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Important Concepts & Short Tricks on Speed, Distance & Time Posted in Quants, Study Notes By GradeStack On October 27, 2015

Dear Readers,

We are providing you Important Short Tricks on Speed, Distance & Time which are usually asked in SSC

Exams. Use these below given short cuts to solve questions within minimum time. These shortcuts will

be very helpful for your upcoming SSC CHSL Exam 2015.

Important formulae and facts of Time and Distance

Speed is a very basic concept in motion which is all about how fast or slow any object moves. We define

speed as distance divided by time.

Distance is directly proportional to Velocity when time is constant.

Speed Distance Time formula mathematically written as:‐ Speed = distance/time

Formula of Time :‐time = distance/ Speed

So Formula of time is, time is equal to distance upon speed.

Formula of Distance:‐Distance = (Speed * Time)

Distance = Rate x Time

To find rate, divide through on both sides by time:

Rate = Distance/Time

Rate is distance (given in units such as miles, feet, kilometers, meters, etc.) divided by time

(hours, minutes, seconds, etc.). Rate can always be written as a fraction that has distance units in

the numerator and time units in the denominator, e.g., 25 miles/1 hour.

So distance is simply speed into time.

Note: All three formulae that formula of speed, formula of time and formula of distance areinterrelated.

Convert from kph (km/h) to mps(m/sec)

For converting kph(kilometre per hour) to mps(meter per second) we use following formula

x km/hr =( x ∗5/18) m/sec

Convert from mps(m/sec) to kph(km/h)

For converting mps(meter per second) to kph(kilometre per hour) we use following formula

x m/sec= X *(18/5) km/h

If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by then to cover the

same distance is :1/a : 1/b or b : a

Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then,

the average speed during the whole journey is :‐ 2xy/(x + y)

Relation between time, distance and speed: Speed is distance covered by a moving object in unit

time: Speed= Distance covered/ Time Taken

Rule : 1: Ratio of the varying components when other is constant: Consider 2 objects A and Bhaving

speed Sa, Sb.

Let the distance travelled by them are Da and Db respectively and time taken to cover these distances

be Ta and Tb respectively.

Let’s see the relation between time, distance and speed when one of them is kept constant

1. When speed is constant distance covered by the object is directly proportional to the time

taken.

ie; If Sa=Sb then Da/Db = Ta/Tb

2. When time is constant speed is directly proportional to the distance travelled. ie; If Ta=Tb

then Sa/Sb=Da/Db

Wednesday, December 16, 2015 12:21 PM

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3. When distance is constant speed is inversely proportional to the time taken ie if speed

increases then time taken to cover the distance decreases. ie; If Da=Db then Sa/Sb= Tb/Ta

Rule 2: We know that when distance travelled is constant, speed of the object is inversely

proportional to time taken

1. If the speeds given are in Harmonic progression or HP then the corresponding time taken will be in

Arithmetic progression or AP

2. If the speeds given are in AP then the corresponding time taken is in HP

Distance Constant

If distance travelled for each part of the journey, ie d1=d2=d3=…=dn=d, then average speed of the

object is Harmonic Mean of speeds.

Let each distance be covered with speeds s1,s2,…sn in t1,t2,…tn times respectively.

Then t1 =d/s1

t2 = d/s2

tn =d/sn

then, Average Speed= [(d + d + d+ … ntimes)]/ [d/s1 + d/s2+ d/s3+ … d/sn

Average Speed= (n)/[(1/s1 + 1/s2+ …. 1/sn)]

Time Constant

If time taken to travel each part of the journey, ie t1=t2=t3=…tn=t, then average speed of the

object is Arithmetic

Let distance of parts of the journey be d1,d2,d3,…dn and let them be covered with speed s1,s2,s3,…sn

respectively. Then d1=s1 t , d2=s2t, d3=s3t, … dn=snt

then , Average Speed= [(s1/t+ s2/t+ …. sn/t)/(t + t+ … ntimes)]

Average Speed=( s1+ s2+s3+ … + sn)/n

Relative Speed

If two objects are moving in same direction with speeds a andb then their relative speed is|a‐b|

If two objects are moving is opposite direction with speeds a and b then their relative speed is

(a+b)

Some Question on Above formulas

Ques 1: – A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?

Sol:: 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.

Ques 2: – A car travels along four sides of a square at speeds of 200, 400, 600 and 800 km/hr. Find

average speed.?

Sol: Let x km be the side of square and y km/hr be average speed

Using basic formula, Time = Total Distance / Average Speed

x/200 + x/400 + x/600 + x/800= 4x/y

=25x/ 2400 = 4x/ y

= y= 384

Average speed = 384 km/hr

Ques 3: A motor car does a journey in 10 hrs, the first half at 21 kmph and the second half at 24kmph.

Find the distance?

Sol:

Distance = (2 x 10 x 21 x 24) / (21+24)

= 10080 / 45

= 224 km.

Ques 4:A boy goes to school at a speed of 3 kmph and returns to the village at a speed of 2 kmph. If he

takes 5 hrs in all, what is the distance between the village and the school?

Sol : 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


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Required distance = 6 km.

Ques 5: Walking ¾ of his speed, a person is 10 min late to his office. Find his usual time to cover the

distance?

Sol : Usual time = Late time / {1/ (3/4) – 1)

= 10 / (4/3 ‐1 )

= 10 / (1/3)

= 30 minutes.

We hope that the post would have cleared all your doubts related to the topic.

The terms time and distance are related to the speed of a moving object.

Time and Distance: 8 Shortcuts & Tricks explained with solved examples


Speed: Speed is defined as the distance covered by an object in unit time.

Speed = Distance/Time

Some Important Facts:

* Distance travelled is proportional to the speed of the object if the time is keptconstant.

* Distance travelled is proportional to the time taken if speed of object is kept constant.

* Speed is inversely proportional to the time taken if the distance covered is keptconstant.

* If the ratio of two speeds for same distance is a:b then the ratio of time taken to coverthe distance is b:a

Relative Speed:

i. If two objects are moving in same direction with speeds of x and y then their relativespeed is (x - y)

ii. If two objects are moving in opposite direction with speeds of x and y then theirrelative speed is (x + y)

Unit Conversion:

i. To convert 'X' Km/hr into m/s

- Multiply X with 5/18

ii. To convert 'x' m/s into Km/hr

- Multiply x with 18/5

Some Important Shortcut Formulas

Trick-1:

If some distance is travelled at x km/hr and the same distance is travelled at y

km/hr then the average speed during the whole journey is given by

Average speed = (2*x*y)/(x+y)

Ex: A peson travelled a distance with 8 km/hr and return the journey with 4km/hr.then,the average speed during the journey ?

a. 5 km/hr b. 16/3 km/hr

c. 6 km/hr d. 12 km/hr


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Trick-2:

If a person travels a certain distance at x km/hr and returns at y km/hr, if the time

taken to the whole journey is T hours then the one way distance is given by

D=[T*x*y]/(x+y)

Ex: If a person travels certain distance at 6 Km/hr and returns at 8 Km/hr, if the timetaken to the whole journey in 7 hours.then,the one way distance ?

a. 16 b. 24 c. 8 d. 32

Sol: D= (7*8*6)/(8+6)= 24 km

Trick-3:

If a car does a journey in ‘T‛ hrs, the first half at ‘x‛ km/hr and the second half at

‘y‛ km/hr. The total distance covered by the car is :

(2 * Time * x * y )/ (x + y).

