Suppose two trains are moving in the same directions at X m/s and Y m/s then their relative speed

Distance (D) = Speed (S) Time (T) X kmph = X 5 18 m/s X m/s = X 18 5 kmph If the ratio of the speeds A & B is a : b, then the ratio of the times taken by them to cover the same distance is = 1 a 1 b = b : a Suppose two trains are moving in the same directions at X m/s and Y m/s then their relative speed = (X Y) m/s Suppose two trains are moving in the opposite directions at X m/s and Y m/s then their relative speed = (X + Y) m/s If two trains of length A meters and B meters are moving in same directions at X m/s and Y m/s then the time taken by the trains to cross each other = A+B X Y seconds If trains move in opposite direction s = A+B X+Y seconds

1 Samir drove at the speed of 45kmph from home to resort. Returning over the same route he got stuck in traffic and took an hours longer. Also he could drive only at the speed of 40 kmph. How many kilometers did he drive each way? Let distance = D D = 360 km (answer) D 45 D 40 = 1 2 A 320 meter long train moving an average speed of 120 kmph crosses a platform in 24 seconds. A man crosses the platform in 4 minutes. What is the speed of the man in m/s. Platform length = X- meters 320 + X = 24 120 5 18 X = 480 meters Therefore, man s speed = 480 4 60 = 2 m/s (answer) 3 A driver was supposed to drive at a uniform speed to cover a distance of 180 km, he was 54 minutes late. To cover this lost time he had to increase the speed by 10 kmph, what is the original speed Let original speed = X- kmph

X = 40 kmph (answer) 180 X 180 X + 10 = 54 60 4 A car covers a certain distance taking 7 hours in forward journey. During the return journey speed was increased by 12 kmph and it takes 5 hours. What is the total distance. Let the speed = X- kmph 7 X = 5(X + 12) X = 30 kmph Therefore, Distance => 7X = 7 30 = 210 km (answer) 5 Two trains of equal length running in opposite directions, pass a pole in 18 & 12 seconds. The train will cross each other in.. Train length = X- meters 1 st train speed = X/18 m/s 2 nd train speed = X/12 m/s Relative speed => X/18 + X/12 = 5X/36 m/s Total distance => X + X = 2X meters Therefore, Time = Distance Speed = 2X 5x 36 = 14.4 seconds (answer) 6 A thief seeing a policeman from a distance of 200 meters starts running with a speed of 8 kmph. The policeman gives chase immediately with a speed of 9 kmph and the thief is caught. The distance run by the thief is.

Relative speed => (9-8) = 1 kmph = 1 5 = 5/18 m/s 18 Time taken by policeman to chase thief = 200 5 18 = 720 seconds Distance run by the thief => 8 5 18 720 = 1600 meters (answer) 7 A train moving at a rate of 36 kmph crosses a standing man in 10 seconds. It will cross a platform 55 meters long in.. Speed => 36 5 = 10 m/s 18 Distance of train => 10 10 = 100 meters Total distance => 100 + 55 = 155 meters Time = 155 10 = 15.5 seconds (answer) 8 A person travels 285 km in 6 hours. In the first part of the journey he travels at 40 kmph by bus. In the second part, he travels at 55 kmph by train. The distance travelled by train is.. Train distance = X, Bus distance = 285 X X = 165 km (answer) X 55 + 285 X = 6 40 9 A train overtakes two persons who are walking in the same directions in which the train is running at the rate of 2 kmph and 4 kmph and passes them completely in 9 seconds and 10 seconds respectively. The length of the train is.

