# Time Speed and Distance Formulas, Tricks with Examples

## Time Speed and Distance Formulas, Tricks with Examples

Hello Students,

Time, speed, and distance are important topics in mathematics, physics, and engineering, and they are often tested in competitive exams such as the GRE, GMAT, SAT, CAT, and other aptitude tests. These concepts are used in real-world scenarios, such as calculating travel time, determining the speed of a moving object, and solving problems related to fuel efficiency, among others.

These formulas are fundamental and can be used to solve a wide range of problems related to time, speed, and distance. For example, if you know the speed of a vehicle and the distance it has traveled, you can use the first formula to calculate the time it took to cover that distance.

Another important trick to remember is that the distance covered is the product of the speed and time, and the time taken is inversely proportional to the speed. This means that if the speed of a vehicle is increased, the time taken to cover a fixed distance will decrease, and if the speed is decreased, the time taken will increase.

The following are some of the formulas and tricks for time, speed, and distance that are commonly used in competitive exams:-

1. Speed = Distance / Time
2. Time = Distance / Speed
3. Distance = Speed x Time

For example, let’s say that a car covers a distance of 120 km in 2 hours. Using the first formula, we can calculate the speed of the car as follows:

Speed = Distance / Time = 120 km / 2 hours = 60 km/hour

Similarly, if a train covers a distance of 800 km at a speed of 80 km/hour, we can use the second formula to calculate the time taken as follows:

Time = Distance / Speed = 800 km / 80 km/hour = 10 hours

These formulas and tricks are very useful in solving problems related to time, speed, and distance, and they can save a lot of time during competitive exams. It is important to practice using these formulas and tricks regularly to improve your problem-solving skills and increase your chances of success in exams.

### Related Topics:-

1. सरलीकरण (Specification)
2. आसन्न मान (Approximate Value)
3. समीकरणों की समानता (Equality of Equations)
4. घातांक एवं करणी (Indices and Surds)
5. लघुत्तम समापवर्तक एवं महत्तम समापवर्तक (LCM and HCF)
6. संख्या पद्धति (Number System)
7. प्रतिशतता (Percentage)
8. लाभ, हानि और बट्टा (Profit, Loss and Discount)
9. साधारण ब्याज (Simple Interest)
10. चक्रवृद्धि ब्याज (Computed Interest)
11. अनुपात एवं समानुपात (Ratio and Proportion)
12. साझेदारी (Partnership)
13. मिश्रण (Mixture)
14. मिस्र समानुपात (Mixture)
15. मिश्र समानुपात (Mixed Proportion)
16. उम्र सम्बन्धी प्रश्न (Problems on Age)
17. समय तथा काम (Time and Work)
18. नल तथा टंकी (Pipes and Cistern)
19. औसत (Average)
20. समय तथा दूरी (Time and Distance)
21. रेलगाङी से सम्बंधित प्रश्न (Problems Based on Trains)
22. नाव एवं धारा (Bots and Streams)
23. क्षेत्रफल (Area)
24. पृष्ठ क्षेत्रफल तथा आयतन (Surface Area and Volume)
25. आंकणो की पर्याप्तता (Data Sufficiency)
26. संख्या श्रेणी (Number Series)
27. क्रमचय एवं संचय (Permutation and Combination)
28. प्रायिका सिद्धांत (Probability Theory)
29. एकघातीय एवं द्विघातीय समीकरण (Linear & Quadratic Equations)
30. ग्राफ (Graph)
31. सारणीकरण (Tabulation)
32. लघुगणक (Logarithms)
33. श्रेणी (Progression)

### समय चाल और दूरी Formulas

किमी (km) / घंटा (Hour) से मीटर (Meter) / सेकंड (Second) में बदलने के लिए, 5 / 18 से गुणा
इसलिए, 1 किमी (km) / घंटा (Hour) = 5 / 18 मीटर (Meter) / सेकंड (Second)
M / sec से किमी (km) / घंटे (Hour) में बदलने के लिए, 18/5 से गुणा
इसलिए, 1 m / sec = 18/5 किमी (km) / घंटा (Hour) = 3.6 किमी (km) / घंटा (Hour)
1 किमी (km) / घंटा (Hour) = 5 / 8 मील (Mile) / घंटा (Hour)
1 गज (Yard) = 3 फीट (Feet)
1 किलोमीटर (km) = 1000 मीटर (Meter) = 0.6214 मील (Mile)
1 मील (Mile) = 1.609 किलोमीटर (km)
1 घंटा (Hour) = 60 मिनट (Minute) = 60 * 60 सेकंड (Second) = 3600 सेकंड (Second)
1 मील (Mile) = 1760 गज (Yard)
1 गज (Yard) = 3 फीट (Feet)
1 मील (Mile) = 5280 फीट (Feet)
In Simple Words, Memorable Facts About Speed, Distance and Time
1 किमी (km) प्रति घंटा (Hour) = 5 / 18 मीटर (Meter) / सेकेण्ड (Second)
1 मीटर (Meter) प्रति सेकेण्ड (Second) = 18 / 5 किमी (km) / घंटा (Hour)
1 किमी (km) प्रति घंटा (Hour) = 5 / 8 मील (Mile) / घंटा (Hour)
1 मील (Mile) प्रति घंटा (Hour) = 22 / 15 फुट (Feet) / सेकेण्ड (Second)

### Time And Distance – Aptitude Questions and Answers

Q.1. A students walks from his house at 5/2 km/h and reaches his school late by 6 min. Next day, he increase his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house?