Ex: A motorcar does a journey in 10 hrs, the first half at 21 kmph and the second half at24 kmph. Find the distance?

a. 124 b. 224 c. 225 d. 125

Sol: Distance = (2 x 10 x 21 x 24) / (21+24) = 10080 / 45 = 224 km.

Trick-4:

If the same distance is covered at two different speeds S1 and S2 and the time

taken to cover the distance are T1 and T2 ,then the distance is given by

D = [(s1*s2)/(s1-s2)]*(t2-t1)

Ex: A person travelled a distance with two different speeds 5 Km/hr and 6 km/hr and timetaken to cover distance are 3 hrs and 4 hrs.the distance travelled by person ?

a. 30 km b. 20 km c. 15 km d. 38 km

Sol: D= [(5*6)/(5-6)]*(4-3) =30km

5*3*2Trick-5:

A distance covers in some time with S1 speed.if it takes T hr more to cover same

distance with S2 speed. So, the distance is

D = (T*S1*S2)/(S1-S2)

Ex: A person covers a distance with 5 km/hr in some time.if he moves with 3 km/hr speedhe covers the distance in 2 hr more.the distance travelled by person ?

a. 10 km b. 15 km c. 18 km d. 21 km

Sol: D= (2*5*3)/(5-3) = 15 km

Trick-6:

If a distance traveled with S1.then,it takes T hrs late.same distance traveled with

S2.then, it takes T hrs earlier.so,the distance is

D = (2*S1*S2*T)/(S2-S1)

Ex: A person travelled a distance with 5 km/hr then, he will take 2 hrs more.if he travelswith 7.5 km/hr then,he will reach 2 hrs earlier.the distance traveled by person ?

a. 55 km b. 60 km c. 65 km d. 70 km

Sol: D = (2*5*7.5*2)/(7.5-5) = 60 km


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Trick-7:

If a body covers part of the journey at speed p km/hr and the remaining part of the

journey at a speed q km/hr and the distances of the two parts of the journey are in

the ratio m : n, then the average speed for the entire journey is

= (m+n) pq / (mq+np).

Ex: If a Manish covers part of the journey at speed 2 km/hr and the remaining part of the journey at a speed 4 km/hr and the distances of the two parts of the journey are in theratio 3 : 2, then the average speed for the entire journey ?

a. 5 b. 2.5 c. 10 d. 7.5

Sol: = [(3+2)*2*4] / [(3*4)+(2*2)]

= 2.5 km/hr

Trick-8:

A train travelling at a speed of 'S1' kmph leaves A at 't1' hrs. and another train

travelling at speed 'S2' kmph leaves A at 't2' hrs in the same direction. Then the

meeting point's distance from starting is given by = (S1 xS2 X Difference in time)

/Difference in speed.

Ex: A train travelling 25 kmph leaves Delhi at 9 a.m. and another train travelling 35 kmphstarts at 2 p.m. in the same direction. How many km from will they be together ?

a. 437.5 b. 137.5 c. 237.5 d. 337.5

Sol : Meeting point's distance from the starting point = [25 x35 x(2p.m. - 9 a.m)] / (35-25) = (25 x35x 5) / 10 = 4375 / 10 = 437.5 km .


Some Important Facts Distance travelled is proportional to the speed of the object if the time is kept

constant. Distance travelled is proportional to the time taken if speed of object is kept

constant. Speed is inversely proportional to the time taken if the distance covered is kept

constant. If the ratio of two speeds for same distance is a:b then the ratio of time taken to

cover the distance is b:a

Relative Speed If two objects are moving in same direction with speeds of x and y then their relativespeed is (x - y)If two objects are moving is opposite direction with speeds of x and y then their relativespeed is (x + y)

Unit Conversion


Rule 1: If some distance is travelled at x km/hr and the same distance is travelled at ykm/hr then the average speed during the whole journey is given by


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Example

Rule 2: If a person travels a certain distance at x km/hr and returns at y km/hr, if thetime taken to the whole journey is T hours then the one way distance is given by

Example Rule 3: If two persons A and B start their journey at the same time from two points P andQ towards each other and after crossing each other they take a and b hours in reaching Qand P respectively, then

Example

Rule 4: If the same distance is covered at two different speeds S1 and S2 and the time

taken to cover the distance are T 1 and T 2, then the distance is given by

Quant – Tricks for Time‐Speed‐Distance‐Trains By Shubhra -

October 1, 2015

Hello Aspirants, Here we are providing you some tricks which can be helpful to solve Time Speed & Distance Questions in Quant Section.

Type 1: When you are given two different speeds (s1 and s2) for travelling througha certain distance, and total time (t) for these two journeys: Formula:

Distance = (S1*S2)/[(S1+S2)t]

Ex 1: A boy/train goes from A to B at 3 km/hr, back from B to A at 2 km/hr. Totaltime for these two journeys is 5 hours, then distance from A to B is given by: Sol: Distance = Product of speeds/Addition of speeds * Time D = 3*2/(3+2) * 5 = 6 km D = 6km

Ex 2: A boy/train travels from A to B. He covers half distance at 3 km/hr, and otherhalf at 2 km/hr. Total time taken is 5 hours, then distance from A to B is given by: Sol:

So this time total distance from A to B = (2* Product of speeds)/(Addition of speeds * Time) D = 2 * [3*2/(3+2) * 5 ]= 12 km D = 12km

Ex 3: A boy/train travels from A to B. He covers 2/3rd distance at 2 km/hr, andrest at 3 km/hr. Total time taken is 24 hours, then distance from A to B is given by: Sol: S1 = 2 km/hr, S2 = 3 km/hr, Total time, T = 24 hrs Distance left = 1 – 2/3 = 1/3 Reciprocate both distances, i.e. D1 = 3/2 and other D2 = 3/1 Distance from A to B = D1*S1 * D2*S2/ (D1*S1 + D2*S2) * T i.e. 3/2*2 * 3/1*3/ (3/2*2 + 3/1*2) * 24 = 54 km

http://www.affairscloud.com/author/shubhra/


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D = 54km

Type 2: When you are given two different speeds (s1 and s2) for travelling througha certain distance, and total difference in time (t) is given for these two

journeys: Formula:

Distance = (S1*S2)/[(S1-S2)t]

Ex 1: A boy/train goes from A to B. If speed is 30 km/hr, he/it is late by 10minutes. If speed is 40 km/hr, he/it reaches 5 minutes earlier, then distance from Ato B is given by: Sol: Difference in time = 10 – (-5) = 15 minutes or 1/4 hr……..earlier(-5) Distance = Product of speeds/Difference of speeds * Difference in time = 30*40/(40-30) * 1/4 = 30 km D = 30km