Let train speed = S, Distance = D And, D S 4 = 10 3600 From above equations, we get D = 50 meters (answer) D S 2 = 9 3600 10 Ramana started a journey at 1 pm at 30 kmph. Karthik started from the same spot and in the same direction at 1.40 pm at 40 kmph. Kartikovertook Ramana in. Since Kartik starts 40 minutes after Ramana Distance covered by Ramana in 40 minutes => 30 40 = 20 km 60 Relative speed => 40-30 = 10 kmph Therefore, time = 20 10 = 2 hours (answer) 11 A train crosses a man on the platform in 8.5 seconds and crosses the platform of 240 meters in length in 20.5 seconds. What is the length of the train.. Train length = X, Train speed = Y X = 8.5 seconds (1) Y X + 240 Y 20.5 seconds.. (2)

From equations 1 & 2 X = 170 meters (answer) 12 A motor car does a journey in 17.5 hours covering the first half at 30 kmph and the second half at 40 kmph. Find the distance of the journey. Distance (D) = S T D = 600 km (answer) = 2 X Y X + Y T 2 30 40 = ( 30 + 40 ) 17.5 13 A car starts from A for B travelling 20 kmph. 1 ½ hours later another car starts from A & travelling at the rate of 30 kmph reaches B, 2 ½ hours before the first car. Find the distance from A to B.. Distance = D D = 240 km (answer) D 20 D 30 = 3 2 + 5 2 14 If a man walks 20 km at 5 kmph, he will be late by 40 minutes. If he walks at 8 kmph, how early from the fixed time will he reach Time (T) = 20 5 = 4 hours Fixed time = 4 hours 40 minutes = 3.20 minutes Time taken to cover 20 km at 8 kmph = 20 8 = 2.30 minutes Therefore, required time => 3.20 minutes 2.30

minutes = 50 minutes (answer) 15 In covering a certain distance, the speed of A & B in the ratio of 3: 4. Atakes 30 minutes more than B to reach the destination. The time taken by A to reach the destination Distance =D Let A s speed = 3X kmph B s speed = 4X kmph D 3X D 4X = 30 60 D 3X = 2 hours Therefore, time taken by A to reach the destination = 2 hours (answer) 16 A train covers certain distance between two places at a uniform speed. If the train moved 10 kmph faster, it would take 2 hours less. And, if the train was slower by 10 kmph, it would take 3 hours more than the scheduled time. Find the distance covered by the train If the distance be X- km and usual speed be Y- kmph Then usual time = X/Y hours x y + 10 = x y 2 And, x y 10 = x y + 3 x = 3y(y 10) 10. (2) x = y(y + 2) (1) 5

From equations (1) and (2) We get, x = 600 km (answer) 17 A train passes two bridges of lengths 800 m & 400 m in 100 seconds & 60 seconds respectively. The length of the train is. Train length = X- meters As given, speed = x+800 100 Speed = x+400 60 From equations 1 & 2 X = 200 meters (answer) (1). (2) 18 Two cars start at the same time from one point and move along two roads at right angle to each other. Their speeds are 36 kmph& 48 kmph respectively. After 15 seconds the distance between them will be Let O be the starting point. The car running at 36 kmph is moving along OB & that at 48 kmph moving along OA. Also let they reach at B & A after 15 seconds respectively OA = 48 5 15 = 200 meters 18 OB = 36 5 15 = 150 meters 18 Required distance AB = (200) 2 + (150) 2 = 250 meters (answer) 19 Two trains start from stations A & B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the

second train has travelled 120 km more than the first. The distance between A & B is Relative speed => 60-50 = 10 kmph Time (T) = 120 = 12 hours 10 Therefore, distance of AB => 50 12 + 60 12 = 1320 km (answer) 20 A train running at 7/11th of its own speed reached a place in 22 hours. How much time could be saved if the train would run at its own speed Since the train runs at 7/11 th of its own speed The time it takes is 11/7 th of its usual time Let the usual time taken be T- hours Then we can write, 11 7 T = 22 T = 14 hours Therefore, time saved => 22-14 = 8 hours (answer) 21 A man can reach a certain place in 30 hours. If he reduces his speed by 1/15th he goes 10 km less in that time. Find his speed in kmph. Let the speed of the man = X- kmph 30 X 30(X x 15 ) = 10 X = 5 kmph (answer) 22 Two trains, Kolkata & Mumbai starts at the same time from stations Kolkata & Mumbai respectively towards each other. After passing each other they take 12 hours & 3 hours to reach Mumbai & Kolkata respectively. If the

