मित्र A और B एक ही मित्र C से मिलने के लिए उसी बिंदु से चलना शुरू करते हैं। लेकिन मित्र C, A और B से मिलने के लिए भी आगे बढ़ रहा है, यह दिया जाता है कि A और B की सापेक्ष गति 8 मीटर/सेकंड है और B और C की सापेक्ष गति  12 मीटर/सेकंड है। यदि B की गति A और C की गति से अधिक है, तो A और C की सापेक्ष गति (मीटर/सेकंड) में ज्ञात कीजिए?

Option “A” is correct.

Let the speed of A, B and C be a m/s, b m/s and c m/s respectively.

Relative speed of A and B

⇒ b – a = 8      …. (1)

Relative speed of B and C

⇒ b + c = 12     …. (2)

 Subtracting equation 1 form 2

⇒ b + c – b + a = 12 – 8 = 4

⇒ c + a = 4 m/s

125. Two cars start together in the same direction from the same place. The first goes with a uniform speed of 10 km/h. The second goes at a speed of 8 km/h in the first hour and increases its speed by 1/2 km/hr with each succeeding hour. After how many hours will the second car overtake the first one, if both go non-stop?

दो कारें समान स्थान से समान दिशा में चलना प्रारंभ करती हैं। पहली 10 किमी/घंटा की एकसमान गति से चलती है। दूसरी पहले घंटे में 8 किमी/घंटा की गति से जाती है और प्रत्येक अगले घंटे में अपनी गति में 1/2 किमी/घंटा की वृद्धि करती है। यदि दोनों बिना रुके चलती हैं, तो कितने घंटों के बाद दूसरी कार पहली कार से आगे निकलेगी?

Option “A” is correct.

Let the second car overtakes the first car in x hours.

The first car goes with a uniform speed of 10 km/h

∴ Distance covered by the first car in x hours = 10x km.

Second car

In first hour speed = 8 km/h

In second hour speed = (8 + 1/2) = 8.5 km/h

In third hour, second cars speed = (8.5 + 0.5) = 9 km/h

∴ Distance covered by second car in x hours,

⇒ (8 + 8.5 + 9 + …. Till x no. of terms) [it is in AP where first term = 8 and common difference= 0.5]

⇒ x/2× {2 × 8 + (x – 1) × 0.5} km

According to question,

⇒ 10x = x/2 × {2 × 8 + (x – 1) × 0.5}

⇒ 10x × 2/x = 16 + 0.5x – 0.5

⇒ 20 = 15.5 + 0.5x

⇒ 0.5x = 4.5

⇒ x = 9

∴ After 9 hours the second car will overtake the first car.

126. Two cyclists X and Y start at the same time from place A and go towards place B at a speed of 6 km/h and 8 km/h, respectively. Despite stopping for 15 minutes during the journey, Y reaches 10 minutes earlier than X. The distance between the places A and B is:

दो साइकिल चालक X और Y स्थान A से समान समय पर साइकिल चलाना शुरू करते हैं और क्रमशः 6 किमी/घंटा और 8 किमी/घंटा की गति से स्थान B की ओर जाते हैं। यात्रा के दौरान 15 मिनट के लिए रुकने के बावजूद Y, X से 10 मिनट पहले स्थान B पर पहुँचता है। तो स्थान A और B के बीच की दूरी क्या है?

Option “C” is correct.

Speed of X = 6 km and speed of Y = 8 km

Speed ratio of X and Y = 6 : 8 = 3 : 4

Time ratio of X and Y = 4 : 3

As we know, Time is inversely proportional to speed.

The time difference = 15 min + 10 min = 25 min

4 – 3 unit = 1 unit

1 unit = 25 min

Time taking by X = 4 × 25 = 100 min = 100/60 hr

∴ Distance covered by X in 100/60 hr = 100/60 × 6 = 10 km.

127.A priest from a temple rang the bell at an interval of 60-sec but a man hears it in 55-sec interval if the distance between the temple and the man is 900 m find the time in which he will arrive in the temple given that the speed of sound of the bell is 330 m/sec.

एक मंदिर का एक पुजारी 60 सेकंड के अंतराल पर घंटी बजता है लेकिन एक व्यक्ति को यह 55 सेकंड के अंतराल पर सुनाई देती है यदि उस व्यक्ति और मंदिर के मध्य की दूरी 900मी है और यह दिया गया है कि ध्वनि कि गति 330मी/सेकंड है तो ज्ञात कीजिये वह व्यक्ति कितने समय में मंदिर पहुँच जायेगा?

Option “A” is correct.

                              Sound            Man         

Time (in sec)            5                  55

                                   1       :          11

Speed                       11      :           1

Speed of sound = 11R = 330 m/sec

So the speed of man = 1R = 30 m/sec

∴ Man arrives to the temple in = 900/30 = 30 seconds

128. The driver of an ambulance sees a bus 40 m ahead of him, after 20 seconds, the bus is 60 meter behind. If the speed of the ambulance is 30 km/h what is the speed of the bus?

एम्बुलेंस का चालक अपने 40 मीटर आगे किसी बस को देखता है, 20 सेकेंड के बाद, बस उसके 60 मीटर पीछे आ जाती है। यदि एम्बुलेंस की गति 30 किमी/घंटा होती है तब बस की गति क्या है? 

Option “B” is correct.

Let speed of bus be x km/hr

Speed of ambulance = 30 km/hr

⇒ Relative speed = (30 – x) km/hr

Total distance covered in 20 seconds = 40 + 60 = 100 m

According to the question

(30 – x) × 5/18 = 100/20

⇒ 30 – x = 5 × 18/5

⇒ 30 – x = 18

⇒ x = 30 – 18

⇒ x = 12 km/hr

∴ Speed of bus is 12 km/hr

129. A train covers certain distance between two places at a uniform speed. If the train moved 10 km/hr faster it would take 2 hours less. And, if the train were slower by 10 km/hr, it would take 3 hours more than the scheduled time. Find the distance covered by the train.