Ex 2: A train leaves Delhi at 9 AM at 25 km/hr, another train at 35 km/hr leavesDelhi at 2 PM in same direction. How many kilometres from Delhi they will betogether? Sol: Difference in times = 2 PM – 9 AM = 5 hrs Distance from Delhi = 25*35/(35-25) * 5 D = 17.5km

Type 3: When different speed and time are given to find the original time. Formula: Original time = (Reciprocal of given speed*t) – t(late) Original time = (Reciprocal of given speed*t) + t(early)

Ex 1: Walking at 3/4th of usual speed, person is late by 10 minutes. Find the usualtime. Sol: Let t be usual time. Reciprocate speed for time, i.e. 4/3 * t Person is late by 10 minutes, so (4/3 * t) – t = 10 t = 30 min

Type 4: To find the time of halts when the speed with stoppages (s1)and speedwithout stoppages (s2) is given. Formula: Halt time = (s2-s1)/s2

Ex 1: A man/train leaves from A to B. Speed with stoppages = 60 km/hr, speedwithout stoppages = 80 km/hr. Find the time man/train stop? Sol: Minutes per hour the man/ train stops = (80-60)/80 = 1/4 hr = 15 min Time = 15 min

Type 5 : A & B start at same time towards each other. After meeting each other,they reach their destinations after ‘a’ and ‘b’ hrs respectively.

Formula: A’s Speed : B’s Speed = √b : √a

Ex 1: Two, trains, one from Delhi to Mumbai and the other from Mumbai to Delhi,start simultaneously. After they meet, the trains reach their destinations after 4hours and 9 hours respectively. The ratio of their speeds is: Sol: A:B = √b : √a = 3 : 2 A:B = 3:2

Type 6 : A man takes ‘a’ hrs to walk to a certain distance and ride back.

1. If riding both ways takes him b hrs, then Walking both ways takes him [a + (a-b)] hrs


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2. If walking both ways takes him b hrs, then Riding both ways takes him [a – (b-a)] hrs

Ex 1: A man takes 6 hrs to walk to a certain distance and ride back. If riding bothways takes him 4 hrs, then Walking both ways takes him how much time? Sol: T = [a + (a-b)] hrs T = 6 +2 T = 8hrs

Ex 2: A man takes 5 hrs to walk to a certain distance and ride back. If walking bothways takes him 7 hrs, then Riding both ways takes him how much time? Sol: T = [a – (b-a)] hrs T = 5 – 2 T = 3hrs

Type 7 : A man/train leaves from A at x1 AM and reaches Q at y1 AM. Anotherman/train leaves from Q at x2 AM and reaches P at y2 PM. When they willmeet? Formula:

They will meet at = (x1+y2)/2

Ex 1: A man/train leaves from P at 6 AM and reaches Q at 10 AM. Anotherman/train leaves from Q at 8 AM and reaches P at 12 PM. When they will meet

Sol:Since both takes 4 hrs to reach their destination, so they will exactly at the middleof 6 AM and 12 PM, i.e. at 9 AM.

Shortcut Tips To Solve Time and Distance Problems With Examples iruvuru sai Aptitude No comments

Shortcut Tips To Solve Problems On Time and Distance

problems on Time and Distance are Usually asked in many of the CompetitiveExaminations. Since these are the important section in which Candidates are going to be tested on their Problem Solving Skills.Some might be knowing how to solve these kind of problems,but they may face

the problem of time management during Examination.Keeping this in view,we are providing you withsome simple tricks which helps you to solve these problems within less than a minute.usually these Timeand distance Questions are lengthy.By looking at the question itself one feels that it is waste of attempting the question.But there lies a very simple logic in these questions and you just need to find thelogic and apply the simple formulae. All you need to do is to Understand the concept of the givenQuestion and apply the terms in the Formulae which we are providing here based on the questionrequirement.Practice this approach,and you can play with these type of Questions in theexaminations……

These are the very basic formulae in Time and Distance which one need to know: Basic Formulae :

1.Velocity (speed) =Distance / Time 2.Distance= Speed × Time 3.Time= Distance / Speed Based on the above Formulae ,we can obtain these Relationship

T1 ×S1 = T2× S2 when distance remains constant D1/S1 = D2/S2 When Time remains constant D1/T1 = D2/T2

There you are. With this basic idea,we can solve many problems depending on the question.Oneneed to study the question very carefully,there will be a simple logic involved.once you find the logic,youcan solve them within a few seconds.

http://www.sparkyourcareer.in/2014/03/shortcut-tips-to-solve-time-and.html#comment-formhttp://www.sparkyourcareer.in/search/label/Aptitudehttp://www.sparkyourcareer.in/2014/03/shortcut-tips-to-solve-time-and.html


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Rule 1: 1. Two Vehicals are travelling in any direction but at the end meet at some place (Important thing to

be noted is that time is constant for both the vehicles i.e. they take same time to reach the location).Assumption in this case is that both the vehicals are started at the same time. In case,if they start atdifferent times, we can still solve it by considering the time from which the later vehicle started.Concept to be followed is D1/S1 = D2/S2.

2. Two cars are traveling from one location and stopping at another location or coming back to the samelocation [Important thing to be noted is that the distance travelled by both the cars should be same].Concept to be used is T1*S1 = T2*S2.

Example 1:

Abhinav uses his car to go to office which is 50 km from his place. It takes him 1 hour to reach office if he travels at usual speed of 50Km per hour. Suppose, on a particular day he managed to reach office in30 min,what is the speed with which he drove his car?

Solution: Study the question carefully.If we observe the above question,distance is constant Applying the concept,when distance is constant

T1×s1 = T2×s2 1hr×50 = ½ hr ×s2 S2 =100 km per hr

Hence the speed with which he drove the car is 100 km per hr.

Unit conversion :

For converting km/hr into m/sec ,Multiply with 5/18

For Converting m/sec into km/hr Multiply with 18/5 Relative Speed

1. If two objects are moving in same direction with speeds of x and y then theirrelative speed is (x - y)

2. If two objects are moving in opposite direction with speeds of x and y then theirrelative speed is (x + y) RULE 2: If a person travels a certain distance at x km/hr and returns at y km/hr, if the time taken to the whole

journey is T hours then the one way distance is given by One way distance =T (xy/x+y)

Example 2: Binnu goes to office at the speed of 15 km/hr and returns to her home at the speed of 10km/hr.If she takes 4 hours in all, What is the distance between her Office and home? Explanation : consider, x=15 km/hr Y=10 km/hr and T=4 hrs One way distance =T (xy/x+y)

= 4 (1500/25) =240

So,the distance between the Office and home is 240 km.