Kolkata train is moving at the speed of the 48 kmph, the speed of the Mumbai train is. Formula, S 1 T 1 = S 2 T 2 S 2 = 96 kmph (answer) 48 12 = S 2 3 23 Two trains starts at the same time from Mumbai & Pune and proceed towards each other at the rate of 60 kmph& 40 kmph respectively. When they meet it is found that one train has travelled 20 km more than the other. Find the distance between Mumbai & Pune Both will meet at the time => T = D 100 Also 60T 40T => 20T = 20 T = 1 So, D = 100T => 100 1 = 100 km (answer) 24 A boy takes as much time in running 12 meters as a car takes in covering 36 meters. The ratio of the speeds of the boy & the car is Here time is same, so Boy s speed = X- kmph & Car s speed = Y kmph Therefore, 12 = 36 x y X: Y = 1: 3 (answer) 25 A & B are two stations. A train goes from A to B at 64 kmph and returns to A at a slower speed. If its average speed for the whole journey is 56 kmph, at what speed did it return?

2 x y x + y = 56 (from formula) Y = 49. 77 kmph (answer) 2 64 Y 64 + Y = 56 26 Amit started cycling along the boundaries of a square field from cover point A. After half an hour, he reached the corner point C, diagonally opposite to A. If his speed was 8 kmph, what is the area of the field in square. Distance covered in 1 hour = 2X 2 = 1 2 8 2X = 4 => X= 2 Area => X 2 = (2) 2 = 4 sq. km (answer) 27 A train travelling at 48 kmph completely crosses another train having half it length and travelling in opposite directions at 42 kmph in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is

1 st train length = X- meters 2 nd train length = x 2 meters Relative speed (48 + 42) = 90 kmph = 90 5 18 = 25 m/s And, X+ X 2 25 = 12 => X = 200 meters And length of the platform = Y- meters Speed of the first train = 48 5 18 m/s 200 + Y = 45 40 3 Y = 400 meters (answer) = 40 3 28 A man is walking at a speed of 10 kmph. After every km, he takes a rest for 5 minutes. How much time will he take to cover a distance of 5 km Total time for rest = 20 minutes (from question) Time = 5 10 = 1 2 hour or 30 minutes Therefore, total time => 30 minutes + 20 minutes = 50 minutes (answer) 29 In a one kilometer race, A can beat B by 30 meters while in a 500 meter race B can beat C by 25 meters. By how many meters will A beat C in a 100 meter race.. When A runs 1000 meter, B runs 970 meter When B runs 500 meter, C runs 475 meter When B runs 97 meter, C runs =? 475 500 97 = 92.15 meter A will beat C by (100-92.15) = 7.85 meter (answer)

30 A car covers 1/5 of the distance from A to B at the speed of 8 kmph, 1/10 of the distance at 25 kmph and remaining at the speed of 20 kmph. Find the average speed of the whole journey Let the total distance = 1 km Time = distance/ time Therefore, total time = = 1 5 8 + = 8/25 hours 1 + 1 10 25 1 5 1 10 20 Therefore, average speed = Total distance/ total time = 1 8 125 = 15. 625 kmph (answer) 31 A boy started from his house by bicycle at 10 am at a speed of 12 kmph. His elder brother started after 1 hour 15 minutes by scooter along the same path and caught him at 1.30 pm. The speed of the scooter will be (kmph).. Distance covered by cycling in 3 ½ hours = Distance covered by scooter in 2 ¼ hours Let speed of the scooter = X 12 7 2 = x 9 4 X = 18 3 2 kmph (answer) 32 Buses start from a bus terminal with a speed of 20 kmph 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..

