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## Speed Time and Distance Methods shortcut tricks

Speed Time and Distance Methods tricks are very important thing to know for your exams. Competitive exams are all about time. If you know how to manage time then you will surely do great in your exam. Most of us skip that part. Few examples on **speed time and distance shortcuts**

is given in this page below. These shortcut tricks cover all sorts of tricks on Speed Time and Distance. Visitors are requested to carefully read all shortcut examples. These examples here will help you to better understand shortcut tricks on Speed Time and Distance Methods.

Before starting anything just do a math practice set. Write down twenty math problems related to Speed Time and Distance Methods on a page. Using basic math formula do first ten maths of that page. You also need to keep track of timing. Write down the time taken by you to solve those questions. Now read our examples on speed time and distance shortcut tricks and practice few questions. After finishing this do remaining questions using Speed Time and Distance shortcut tricks. Again keep track of Timing. The timing will be surely improved this time. But this is not all you need. You need to practice more to improve your timing more.

### Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. That doesn’t mean that other sections are not so important. But only math portion can leads you to a good score. Only practice and practice can give you a good score. The only thing you need to do is to do your math problems correctly and 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. For those we prepared this speed time and distance shortcut tricks. We always try to put all shortcut methods of the given topic. But it possible we miss any. We appreciate if you share that with us. Your little help will help others.

Now we will discuss some basic ideas of **Speed Time and Distance**. On the basis of these ideas we will learn trick and tips of shortcut speed time and distance. If you think that **how to solve speed time and distance questions using speed time and distance shortcut tricks**, then further studies will help you to do so.

### Speed Time and Distance

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

Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks are comes into action.

### Formula

**Distance = Speed x Time**

Generally use this formula we can find the distance of any running train, car etc. If Speed of train or car is given with time and using this multiplying this two we can find the distance covered by train. The unit of distance is kilometers, meters, miles, etc.

To find the speed we can divide Distances by Time.

The unit of Speed has written in fractions that is Km/hr ( written as Distance unit in numerator and Time units in the denominator ), Suppose 35 Km/hr.

pls tell the question that where the two train go in same direction one is much more speed than the other. whose speed are different find length of the train, speed, time to cross the train etc

First train length 50mt.

Second Train lenth =X mt (wich we have find)

Speed of First Train = 50Km/hr

Speend of 2nd train= 30 km/hr

Time=36 Seconds ( fast train take to cross the slower train)

Formula is -Distance = Time *Speed

=> 50+x = 36*(50-30)*5/18 => 50+x= 200

and x= 150mr ans.

just subtract 2 speeds if same direction,direction can always be added only and calc time as usual

Relative speed of two trains means the substraction of two speeds of those trains = speed of fast train – speed of slower train =x -y (let)

Now , addition of two train’s length = a + b (let)

So time for over take = (a +b)÷ (x+y)

Time taken to catch/cross = length/ relative speed

For relative velocity,

A) if it is travel in same direction =

Relative speed = u-v

B) if it is in opposite direction,

Relative speed = u+v

Length of each train should be added.

Depends upon the given data, you can calculate whatever you want.

so nice example

Please provide me the answer of this question. A non stop train starts from Delhi at 6:00am and reaches Agra at 10:00am. The other nonstop train starts from Agra at 8:00am and reaches Delhi at 11:30am. If the distance between Agra and Delhi is 200 km, than what time the two train meet each other?

whats the answer?

Trains meets on 9:42 am

after 1hour 52mins.

total distance-200km…

speed of train from delhi(d)-200/4=50kmph

speed of train from agra(a)-200/(7/2)=(400/7)kmph

from 6:00 to 8:00 train from (d) would hav travelled =50*2=100km

now at 8:00 train at (a) also starts ….now distance left is 100km and both are travelling opp to each other:

time for their encounter: 100/(50+(400/7)) =14/15 hrs=42min

therefore total time from 6:00=2hrs+42min=8:42

ans must be 8:42 and not 9:42

at 8:56 they both will meet

as train x has sp=50kmph n y has sp=57+(1/7)

both at 8 am will at 100 km covered by x n 100km left where as y will start

so time to meet would be 100/(50+57+(1/7))=(14/15)*60 = 56 min

hence 8:56 am they’ll meet

Hi can you pls explain me brifely i dint understand

yah 14/15 kaisay aaya explain plz

Distance =200 KM

Time taken by train A= 4hr

Time taken by train B =3h 30 mins(7/4hr)