Rule 3 : If two persons A and B start their journey at the same time from two points P and Q towards each other


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and after crossing each other they take a and b hours in reaching Q and P respectively, then

Example 3 : Anchun and deepu started their journey from two different places towards each other’s place.After c rossing each other,they completes their journey in 1 and 4 hours respectively. Find thespeed of deepu if speed of Anchun is 40 km/hr. Explanation : Let us consider A=Anchun and B= deepu A=1 and b=4

40/deepu's speed =√4/√1 Hence, deepu’s speed = 80 km/hr

Rule 3 :If the same distance is covered at two different speeds S1 and S2 and the time taken to cover the

distance are T1 and T2, then the distance is given by

Example 4 : Two buses travel the same distance at the speed of 60 kmph and 30 kmph. Find the distancewhen the time taken by both trucks are 2hr and 1 hr respectively Explanation : Let us substutute the given terms in the formula S1=60 kmph S2=30 kmph T1-T2= 1 Then ,applying these in the formula we get

=(1800/30)× (1 )

= 60 km

Example:5 :- A person travels a total journey of 240 kms. He travels his half distance at a speed of 40km/h and half of the remaining at a speed of 30 km/h and the last remaining at a speed of 20 km/h. whatis his average speed during the whole journey?

Let us Solve this Using simple Trick

D1=first half distance of 240 kms=120 km, speed=x=40km/h D2=half of the remaining(120)=60 km, speed=y= 30km/h D3=the last remaining=60 km, speed=z=20km/h

Here is a tricky rule for solving this type of question very quickly.

Rule 5 :- If three different distances D1, D2, D3 are travelled at 3 different speeds i.e. x,y,zrespectively, then Average speed=(D1+D2+D3)/{(D1/x)+(D2/y)+(D3/z)} km/h

Which is obtained from the basic formulae, speed= distance/time Now,applying the given terms in the formulae we get


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Average speed= (120+60+60)/ {(120/40)+(60/30)+(60/20)} km/h

Average speed= 240/(3+2+3)=240/8 km/h

Average speed=30 km/h.

Example 6: In travelling from A to B, a train covers a distance of 40 km with a speed of 20km/hr andnext 60 km with a speed of 30 km/hr .Calculate the average speed of train during the whole journey? Rule 6:

Explanation : Now here is a shortcut trick rule for this question.

This is similar to the above example, When two distances are covered with two different speeds,then theaverage speed is obtained by using this formula

Then average speed={(d1+d2)XY} /{(d1Y+d2X)}

Explanation for the above Formula:

Time= distance/ speed =( d1/x + d2/y)

Speed = Distance/ Time= ( (d1+d2) / (d1/x + d2/y) )

If d1=first distance=40 km

And d2=second distance=60 km And X=first speed=20 km/hr And Y=second speed=30 km/hr

Now,putting the values of d1, d2,X and Y from the given example in the above formula, we get

Average speed= {(40+60)20x30}/{(40x30)+(60x20)}

Average speed=(100x20x30)/(1200+1200) =(100x20x30)/2400 =(100x6)/24 =25 km/hr

Example 7: A student reaches his home from school late by 20 minutes from his scheduled time if hewalks at a speed of 6 km/h,but if he walks at a speed of 8 km/h, he will reach his home 5 minutes early.What will be the distance from school to home.

Rule 7 :


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Explanation : If a person reaches his destination in t1 time early by walking at a speed of x and t2time later by walking at a speed of y, then the distance between both the places is calculated by usingthis formula

= {x y (t1+t2)}/{(x-y)60}

Now substute the terms in the above formula Now t1=20 minutes, t2=5 minutes And x=6 km/h, y= 8 km/h

The distance will be then

={(20+5) 6X8}/(8-6)60 = (25X48)/2X60 = 10 km is the answer

Example 8 : A man travelled from the Village to station at the rate of 20 kmph and walked back at therate of 5kmph.Find the average speed during the whole journey?

Rule 8 : Explanation : Suppose a man covers a certain distance at x km/hr and an equal distance at ykm/hr.Then,the average speed during the Whole journey is given by the formula

{ 2xy/(x +y) } km/hr On substuting the terms in the above formula we get

X = 20 kmph Y= 5 kmph Average Speed = 2xy/ (x+y)

=2*20*5/(20+5) =200/25

=8 km/hr

Time and Distance

Time and Distance Speed = Distance/Time

Distance = Speed × Time

Time = Distance/Speed

If the new speed is x/y of the original speed, then the change in time to cover the same distannce

is given by

Change in time =

(yx−1)

X Original Time

To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)

To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)


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1. If some distance is travelled at x km/hr and the same distance is travelled at y km/hr then the

average speed during the whole journey is given by:

Average Speed=

2. If a person travels a certain distance at x km/hr and returns at y km/hr, if the time taken to the

whole journey is T hours then the one way distance is given by

One Way Distance

3. If two persons A and B start their journey at the same time from two points P and Q towards each

other and after crossing each other they take a and b hours in reaching Q and P respectively, then

4. If the ratio of the speeds of A and B is a : b, thenthe ratio of the times taken by them to cover the

same distance is

5. If the same distance is covered at two different speeds S1 and S2 and the time taken to cover the

distance are T1 and T2, then the distance is given by

Questions Asked from this Chapter:

Question Type 1: Based on basic concept of Time & Distance

Example:

A man crosses a road 250 metre wide in 75 seconds. His Speed in Km/hr is:

a) 10b) 12

c) 12.5

d) 15

Solution:


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Question Type 2: Based on Train and Platform/Bridge Example:

A train 200 metre long running at 36 Kmph takes 55 seconds to cross a bridge. The length of the bridge

is:

a) 375 m

b) 300 m

c) 350 m

d) 325 m

Solution:

Question Type 3: Based on Train and a man/Pole/Signal

Example:

A train 100 m long is running at the speed of 30 Km/hr. The time(in seconds) in which it will pass a man

standing near the railway line is:a) 10

b) 11

c) 12

d) 15

Solution:

Question Type 4: Based on Two trains crossing each other in opposite direction

Example:

Two trains 105 metres and 90 metres long, run at the speed of 45 Km/hr and 72 Km/hr respectively, in

opposite directions on parallel tracks. The time which they take to cross each other, is:

a) 8 seconds

b) 6 seconds

c) 7 seconds

d) 5 seconds

Solution:


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Question Type 5: Based on Two trains travelling in same direction at different speed

Example:

Two trains of equal length are running on parallel lines in the same direction at 46 Km/h and 36 Km/h.

The faster train passes the slower train in 36 seconds. The length of each train is:

a) 82 m

b) 50 m

c) 80 m

d) 72 m

Solution:

Question Type 6: Based on a train/man/car changes his speed, then he reaches his destination

before latter

Example:

The student go to school at the rate of 2.5 Km/h and reaches 6 minutes late. If he travels at the speed of

3 Km/h, he is 10 minute early. The distance between the school and his house is:

a) 5

b) 4

c) 3

d) 1

Solution:

Question Type 7: Based on any Train crossing both platform and man/pole at same time

Example:


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A train passes a platform 110 m long in 40 seconds and a boy standing on the platform in 30 seconds.

The length of the train is:

a) 100 m

b) 110 m

c) 220 m

d) 330 m

Solution:

Question Type 8: Based on a car travelling with x/y speed of its usual speed

Example: Walking at 6/7th of his usual speed a man is 25 minutes late. His usual time to cover this distance is:

a) 2 hours 30 minutes

b) 2 hours 15 minutes

c) 2 hours 25 minutes

d) 2 hours 10 minutes

Solution:

Question Type 9: Based on Average Speed

Example:

A train moves with a speed of 30 Kmph for 12 minutes and for next 8 minutes at a speed of 45 Kmph.