Distance covered in 10 minutes at 20 kmph = Distance covered in 8 minutes at (20 + X) kmph Here, X = Speed of the man 20 10 8 = (20 + X) 60 60 X = 5 kmph (answer) 33 A railway half- ticket costs half the full rate. But the reservation charge on the half- ticket is the same as that on full ticket. One reserved full ticket for a journey between two stations is 525/- and the cost of one full and one half reserved tickets is 850/-. What is the reservation charge Let the half- ticket = X/- Reservation charge = Y/- Therefore, full ticket will be = 2X 2 X+ Y = 525.. (1) 2 X + Y + X + Y = 850 3X + 2Y = 850 (2) From equations 1 &2 X = 200/- & Y = 125/- Therefore, reservation charge Y = 125/- (answer) 34 A man travels 35 km partly at 4 kmph and at 5 kmph. If he covers former distance at 5 kmph and later distance at 4 kmph, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is Suppose the man covers first distance in X- hours and second distance in Y- hours Then, 4 X + 5Y = 35 (1) 5 X + 4Y = 37. (2) From equations 1 & 2

X = 5- hours & Y = 3- hours Therefore, total time taken => 5 + 3 = 8 hours (answer) 35 Two persons, Ajith&Lalitha start at the same time from Secunderabad& Vijayawada and proceed towards each other at 45 kmph and 54 kmph respectively. When they meet, it is found that one of them has travelled 72 km more than the other. The distance between the places.. 9 km (54-45) difference arises in the 99 km (45 + 54) distance So, 72 km difference will arise in 792 km distance i. e 9 99 72? 72 99 9 = 792 km (answer) 36 The distance between two cities is 800 km. A motor car starts from the first city at the speed of 30 kmph. At the same time, another car starts from the second city towards the first city at the speed of 50 kmph. The distance of the point from the first city where both the cars meet is Since time of travel for both cars is same So, V 1 = 30 kmph& V 2 = 50 kmph Time = distance/speed D1 V1 = D2 V2 D 1 800 = 30 50 = 300 km (answer) 37 A train can travel 50% faster than a car. Both starts from point A at the same time and reach point B 75 km away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is..

Let the speed of the car = X- kmph Then speed of the train => X 150 100 = 3x 2 kmph X = 120 kmph (answer) 75 X 75 3X 2 = 125 10 60 38 In a KM race A can beat B by 80 meters and B can beat C by 60 meters. In the same race, A can beat C by.. While A runs 1000 meters, B runs => 1000-80 = 920 meters And while B runs 1000 meters, C runs => 1000-60 = 940 meters While B runs 920 meters, C runs => 920 940 100 = 4324 5 meters While A runs 1000 meters, C runs 4324 meters Therefore, A can beat C by = 1000-4324 5 = 135 5 1 meters (answer) 5 39 In a 100 meter race, A runs at 5 kmph. A gives B a start of 8 meter and still beats him by 8 seconds. Find out the speed of B.. Time taken by A to cover 100 meter = 100 5 ( 5 18 ) = 72 seconds Therefore B covers (100-8) or 92 meters in (72 + 8) or 80 seconds Therefore, speed of B = = 4.14 kmph (answer) 92 80 ( 18 5 ) 40 A can run 224 meters in 28 seconds and B in 32 seconds. By what distance A beat B..

Clearly A beats B by 4 seconds (32-28=4) Now find out how much B will run in these 4 seconds Speed of B => 224 32 = 7 m/s Distance covered by B in 4 seconds => speed time = 7 4= 28 meters (answer) 41 The speed of a car is increased by 2 km after every hour. If the distance travelled in the first hour was 35 km, then what was the total distance travelled in 12 hours This is the problem of arithmetic progression with the first term a = 35 Common difference d = 2 And, total no. of terms (n) = 12 The sum of this series will be the total distance travelled Sum S n = n 2 [2a + (n 1) d] = 12 2 [2 35 + (12 1) 2] = 552 km (answer) 42 A man can reach a certain place in 30 hours. If he reduces his speed by 1/15th, he goes 10 km less in that time. Find his speed per hour. Let speed = S 30 S 30 ( 14 s 1 ) = 10 (1 = 14 ) 15 15 15 S = 5 kmph (answer) 43 When a person reduces his speed from 42 kmph to 36 kmph he takes 20