(A) 5/4 km

(B) 7/4 km

(C) 9/4 km

(D) 11/4 km

Ans . B
Solution
Given, initial speed, a = 5/2.

And new speed, b= (5/2)+1 = 7/2.

Difference between two the times

= 6+6=12 min

According to the question,

X/(5/2) – X/(7/2) = 12/60→ 2X/5 – 2X/7 = 1/5

→14X-10X= 7 → X=7/4 km

Q.2. A thief is a spotted by a policeman from a distance of 200 m. When the policeman starts chasing, the thief also starts running. If the speed of the thief be 16 km/h and that of the policeman be 20 km/h, how far the thief will have run before he is overtaken?

(A) 800 m

(B) 850 m

(C) 700 m

(D) 650 m

Ans . A
Solution
= 16×(5/18) = 40/9 m/s.

And b = 20 km/h = 20×(5/18) = 50/9 m/s

Required distance = d×{a/(b-a)}

=

= 200×(40/10)=800m.

Q.3. A certain distance is covered at a certain speed. If half of this distance is covered in 4 times of the time, find the ratio of the two speed.

(A) 1:8

(B) 1:4

(C) 4:1

(D) 8:1

Ans . D
Solution
Then, First Speed = X/Y km/h

Again, X/2 km be covered in 4Y h.

New speed = (X/2)×(1/4Y)= X/8Y km/h

Hence, required ratio of speeds

= X/Y : X/8Y = 1: 1/8 = 8:1

Q.4. Sunil drives a motorcycle and covers a distance of 715 km at covered in 4 times of the time, find the ratio of the two speed.

(A) 65 km/h

(B) 55 km/h

(C) 60 km/h

(D) 36 km/h

Ans . B
Solution

→ x2+10=

→ x2+65x – 55x – 3575 = 0

→ x (x – 55) (x+65) = 0

→ (x – 55)(x+65) = 0

→ x = 55 or x = – 65

Ignoring the negative value, we get X = 55

⸫ Original speed of the motorcycle = 55 km

#### Time Speed and Distance

Q.5. Amit walks at a uniform speed of 4 km/h and 4 h after his starts, Brijesh cycles after him at the uniform rate of 20 km/h. How far from the starting point will Brijesh catch Amit?

(A) 15 km

(B) 18 km

(C) 13 km

(D) 20 km

Ans . D
Solution
= Distance travelled by Brijesh in X h.

According to the question,

4(x+4) = 20x

→ (4x+16) = 20x

→ 16x = 16

→ x= 16/16=1

⸫ Distance travelled by Brijiesh in 1 h = 1 × 20 = 20 km

Q.6. A train leaves Manipur at 6 am and reaches Dispur at 10 am. Another train leaves Dispur at 8 am and reaches Manipur at 11:30 am. At what time do the two trains cross each other?

(A) 7 : 56 am

(B) 7 : 56 pm

(C) 8 : 56 am

(D) 8 : 56 pm

Ans . C
Solution
Average speed of train from Manipur = x/4 km/h

Let, they meet y h after 6 am.

Then, according to the question,

()+

→ (y/4)+2(y-2)/7=1

→ 7y+8(y-2)=28

→15y=44

→ y=(44/15)h= 2h 56 min

Clearly, trains meet 2 h 56 min after 6 am.

Hence, the train meet at 8 : 56 am.

Q.7. The distance between two stations X and Y is 450 km. A train L starts at 6 pm from X and Y is at average speed of 60 km/h. Another train M starts from Y at an average speed of 60 km/h. Another train M starts from Y at 5: 2 pm and moves towards x at an average speed of 80 km/h. How far from X will the two trains meet and at what time?

(A) 170 km, 8 : 50 pm

(B) 150 km, 7 : 50 pm

(C) 170 km, 6 : 50 pm

(D) 150 m, 9 : 50 pm

Ans . A
Solution
Then, (Time taken by M to cover

(450 – a) km) – (Time taken by L to cover a km) = 40/60.

→ (450-a)/80 – a/60 = 40/60.

→ (450-a)/80 = 40/60+a/60.

→ (450-a)/8 – (40+a)/6 = 0

→ 3(450-a) – 4(a+40)=0

→ 7a= 1190

→ a = 1190/7 = 170

⸫ Time taken by L to cover 170 km = (170/60) h = 2 h 50 min.

Q.8. Two friends X and Y walk from A to B at distance of 39 km, at 3 km an hour and km an hour respectively Y reaches B, returns immediately and meet X at C. Find the distance from A to C.

(A) 26 s

(B) 25 s

(C) 15 s

(D) 27 s

Ans . A
Solution
= 144 km/h

= 144×(5/18)m/s

= 8×5 = 40 m/s

Distance covered in passing each other = (512+528) m

= 1040 m

⸫ Required time = 1040/40 = 26s.

Q.9. A train passes two persons who are walking in the direction opposite to the direction of train at the rate of 10 m/s and 20 m/s respectively in 12s and 10s respectively. Find the length of the train.

(A) 500 m

(B) 900 m

(C) 400 m

(D) 600 m

Ans . D
Solution
a = 10 m/s and b = 20 m/s

Now, length of the train

= Difference in speeds × T1 × T2/ T1- T2

= (20-10)×12×10/12-10

= (10×12×10)/2

= 600 m.

#### Time Speed and Distance Question and Answer

Q.10. From stations M and N, two trains start moving towards each other at speed 125 km/h and 75 km/h respectively. When the two trains meet each

(A) 190 km

(B) 200 km

(C) 145 km

(D) 225 km

Ans . B
Solution
Now, distance between the stations M and N

=

=

= (200/50)×50 = 200 km.