एक ट्रेन एक समान गति से दो स्थानों के बीच निश्चित दूरी तय करती है। अगर ट्रेन 10 किमी/घंटा तेज  गति से आगे बढ़ती है तो 2 घंटे कम लगते हैं। और, यदि ट्रेन 10 किमी/घंटा धीमी थी, तो उसे निर्धारित समय से 3 घंटे अधिक लगेंगे। ट्रेन द्वारा तय की गई दूरी ज्ञात कीजिये।

Option “B” is correct.

Let distance be d, speed of train be x km/hr

According to the question

d/x – d/(x + 10) = 2

⇒ d [(x + 10 – x)/x (x + 10)] = 2

⇒ 10d/x(x + 10) = 2

⇒ d = 2x (x + 10)/10     —-(1)

d/(x – 10) – d/x = 3

⇒ d [(x – x + 10)/x(x – 10)] = 3

⇒ 10d/x(x – 10) = 3

⇒ d = 3x (x – 10)/10     —-(2)

From equation (1) and equation (2)

2x (x + 10)/10 = 3x (x – 10)/10

⇒ 2 (x + 10) = 3 (x – 10)

⇒ 2x + 20 = 3x – 30

⇒ 3x – 2x = 30 + 20

⇒ x = 50

Speed of train is 50 km/hr

From equation (1)

d = 2 × 50 × (50 + 10)/10

⇒ d = 100 × 60/10

⇒ d = 600 km

∴ The distance covered by train is 600 km.

130. Two spots A and B are at a distance of 20 km. Paul started from spot A towards B and at the same time, King started from spot B towards A. They meet each other after 1 hour. After that, Paul reduced his speed by 4 km/h and King increased his speed by 4 km/h. They reached their destinations simultaneously. Find Paul’s initial speed (in km/h).

दो स्थल A और B, 20 किमी की दूरी पर हैं। पॉल ने स्थल A से B की ओर चलना शुरू हुआ और उसी समय, किंग ने B से A की ओर चलना शुरू किया। वे 1 घंटे के बाद एक दूसरे से मिलते हैं। उसके बाद, पॉल ने अपनी गति 4 किमी/घंटे कम कर दी और किंग ने अपनी गति 4 किमी/घंटे बढ़ा दी। वे एक साथ अपने गंतव्य तक पहुंचे। पॉल की प्रारंभिक गति (किमी/घंटे में) ज्ञात कीजिये।

Option “A” is correct.

Let the speed of Paul is M km/h and speed of King is N km/h

Time = Distance/Speed

1 Hour = 20 km/(M + N)

M + N = 20 km/h      —-(1)

Distance travelled by Paul in 1 hour with M km/h speed is M km and Distance travelled by King in 1 hour with N km/h speed is N km

It means M + N = 20 km

Which means that Paul travelled M km, so left distance for him to travel is N km and Similarly N left distance to travel is M km

N/(M – 4) = M(N + 4)

N × (N + 4) = M × (M – 4)  

N2 + 4N = M2 – 4M

M–  N2 = 4M + 4N

M – N = 4 km/h    —-(2)

From 1 and 2

2 M = 24 km/h

M = 12 km/h

∴ Paul’s initial speed is 12 km/h

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Speed Distance Time questions : Understanding the nuances of speed, distance, and time is crucial in both academic and practical realms. Throughout this article, we have navigated the vast realm of Speed Distance Time questions, unveiling their importance and intricacies. If there’s one takeaway from this extensive discussion, it’s the undeniable significance of mastering this topic.

Often, students find themselves grappling with time speed and distance questions. These are not mere mathematical exercises but vital tools that describe motion, travel, and even the very fabric of our day-to-day activities. From calculating the time taken to travel between two cities to understanding how fast a car should move to reach a destination at a particular time, these concepts touch various aspects of our lives.

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For those with an aptitude inclination, the segment on time speed and distance aptitude offers a goldmine of information. Aptitude tests, often used in job assessments and competitive exams, frequently feature these questions. Hence, being proficient in this area is not just a matter of academic excellence but can be a stepping stone to career opportunities.

Moreover, it’s not just about knowing the formulae. The real skill lies in application. By going through speed time and distance problems with solutions, learners can recognize common patterns, traps, and shortcuts. These insights can be invaluable, especially under timed test conditions.

Now, you might wonder, what is the relation between speed, distance, and time? Simply put, they are interconnected pillars of motion. The distance covered is a result of how fast you move (speed) and for how long (time). This relationship is encapsulated in the formula: Distance (d) = Speed (v) × Time (t). While this may appear straightforward, the multitude of Speed Distance Time questions derived from this formula can range from simple to highly complex, demanding not just mathematical acumen but logical reasoning too.

As we wrap up, it’s essential to reiterate the significance of continuous practice. The more speed time and distance questions you tackle, the better you become. And while theoretical knowledge is essential, it’s through consistent engagement with problems and their solutions that true mastery is achieved.

In summary, this article has been a deep dive into the world of Speed Distance Time questions, touching upon their relevance, complexity, and the tools needed to conquer them. Whether you’re a student aiming for academic excellence or a professional gearing up for aptitude tests, this compilation serves as a reliable guide. Remember, the journey of understanding speed, time, and distance is not just about solving problems; it’s about understanding a fundamental aspect of our world. With the knowledge gained, one is not just prepared for exams but also better equipped to navigate the challenges of everyday life.