Find the average speed of the train?

a) 37.5 Kmphb) 36 Kmph

c) 48 Kmph

d) 30 Kmph

Solution:


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Question Type 10: Based on Ratio

Example:

The speed of two train are in the ratio 6:7. If the second train runs 364 Km in 4 hours, then speed of the

first train is:

a) 60 Km/h

b) 72 Km/h

c) 78 Km/h

d) 84 Km/h

Solution:

Question Type 11: Based on Races

Example:

In a race of 200 metres, B can give a start of of 10 metres to A, and C can give a start of 20 metres to B.

The start that C can give to A, in the same race is:

a) 30 m

b) 25 m

c) 29 m

d) 27 m

Solution:

A thief is spotted by a policeman from a distance of 100 metres. when the policeman starts the chase,

the thief also starts running. If the speed of the thief be 8km/hr and that of policeman 10km/hr, how far

the thief will have run before he is overtaken?

their time will same so by equating


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distanceby theif/speed of theif = distance by police/speed of police

(.1+x)/8 = x/10

by simplifying x = .5 km

hence theif will move .1+.5=.6 km = 600mt

Gateway to SBI Mains: Time, Speed & Distance ConceptsToday we'll cover Time, Speed and Distance basic concepts. The type of questions varies

from simple to quite complex. But No need to worry if you know the basics.

There are many formulas deducible based solely on the basic concepts. Today, we'll try to

know what these formulas says in general so that we don't burden ourselves remembering

them.

Time never stops!

If I'm typing all this, and I stop in the middle, then only my average speed of typing goes down.

Time never stops. With each second that pass, Present becomes past and future becomes

present. This is the first thing you need to know.

The second thing is speed.

In time-distance problems, if we take Distance as a constant thing, then speed and time becomes

variables. We change speed, and thereby we changes the time taken.

Suppose, Delhi to Agra is 120 km. And my motorcycle covers 40 km in one hour. So, how much

time I will take to reach Agra?

Simple! 3 hrs. time.

But my friend's car covers 60 km in an hour. He will take how much time?

Simple! 2 hrs. time.

Means to say, my friend will reach Agra 1 hour before me.

So, keeping the distance constant, we got two times for two speeds. The time taken is inversely

proportional to speed.

Basic formula we used here for calculation of time taken is:

Time taken = Distance/Speed

And using this formula, w e can calculate speed, or, distance, if two other things are known

Speed = Distance/Time

Distance = Speed * Time

Feel this in mind before we go further...

Let’s now entertain the concept of average speed!

Question: I travel half of my journey by Bus with speed of 60kmph and the rest half on my friend's

motorcycle with speed of 80kmph. What is my average speed of total journey?

Average speed is that speed which covers the total distance in the total time (that is, the total time


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taken to cover the distance if I go by variable speeds)

Average speed = Total distance / Total time taken

Now, here in this question, 'speed' is variable (means changing). Distance is taken constant. So,

Time taken will also be variable depending upon the speeds.

Time = Distance/speed, T= D/S

Let total distance be 2D, so that for 1st speed we have half distance 'D', and for second speed we

have second half distance 'D'

S1 = 60kmph

S2 = 80kmph

So, we have

T1 = D/60

T2 = D/80

Now, average speed = total distance / total time

Total distance = distance for 1st time + Distance for 2nd time = 2D

Total time = D/60 + D/80 = [7D/240]

Average speed w ill be = [2D] / ( [7D/240] ) = 480/7 kmph

Let’s derive this formula

Let 1st speed (60) = X

Let 2nd speed (80) = Y

T1 = D/X

T2 = D/Y

Total Distance = 2D

Total time = D/X + D/Y = [Y+X]*D/[XY]

Average Speed will be = [2D] / ( [Y+X]*D /[XY] ) = [2XY]/[X+Y]

Note: This formula we have derived taking the distances for both the speed as equal. So, if in

questions, distances varies, this formula will fail to be applicable.

If you can remember the formula, then its fine, but if not, it’s still is fine. Problem is to just find the

average speed. Our suggestion is to stick to basic concepts.

Now let's test you:

Quiz Ques 1: Uday travels one third of its journey by train with speed of 60kmph and the rest of

journey by car with speed of 80kmph. Find his average speed of his journey?

(Answer to this and all quiz questions later)

The distance of the college and home of Rajeev is 80km. One day, he was late by 1 hour than

normal time to leave for college, so he increased his speed by 4kmph and thus he reached to

college at the normal time. What is the changed speed of Rajeev?

To solve this question, first feel what the question is saying.

Distance is 80km. It is constant. Only speed is changed.


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Now, let’s say his normal speed is X kmph, Then he will reach the college in 80/X hour time.

(Equation 1)

With [X+4] speed, he will reach the college in 80/[X+4] hour time. (Equation 2)

Now the question says, he is late 1 hour but with X+4 speed, he reaches the college on time.

That means time in (Equation 1) must be 1 hour more than the time in (Equation 2)

Quiz Ques 2: Can you solve further and find the increased speed of Rajeev?

Let’s now solve a very good question which will clear many concepts in a single run!!

The distance between two places P and Q is 700km. Two persons A and B started towards Q and

P from P and Q simultaneously. The speed of A is 30kmph and speed of B is 40kmph. They meet

at a point M which lies on the way from P to Q.

(i) How long will they take to meet each other at M?

(ii) What is the ratio of PM : MQ?

(iii) What is the distance MQ?

(iv) What is the extra time needed by A to reach at Q than to reach at P by B?

(V) What is the ratio of time taken by A and B to reach their respective destinations after meeting

at M?

(vi) In how many hours will they be separated by only 560 km before meeting each other?

(vii) How long will it take to separate then by 280 km from each other when they cross M (time to

be considered after their meeting)?

First of all we'll make a diagram. Diagram making is important for solving questions like these.Diagrams makes us feel the question a little more clearly.

The concept of relative motion is entertained here.

By relative motion, we means the motion of one thing with respect to another thing.

Suppose you're sitting on the pillion seat behind the motorcycle of your friend who is

driving the motorcycle at 40kmph, then the relative speed of you with respect to

motorcycle (or your friend) will be zero because for your friend, you are not moving an

inch. But with respect to a person selling ice cream in the corner shop, your relative

speed will be 40kmph, because for him, you're moving with a speed of 40kmph.

Now, you steal his ice cream, and get ahead. He also had a bike and he's now driving his

bike behind you with a speed of 50kmph. Will he catch you?


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Of course, he will. Coz now, the relative speed of him is 10kmph more with respect to you.

He will catch you sooner.

The concepts when put mathematically is this:

If two bodies A and B are moving with speed Sa and Sb, then relative speed will be

Sa - Sb, if they're moving in the same direction, and

Sa + Sb, if they're moving in the opposite direction.

(i) Now apply this concept.