Speed of train A = 200/4=50 Km/h

Speed of train B = (200)/(7/2)=400/7

Distance coverd by Train A =2*50=100KM

When train B starts then remaing distance is 100KM

Since they are going to meet each other therefore we are adding the speed of the trains.

ie 50+400/7=750/7

Distance to be covered = 100 KM

Time taken to cover that distance = 100/(750/7)

=>700/750 which gives 14/15.

14/15 gives .933 hrs

Converting it in minutes

.933*60=55.98 minutes.

A man in a train notices that he can count 21telephone posts in 2 minutes. If they are known to be 50 mtr apart, then at what speed is the train traveling.

.5km/min

21×50=1050 mt

Speed = 1050÷2×60 mt/sec

Or

Speed = 1050×18÷120×5 km/hr.

Speed = 31.5 km/hr.

kapil,

I think u should not consider 21 posts. U should take only 20.

for example, if it was 3 posts, like below,

I I I

the gap between them will be only 2 which means,

2 X 50 = 100 mt.

Nice

sonal runs 220 kilometeres everyday.how many kilometeres will she run in 3 weeks?

he will die on first day 😛

Nice..sir can you explain some other tricks for a problem

sir, please can u give some tricks to solve problems of logarithmic without using calculator and log table for cracking engineering appti test

walking at 5/7th of your usual speed , you will reach the market 16min late . what is the usual time taken by you to reach the market

2 parts out of 7 parts of your sped = 16 minutes

1 part=8 minutes

so your usual speed is 5*8 minutes=40 minutes

you walked for 7*8=56 minutes.

hence, u were late by 16 minutes.

=16/(7/5 – 1)

=16/(2/5)

=40min

hey guys plz tell me when should we use +ve and -ve operations while solving quadratic equations wrd problems!

5/7th of your usual speed made you late by 16 minutes

that is., 2 parts out of 7 parts is = 16 minutes

1 part of your speed out of 7 parts is 8 minutes.

so, your usual speed is 5parts*8 minutes= 40 minutes

If a policeman travels 12klm at 3kph, how long does it take him to do 14klm?

3.5 hr

Pls solve this questions a train takes 18 sec to pass completely through a station 162m Long and 15 sec through another station 120 m long.the lenght of the train is ? Ans: 90 m

Assuming that speed of train will be same.

Let length of train be X m.

So

(X+162)/18 = (X+120)/15

=>X=90 m.

awesome site.thankQ very much sir…

a boy has to reach 200km by bicycle from point A to B. at the point X km the bicycle has damaged. then he travelled with 3/4th speed he reached 1hr late. assume if bicycle is damaged before 30km then he will reach 12mints late. find the actual speed of bicycle and X value.

a boy has to reach 200km by bicycle from point A to B. at the point X km the bicycle has damaged. then he travelled with 3/4th speed he reached 1hr late. assume if bicycle is damaged before 30km then he will reach 12mints late. find the actual speed of bicycle and X value.

actual speed = 3,34 km/min and x is 150 kms

A training at the speed of 20m/s.crossa ploe in 24secs less than the timeit required cross a plot from thrice is the length the same speed. what is the length of the train?

please solve this sum

a boy and a girl start waling towards each other at 4 am at the speed of 5kmph and 3kmph respectively. They were initially 28km apart. At what time do they meet?

A train delhi to agra start at 5 am and takes 4 hrs to reach and another train agra to delhi start at 7 am and takes 3 and half hours to reach…at which tym they met with each other? Anyone …try

at 7:51 am

if there are three speeds given and we have to find its average how can we do so

The speed of train is 90kmph it crosses a platform of 150m in 10secs.Find the length of train?

A car travels 840Km at slower speed. If car increases its speed by 10Km/hr it will reach 2 hours early. what is the actual speed of the car?