## CLAT Exam: What is Speed, Distance, and Time?

Understand the concept of Speed, Distance, and Time to be able to solve the questions with ease.

**Speed:**

Speed is easily defined as the distance covered by each unit of time. Statistically, it is defined as:

Speed = Distance travelled / Duration

Time Formula by distance and speed: Time = Range / Speed

Range formula in terms of Speed and Time: Range = (Speed Time -X)

It is very important to know the value of the unit while solving these types of problems.

SI Unit Speed: Meter per second

SI Unit Range: Meter

SI Unit Time: Second

All formulas in speed, time, and distance are related. One needs to connect the information provided to the problems intelligently and this can be easily done with good practice. Now let's talk about the different types of Speed, Distance & Time questions.

## Types of Time Speed and Distance Questions in CLAT Exam

Speed Time & Distance is one of the most frequently asked quantitative aptitude topics in entrance exams. This is one of the topics that candidates are familiar with before they start preparing for a competitive exam.

The concepts of speed, time, and distance are the same, but the types of questions asked in the exam may differ.

Most of the 12-word questions asked are based on speed, time, and distance, but candidates are also prepared to ask about data adequacy and data interpretation based on the TDS (time, distance, speed) topic. need to do it.

Introducing speed, distance and time concepts, formulas and rules to get ready and ace the tough competition. Further on in the article, in addition to speed, distance, and time tricks, there are some sample questions. Go through the time speed and distance of all types of questions in the CLAT exam below.

## Speed, Time, Distance Relationship (Formulas) for CLAT

Speed = Distance / Time-This shows how slow or fast the object is moving. It is the covered distance divided by the time it takes to cover that distance.

Speed is directly proportional to distance and inversely proportional to time. Therefore,

Distance = speed x time and

Time = distance / speed. The higher the speed, the shorter the time required and vice versa.

You can use these formulas to solve all the basic problems. However, when using formulas, it is also important to use the units correctly.

**Type 1 - Basic Questions for CLAT**

Now, we will discuss the most frequently used concept in this chapter i.e. questions based on the unit conversion. So, in order to learn this concept, we need to know how these questions are framed in the question paper.

**Question: A train runs at X km/hr and it takes 18 seconds to pass a pole. The length of the train is 180m. Find the speed of the train in km/hr.**

**Solution:**

Speed **= **distance /Time

Therefore,

S=180m/18s= 10m/s

Now we need the answer in km/hr and for that, we will multiply the given speed with (18/5) in order to get the answer in *km*/*h*

Therefore, S= 10*(18/5)=36*km*/*h.*

So here the trick for the same is :

**Convert metre per second (m/sec) to km per hr (km/h)**

For converting**(meter per second)**to**(kilometer per hour)**we use the following formula

* s m*

**/**= S *(18/5)

*sec*

*km*/*h***Convert km per hr (km/h) to metre per second (m/sec)**

For converting kph(kilometer per hour) to MPs(meter per second) we use the following formula

* ** S*** km/hr**=(

*s*∗5/18)

*m*/*sec***Question: A boy covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?**

#### Solution:

Speed =Distance / Time =Distance covered = 600m, Time taken = 2min 30sec = 150sec

Therefore, Speed= 600 / 150 = 4 m/sec = 4m/sec = (4*18/5) km/hr = 14.4 km/ hr

**Type 2 - Relative Speed Questions for CLAT**

Considering 2 objects *A* and *B* having the speed x, y.

- If the ratio of the speeds of A and B is x:y, then the ratio of the times taken by them to cover the same distance is:
**1/x: 1/y or y: x**

**Question: The ratio of the speed of a bike and a motor is 4:5 then what will be the ratio for the time taken by both the vehicles for the same destination?**

**Solution:**

As the destination is the same so distance will be the same for both Car and bike.

Let the Distance be d

And the speeds for both the vehicles be 4s and 5s

Now, t1 = d/4s----(1)

t2 = d/5s----(2)

so,t1/t2 = 5/4= 5:4

**Type 3 - Average Speed Questions for CLAT**

- Average Speed is another very important concept. It is defined as:

**Average Speed** = Total Distance Travelled /Total Time Taken

**Question: Dewansh travels 320 km at 64 km/hr and returns at 80 km/hr. Calculate the average speed of Dewansh?**

**Solution**:

We know that speed = Distance/ time taken

⇒ ∴Total time taken = 320/64 + 320/80 = 9

⇒ Average Speed = (320 + 320)/9

⇒ Average speed = 71.11 km/hr

**Question: A car moving with a uniform speed of 50km/h covers half the distance with this speed. Half of the time of the remaining distance is covered with a speed of 35km/h and the other half time at 10km/h. If the total distance travelled is 90 km then What was the car’s average speed (approximately) during his entire journey?**

Solution:

Half of total distance = 45km, Speed = 50km/h

Time taken = 45km/50 = 0.9 hr = 0.9 x 60 = 54 minutes

Let time taken for remaining 45 km be T

And, distance via speed 10km/hr by D

Then, ATQ, T/2 = (45 - D) / 35 …(1)

And, T/2 = (D) / 10 …(2)

Equating equations 1 & 2,

(45 - D) / 35 = (D) / 10

10(45 - D) = 35D

45D = 450 or,

D = 10 km, T= 2 hour

Total time = 2+.9= 2.9 hour

Average speed= 90/2.9=31km/hr

**Type 4**

- Suppose a Person covers a certain distance at
**x km/hr**and an equal distance at**y km/hr.**

Then, the average speed for the complete Journey: **2xy/(x + y) **

**Question: A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is?**

#### Solution:

Let x and y be the average speed for the same distance in two different times.

Then, average speed = (2xy)/(x + y)

A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour.

The average speed of train = (2 × 20 × 30)/(20 + 30) = 24 km/hr

**Question: A boy goes to school at a speed of 3 km per hr and returns to the village at a speed of 2 km per hr. If he takes 5 hrs in all, what is the distance between the village and the school?**

**Solution**:

Let the required distance be x km.

Then time taken during the first journey = x/3 hr.

and time taken during the second journey = x/2 hr.

x/3 + x/2 = 5 => (2x + 3x) / 6 = 5

=> 5x = 30.

=> x = 6

Required distance = 6 km.

**==========================================**

## Comments

write a comment