A and B, both are moving in opposite direction w ith speeds of 30kmph and 40kmph. So, their

relative speed will be?

Ans: 30+40 = 70kmph.

They will take how much time to reach at point M?

They will cover total Distance = 700km / with speed of 70 kmph = will take Time = 10 hour to

reach at point M

Understand this before you go further to solve the rest of questions!

(ii) It took 10 hour by both of them to reach at M.

With speed 30kmph, A has covered 30*10 = 300km = PM

With speed 40kmph, B has covered 40*10 = 400km = MQ

Ratio of distances PM : MQ = 300 : 400 = 3 : 4

Note: know this that if time is taken constant, the ratio of distances will be equal to the ratio of their

speed. (Just because distance = speed * time)

How?

D1 = S1*T1

D2 = S2*T2

T1 = T2

------> D1/D2 = S1/S2

(iii) Distance MQ = 400 km

(iv) Time taken by A to reach Q = distance/speed = 700km/30kmph = 70/3 hour

Time taken by B to reach P = distance/speed = 700km/40kmph = 70/4 hour

Extra time taken by A will be ---- 70/3 - 70/4 = 70/12 hour

Understand this before you go further!!

(v) When A has reached at point M, A has covered 300km (because PM = 300km) and B has

covered 400km (because MQ = 400km). Now, A has to cover 400km more and B has to cover

300km more. So,


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Time Ta taken by A to cover MQ = distance MQ/speed = 400/30 hour

Time Tb taken by B to cover PM = distance PM/speed = 300/40 hour

Ratio of their time = Ta/Tb = [400/30]/[300/40] = 16/9

If derived (just like we've solved), we will get to know that this ratio [Ta/Tb] of their time is the ratio

of the reciprocal of squares of their speed.

Why not we derive this?

Suppose A travels X km with speed Sa and B travels 700-X km with speed Sb and reaches point

M in time T.

Time taken to reach point M will be equal.

i.e. [X]/Sa = [700-X]/Sb

i.e. [700-X]/[X] = [Sb/Sa]

Now, for A, rest distance to cover is 700-A with speed Sa, and for B, rest distance to cover is X

with speed Sb, they will take time Ta and Tb to reach their destinations.

Ta = [700-X]/Sa

Tb = [X]/Sb

So, ratio of their times will be = Ta/Tb = ([700-X]/Sa) / ([X]/Sb) = ([700-X]/[X]) * ([Sb/Sa])

But we know that [700-X]/[X] = [Sb/Sa]

So, putting this in equation, we gets, Ta/Tb = ([Sb/Sa]) * ([Sb/Sa])

i.e. Ta/Tb = square of [Sb/Sa] ------ (note Sb/Sa and not Sa/Sb)

Sb = 40 kmph, Sa = 30kmph Ta/Tb = square of [40/30] = square of [4/3] = 16/9

(vi) They will be separated by only 560 Km if they have covered 700-560 = 140 km.

With relative speed of 70kmph, they will cover 140km in 2 hour. So, that means, after 2 hour, they

will be separated by 560km

(vii) Again, after crossing at the point M, their relative speed still will be the same. I.e. they will

cover 280km in 280/7 = 4 hour time.

Understood? Now, try to solve this question!!

Quiz Ques 3: A lives at P and B lives at Q. A usually goes to meet B at Q. He covers the distance

in 3 hour at 150kmph. On a particular day, B started moving away from A While A was moving

towards Q, thus A took 5 hours to meet B. What is the speed of B?

Concept of Boats and Stream

The concepts of boats and streams is also based on this relative speed.

When boat goes downstream, the speed of flowing water helps the boat to move faster with more

speed. When boat goes upstream, the speed of flowing water tries to cancel the speed of boat.

The boat moves slower this time.

If, speed of stream = S and speed of boat is B, then


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Downstream speed, D = B + S

Upstream speed, U = B - S

B generally means speed of boat in still water.

Hence, B = [D+U]/2 and S = D-B = [D-U]/2

Let’s apply this concept in this question:

Ques: A man can row 9 kmph in still water. It takes him twice as long as to row up as to row

down, Find the rate of stream of water.

Let distance covered by boat be 'D'

Speed of stream be 'S'

Speed of boat is 9kmph

Downstream time Td = distance/speed = D/[9+S]

Upstream time Tu = distance/speed = D/[9-S]

Tu is twice than Td

D/[9-S] = 2*D/[9+S]

On solving, we will get S = 3 kmph

Understood? Solve this question now:

Quiz Ques 4: A man can row at 10 kmph in still water. If the river flows at 3 kmph and, it takes 12

hours more in upstream than to go downstream for the same distance. How far is the place?

Concept of Races

In races, questions are asked of two or three player race. Questions are like,

Ques: In a race between Ram and Rahim, Ram has won the 1 km race by 100 meters. What is

the ratio of their speeds?

The concepts are no different than those that we have already covered. Just some twists in the

questions. The objective is to find what the question is saying. Answers will follow.

Now, first step and the most important step is to feel in mind's eye what all is happening in the

race. When Ram has just won the race of 1km, Rahim is how far behind him?

Ans is 100 meters behind him.

And Ram has covered 1000 meters, R ahim has covered how much distance?

Ans is 900 meters.

And Ram has covered 1000 meters in the same time it took Rahim to cover 900 meters.

Distance covered are in the ratio of 1000:900

Know the previous concept that when distance is different and time is constant, speed is directly

proportional to the distance.


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So, speed ratio of Ram : Rahim will also be 1000:900 = 10 : 9

This question will cover the remaining logic.

Ques: in a 1000m race, Ravi gives Vinod a start of 40m and beats him by 19 seconds. If Ravi

gives a start of 30 sec to Vinod, then Vinod beats Ravi by 40m. What is the ratio of speed of Ravi

to that of Vinod?

Case1: Now, visualize, in 1000m race, Ravi has given Vinod a start of 40m and beats him by 19

seconds. Means to say, Ravi runs 1000m while Vinod runs only 960m.

Second thing, when Ravi completed his 1000m, Vinod is still running and he runs for 19s more.

When putting it mathematically, if Ravi has completed the 1000m in T1 seconds, Vinod took

T1+19 seconds to complete the 960m.

Case2: Ravi gives Vinod a start of 30s, then Vinod beat Ravi by 40m. Means to say, When the

race is finished, Vinod has run 1000m while Ravi has run only 960m.

Second thing, Vinod has given the start of 30s. When Vinod has completed his 1000m, Ravi is still

behind him 40m (i.e. Ravi has completed 960m)

When putting it mathematically, if Vinod has run 1000m in T2+30 seconds, Ravi has run 960m in

T2 seconds.

Based on all the above facts, we'll find their speeds.

Ravi's speed = 1000/T1 = 960/T2

T1 = [25/24]*T2

Also, Vinod's Speed = 960/[T1 + 19] = = 1000/[T2 + 30]

Solving this by putting T1's value, we gets, T2 = 120s

Required ratio = [960/T2] / [1000/(T2 + 30)] == [960/120] / [1000/150] = 6/5 = Answer.

