Speed Time and Distance example shortcut tricks are very important thing to know for your exams. Time takes a huge part in competitive exams. If you manage your time then you can do well in those exams. Most of us miss that part. We provide examples on Speed Time and Distance shortcut tricks here in this page below. These shortcut tricks cover all sorts of tricks on Speed Time and Distance. Visitors please read carefully all shortcut examples. These examples here will help you to better understand shortcut tricks on speed time and distance.

Before starting anything just do a math practice set. Then find out twenty math problems related to this topic and write those on a paper. Solve first ten math problems according to basic math formula. You also need to keep track of the time. Write down the time taken by you to solve those questions. Now go through our page for speed time and distance shortcut trick. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of timing. You will surely see the improvement in your timing this time. But this is not all you want. You need more practice to improve your timing more.

Few Important things to Remember

You all know that math portion is very much important in competitive exams. That doesn’t mean that other sections are not so important. But if you need a good score in exam then you have to score good in maths. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks.

But, so many people can’t do this. So Speed Time and Distance shortcut tricks here for those people. Here in this page we try to put all types of shortcut tricks on Speed Time and Distance. But it possible we miss any. We appreciate if you share that with us. Your little help will help others.

Speed Time and Distance Example 4

This is the basic theory of Speed Time and Distance which is applied in question to obtain answers here is Speed Time and Distance Methods of example 4 in different form of examples. In maths exam papers there are two or three question are given from this chapter.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

Speed Time and Distance Example #1

Rajib walks 170 meters every day. How many kilometers will he made in 4 weeks?

4.042 km

4.760 km

5.392 km

5.960 km

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve 4 Weeks = 28 Days. 170 x 28 = 4760 meter.

Converting meter to km: 4760 / 1000 = 4.760 km.

Rough Workspace

Speed Time and Distance Example #2

A bus covers the initial 46 km in 42 minutes and remaining 26 km covers in 38 minutes. Find the average speed of the bus.

54 km/hr

59 km/hr

63 km/hr

68 km/hr

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve Total distance covered by the bus: 46 + 26 = 72 km.

Total time taken to cover the distance: 42 + 38 = 80 min.

So, the average speed of the bus is: 72 x 60 / 80 = 54 km/hr.

Rough Workspace

Speed Time and Distance Example #3

Ajay passes a car in 16 seconds. The same car passes a lamp post in 6 seconds. Find respective ratio between speed of Ajay and speed of the car.

7 : 2

2 : 7

3 : 8

8 : 3

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (C)

How to Solve 1 / 16 : 1 / 6 = 3 : 8

So, the respective ratio between speed of Ajay and speed of the car is 3 : 8.

Rough Workspace

Speed Time and Distance Example #4

A fast train covers a distance in 40 min. If it runs at a speed of 45 km/hr on an average. The speed at which the train must run to reduce the time of journey to 30 min will be?

50 km/hr

60 km/hr

70 km/hr

80 km/hr

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve The train covers the distance in 40 minutes. OR, Time = 40 / 60 = 2 / 3 hr.

Given speed = 45 km/hr. We know, Distance = Speed x Time. So, Distance = 45 x (2 / 3) = 30 km.

Now, the train covers the distance in 30 minutes. OR, Time = 30 / 60 = 1 / 2 hr.

So, the new average speed will be: Speed = Distance / Time Speed = 30 x 2 = 60 km/hr.

Rough Workspace

Example #5

A man walking at the speed of 4 km/hr to cover a certain distance in 2 hr 45 min. Running at a speed of 16.5 km/hr the man will cover the same distance in what time?

18 min

25 min

30 min

40 min

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (D)

How to Solve We know that, Distance = Speed x Time So, Time = 2 hr 45 min then, 2 x 60 min + 45 min then, 165 min OR 165 / 60 therefore, 11 / 4 hr.

Distance = 4 x 11 / 4 = 11 km.

So, the time required to cover the distance at the speed of 16.5 km/hr is: Time = Distance / Speed So, Time = 11 / 16.5 therefore, Time = 40 min.

Rough Workspace

Example #6

A train traveling at a speed of 90 km/hr overtakes a bike traveling at the speed of 54 km/hr in 30 seconds. What is the length of the train in meters?

300 Meters

320 Meters

360 Meters

395 Meters

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve The distance travailed by the train overtaking the bike is the same as the length of the train. Also remember that both the objects are moving in same direction.

So, ( 90 – 54 ) = 36 km/hr.

Now, converting km/hr to min/sec: 36 x (5 / 18) = 10 min/sec.

So, distance travailed in 30 seconds: 10 x 30 = 300 meters.

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Hello I find this site quite helpful in maths problem solving. I want to ask a sum. A policeman sees a robber at a distance of 125 meters. The robber is running at speed of 9km/hr and the police chases him down at a speed of 10 km/hr. How much the police has to run to catch the robber

plzz solve my question x and y walks around a circular cource 100km from the same point if they walks at 5kmph and 7kmph respectively in the same direction when will they meet

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thankssssss

this site is very useful

Hello I find this site quite helpful in maths problem solving. I want to ask a sum.

A policeman sees a robber at a distance of 125 meters. The robber is running at speed of 9km/hr and the police chases him down at a speed of 10 km/hr. How much the police has to run to catch the robber

Answer will be 1.3 Km.

//police take a 1.3 Km. to catch the robber.

It is 1.25.

Relative speed=1km/hr

distance=0.125km

time taken=0.125hr

Distance travelled in 0.125hr by police= t*d

=0.125*10=1.25km

i want problems on data interpretation

thanks i have learnt alot form here….thanking you so much…

plzz solve my question

x and y walks around a circular cource 100km from the same point if they walks at 5kmph and 7kmph respectively in the same direction when will they meet

Its very Nice & Helpful