If you understood all this, try solvingthesequestion using the same basic concepts.

Quiz Ques5: In a 1600m race, A beats B by 80m and C by 60m. If they run at the same time, then

by what distance will C beat B in a 400 m race?

Quiz Ques 6: A beats B by 100m in a race of 1200m and B beats C by 200m in a race of 1600m.

Approximately by how many meters can A beat C in a race of 9600m?

Circular Motion Concept

In circular tracks, the radius should be given or the length of the track is given. If length is given

then its okay, if radius is given then put the formula L = 2*pi*radius to find the length of track.

Two or more runners will run this track with unequal speeds. They will run in the same direction or

opposite direction.

We've covered that when two people run in same direction, their relative speed = speed of person

with more speed - speed of person with less speed.

And when they run in opposite direction, their relative speed becomes = speed of 1st person +

speed of 2nd person.


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This same concept is to be used here.

Time taken by them to meet for first time will be = length of trace / relative speed.

Means to say, if A runs 1000m race with speed of 25mps and B runs race with speed of 15mps

and they both are running in opposite direction, then

Time taken = length of track 1000m / relative speed 25+15mps = 1000/40 = 25seconds

If they run in the same direction, they will take = 1000/10 = 100 seconds.

Sometimes, question is asked of their meeting at the same starting point. For, this, LCM of their

time is taken

That is to say, if A runs 1000m with 25mps speed, he will take = 40sec

B runs 1000m with 15mps speed = he will take = 200/3 sec

LCM of 40 and 200/3 = 200

So, they will take 200 sec to meet again at starting point.

This covers the basics of Time and Distance.

Read more: http://www.bankersadda.com/2015/06/gateway-to-sbi-mains-time-

speed.html#ixzz3xLQbS53y

Study Notes and Quiz on Time and Speed For SBI PO 2015

(1).Relation between distance ,time and speed:

Distance = speed x Time

(2).To convert speed of any object from KMPH to MPS multiply the speed by = 1000 / 3600 = 5 /

18

(3).To convert speed of any object from MPS to KMPH multiply the speed by = 3600 / 1000 = 18/

5

(4).If the speed ratio of A and B is a:b then ratio of time to cover certain distance is = 1/a : 1/b = b

: a

(5).If a person covers certain distance with speed x KMPH and return back with speed y KMPH

then his average speed throughout the journey is

Average speed = 2xy/(x+y)KMPH

(6).If a certain distance is covered with 3 diffrent speed x KMPH, y KMPH and z KMPH then

average speed throughout the journey is

Average speed = 3xyz/(xy+yz+zx)KMPH

http://www.bankersadda.com/2015/06/gateway-to-sbi-mains-time-speed.html#ixzz3xLQbS53y


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(7).If 2 different distances covered with speed x KMPH and y KMPH respectively but required

same time the then average speed throughout the journey is

Average speed = (x+y)/2 KMPH

(8).If 2 trains start at the same time from different points suppose A and B respectively toward

each other and after crossing if they take a and b seconds time resp to reach at B and A point

then

(A’s speed) : (B’s speed) = b : a

Formulae based on Train Problems Relative Speed (Train Problems):

(9)If two trains are moving in the same direction with speed x KMPH and y KMPH where x>y in

that case their relative speed is given as :(x-y) KMPH

(10)If two trains are moving in the opposite direction with speed x KMPH and y KMPH in that case

their relative speed is given as: (x+y) KMPH

Quant Quiz on Time and Distance

1.Walking at 7/8th of his usual speed, a man reached his destination 16 minutes later than the

time he usually takes to reach his destination. Find the usual time taken by him to reach his

destination.

(a) 1 hour, 44 minutes

(b) 1 hour, 52 minutes

(c) 1 hour, 36 minutes

(d) 1 hour, 40 minutes

2.A person goes to office by train. He walks to the railwy station closest to his home to catch the

train. One day, he walked at 4 km/hr and missed the train by 5 minutes. The next day, he walked

at 6 km/hr and reached the station 7 minutes before the arrival of the train. find the distance

between his home and the station.

(a) 2.4 km

(b) 1.8 km

(c) 3.6 km

(d) 3 km

3. Ashok covered a distance of 225 km as follows. He covered the first 15 km at 45 km/hr, the

next 120 km at 60 km/hr and the remaining journey at 90 km/hr. Find his average speed for the journey of 225 km.

(a) 65 km/hr

(b) 67.5 km/hr

(c) 70 km/hr

(d) 73.5 km/hr

4. A person went from P to Q, at an average speed of a km/hr, from Q to R at an average speed

of b km/hr, and from R to S at an average speed of the c km/hr. If PQ = QR = RS, then the

average speed of the person for traveling from P to S was

(a) (a + b + c)/ 3

(b) 3abc/(ab + bc + ca)

(c) 3abc/(a + b + c)

(d) 3(ab + bc + ca)/(a + b + c)

5. Car P starts from town X toward town y. Car Q stars from Y towards X. Both the cars start

simultaneously and travel their meet after journeys at uniform speeds. XY = 200 km. Both cars

meet after 2 hours. If P and Q had travelled in the same direction both the cars would have met in

4 hours. Find the speed of P.

(a) 60 kmph

(b) 85 kmph

(c) 75 kmph

(d) 80 kmph

6. Train P overtakes train Q double its length and travelling at half of speed of train P in 36

seconds. Train P crosses train R going in the opposite direction at double its speed in 8 seconds.

If the speed of train P is 72 kmph then the length of train R is ...........


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(a) 330 m

(b) 360 m

(c) 390 m

(d) 420 m

7. A 480 m long train was travelling at 72 km/hr. It took 32 seconds to cross a cyclist travelling in

the same direction as the train. Find the speed of the cyclist.

(a) 12 km/ph

(b) 15 km/ph

(c) 18 km/ph

(d) 9 km/ph

8. A train, 180m long, crossed a 120 m long platform in 20 seconds, and another train travelling at

the same speed crossed an electric pole in 10 seconds. In how much time will they cross each

other when they are travelling in the opposite direction.?

(a) 11 sec

(b). 13 sec

(c) 12 sec

(d) 14 sec

9. On a circular track, time taken by A and B to meet when travelling in the opposite directions is

1/4 of time taken when they travel in the same direction. Find the ratio of their speeds?

(a) 5: 3

(b) 6 : 5

(c) 4 : 3

(d) 3 : 2

10. How long will three persons starting at the same point and travelling at 4 km/hr, 6 km/hr and 8

km/hr around a circular track 2 km long take to meet at the starting point?

(a) 1/2hr

(b) 1hr

(c) 1.5 hrs

(d) 2 hrs

Answers with Explanation

1. (b) Ratio

Speed 8 : 7

Time 7 : 81 = 16

7 = 7 X 16 = 112 min

= 1 hr 52 min

2. (a) Let S1 = 4 kmph, S2 = 6 kmph

Distance = (S1 × S2)/ (S1-S2) X total time in hr

Distance = (4 ×6)/ (6- 4) x (7+5)/60

= (4 ×6)/2 × 1/5 = 2.4 km

3. (b) Average speed = Total distance / Total time

= 225/(15/45 + 120/60 + 90/90) = 67.5 km/h

4. (b) by above concept No. 6

5. (c) Let speed of car P = S1

& speed of car Q = S2

From Ist case:

2S1 + 2S2 = 200 - (i)

From 2nd case, When cars travelled in Same direction

200/ (S1 - S2) = 4

4 S1 - 4 S2 = 200 (ii)

From Equation (i) & (ii)

S1 = 75 kmph

6. (b) For Train P

length = L, Speed = 72 kmph


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For train Q

length = 2L, Speed = 36 kmph

(L + 2L)/(72 - 36)X5/18

L= 120 meter

For train R

Speed = 2 X 72 = 144 kmph

& length = x meter

(120 + x) / (144 + 72)X5/18 = 8

x = 360 meter

7. (c) Let speed of cyclist = x kmph480/(72 - x) X 5/18 = 32

x = 18 kmph

8. (a) Let speed of 1st train = x kmph

(180 + 120)/(x X 5/18) = 20

x = 54 kmph

T/(54 X 5/18) = 10, T = 150 meter

So, (180 + 150) / (54 + 54) X5/18 = 11 sec

9. (a) Let speed of A = x kmph

& speed of B = y kmph & x > y

When they are travelling in same direction, time taken be t

2PiR/ (x - y) = t ................ (i)

When they are travelling in opposite direction

2PiR/(x + y) = t/4 ................. (ii)

From Eq (i) & (ii)

x + y /x - y = 4

By C & D

x/y = (4 + 1)/(4 -1) = 5/3

x : y = 5 : 3

10. (b) Time taken for the three people meet in hours

= LCM (2/4, 2/6, 2/8)

= 1 hours

Read more: http://www.bankersadda.com/2015/05/study-notes-and-quiz-on-time-and-speed.html#ixzz3xLQjGGGW

All About Time And DistanceHello Readers,

Since you all know that Mathematics is one of the most important part of competitive

exams, here we posting Some Quick Study Notes on "Time and Distance". We hope that

they will help you all in the upcoming competitive exams like SBI PO 2015, Insurance

Exams and IBPS CWE 2015. Hope you all like the post..

Introduction:-



Some Important Facts

Distance travelled is proportional to the speed of the object if the time is kept constant.

Distance travelled is proportional to the time taken if speed of object is kept constant.

Speed is inversely proportional to the time taken if the distance covered is kept constant.

If the ratio of two speeds for same distance is a:b then the ratio of time taken to cover the distance

is b:a

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Relative Speed

If two objects are moving in same direction with speeds of x and y then their relative speed is (x -

y)

If two objects are moving is opposite direction with speeds of x and y then their relative speed is (x

+ y)

Unit Conversion


Rule 1: If some distance is travelled at x km/hr and the same distance is travelled at y km/hr then

the average speed during the whole journey is given by

Example

John goes from his home to school at the speed of 2 km/hr and returns at the speed of 3

km/hr. What is his average speed during whole journey in m/sec?

Sol: Let’s say x = 2 km/hr

And y = 3 km/hr, so

Now, average speed in m/sec

= 2.4*(5/18) = .67m/sec

Rule 2: If a person travels a certain distance at x km/hr and returns at y km/hr, if the time taken to

the whole journey is T hours then the one way distance is given by

Example


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Mr Samson goes to market at the speed of 10 km/hr and returns to his home at the speed

of 15 km/hr. If he takes 3 hours in all, what is the distance between his home and market?

Sol:

Let’s say x = 10 km/hr

y = 10 km/hr, and

T = 3 hrs, then

So the distance between home and market is 18 km.

Rule 3: If two persons A and B start their journey at the same time from two points P and Q

towards each other and after crossing each other they take a and b hours in reaching Q and P

respectively, then

Example

Two persons Ram and Lakhan start their journey from two different places towards each

other’s place. After crossing each other, they complete their journey in 1 and 4 hours

respectively. Find speed of Lakhan if speed of ram is 20 km/hr.

Sol:

Let’s say A = Ram and B = Lakhan

a = 1 and b = 4, then

(20/Lakhan speed) = (2/1)

Lakhan’s Speed = 10 km/hr

Rule 4: If the same distance is covered at two different speeds S1 and S2 and the time taken to

cover the distance are T1 and T2, then the distance is given by

Example

Two trucks travel the same distance at the speed of 50 kmph and 60 kmph. Find the

distance when the distance when the time taken by both trucks has a difference of 1

hour. Sol:

Let’s say S1 = 50 kmph,

S2 = 60 kmph

T1 – T2 = 1

Distance = [(50*60)/(60-50)]*1 = 300km

Quiz On Time And Distance:-


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1.Busses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes.

What is the speed of a man coming from the opposite direction towards the bus terminal

if he meets the buses at intervals of 8 minutes?

a.3 km/hr

b.4 km/hr

c.5 km/hr

d.7 km/hr

e.None of these

2.The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and

travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A

at 75 km/hr. At what time do they meet?

a.10 am.

b.10 : 30 am.

c.11 am.

d.11 : 30 am.

e.None of these

3.Two trains are moving on two parallel tracks but in opposite directions. A person sitting

in the train moving at the speed of 80 km/hr passes the second train in 18 seconds. If the

length of the second train is 1000 m, its speed is?

a.100 km/hr

b.120 km/hr

c.140 km/hr

d.150 km/hr

e.None of these

4.In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay

doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is?

a.5 kmph

b.6 kmph

c.6.25 kmph

d.7.5 kmph

e.None of these

5.It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car.

It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the

speed of the train to that of the cars is?

a. 2 : 3

b. 3: 2

c. 3 : 4

d. 4 : 3

e. None of these

Answers with Explanation:-

1.(c)

Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20+x) kmph

20× 10/60=8/60(20+x)

200 = 160 + 8x

8x = 40

x=40/8=5 kmph

2.(c) Distance travelled by first train in one hour

= 60 x 1 = 60 km

Therefore, distance between two train at 9 a.m.

= 330 – 60 = 270 km


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Now, Relative speed of two trains = 60 + 75 = 135 km/hr

Time of meeting of two trains =270/135=2 hrs.

Therefore, both the trains will meet at 9 + 2 = 11 A.M.

3.(b) Let the speed of second train be x m/s.

80 km/h = (80×5)/18 m/s

According to the question 1000/(x+(80×5)/18)=18

100 – 18x + 400

x=666/18 m/s

= 600/18×18/5 km/h = 120 km/h

4.(a)

Let Abhay's speed be x km/hr.

Then, 30/x-30/2x= 3

6x = 30

x = 5 km/hr.

5.(c)

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then, 120/x+480/y= 8 1/x+4/y=1/15 ....(i)

And, 200/x+400/y=25/3 1/x+2/y=1/24 ....(ii)

Solving (i) and (ii), we get: x = 60 and y = 80.

Ratio of speeds = 60 : 80 = 3 : 4.

Read more: http://www.bankersadda.com/2015/05/all-about-time-and-

distance.html#ixzz3xLQrV2hy
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time speed

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