minutes more than his usual time taken. Find the usual time taken by him Actual speed (S 1 ) = 42 kmph After reducing, speed (S 2 ) = 36 kmph Time difference => 20 minutes => 20 = 1/3 hours Time = Distance / Speed By solving we get D = 84 km Therefore, actual time => D S1 = 84 42 60 D S1 D S2 = 1 3 D 42 D 36 = 1 3 = 2 hours (answer) 44 A cyclist covers a distance of 24 km in a certain time with a certain fixed uniform speed. If he increases his speed by 2 kmph, he takes 2 hours less to cover the same distance. Find his original speed. Let original speed of the cyclist = X kmph Distance = 24 km (constant) 1 st case: 2 nd case: 24 X 24 (X+2) Difference: 2 hours X = 4 kmph (answer) Time = Distance Speed 24 X 24 X + 2 = 2

45 Two gunshots were fired at the interval of 12 minutes. One person moving towards the place from where gunshots were fired, listens the two sounds at the interval of 11 minutes. If the speed of sound is 330 m/s, find the speed of the person Distance travelled by sound in (12 11) minutes will be equal to distance travelled by person in 11 minutes X = 30 m/s (answer) 360 1 60 = x 11 60 46 Two friends started walking simultaneously from points A and B towards each other. 144 minutes later the distance between them was 20% of the original distance. After how much time they meet each other? Let the original distance between A and B is 5x Km Time taken to travel CD = 144 = 36 minutes (answer) 4 47 A dog finds a cat a 25 leaps away. The cat sees the dog coming towards it and starts running with the dog in hot pursuit. In every minute, the dog makes 5 leaps and the cat makes 6 leaps and one lap of the dog is equal to 2 leaps of the cat. Find the time in which the dog will catch the cat. Initial distance = 25 dog leaps Per minute dog makes 5 dog leaps

Per minute cat makes 6 cat leaps = 3 dog leaps Relative speed = 2 dog leaps per minute Then, an initial distance of 25 dog leaps would be covered in 12.5 minutes (answer) 47 A thief steals car at 1.30 pm and drives it at 45 kmph. The theft is discovered at 2 pm and the owner sets off in another car at 50 kmph. He will overtake the thief at. Here there is a gap of half an hour As the theft discovered at 2 pm Distance covered by thief in half an hour = 45 1 2 = 22. 5 km Relative speed, 50 45 = 5 kmph Relative speed in same direction is the difference in speed Time = distance = 22.5 = 4. 5 hours Relative speed 5 So, 2 pm + 4.5 hours = 6.30 pm (answer) 48 Hari and Ravi started a race from opposite ends of the pool. After a minute and a half, they passed each other in the centre of the pool. If they lost no time in turning and maintained their respective speeds, how many minutes after starting did they pass each other second time?

First meeting 3 minutes 2 In second meeting, they will have to cover double distance than their first meeting Therefore, total time = 3 + 3 2 2 2 = 4 1 minutes (answer) 2 49 Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in the train approaching the place hears the second shot 10 minutes after the first. The speed of the (in kmph), supposing that sound travels at 330 metres per second Let the speed of the train be x m/s Then, Distance travelled by the train in 10 minutes = Distance travelled by sound in 30 seconds x 10 60 = 330 30 x = 16.5 Therefore, speed of the train = 16.5 m/s = 16.5 18 kmph = 59.4 kmph (answer) 50 The Taxi charges in a city contain fixed charges and additional charge per km. The fixed charge is for a distance of upto 5 km and additional charge per km thereafter. The charge for a distance of 10 km is 350/- and for 25 km is 800/-. The charge for a distance of 30 km is.. Fare of 10 km ----- 350/- Fare of 25 km ------ 800/- 5

Therefore, additional charge per km = 800 350 15 = 30/- per km So, charge upto 5 km = 350 (5 30) = 200/- Fare of 30 km = 5 km + 25 km = 200 + 25 30 = 950/- (answer) 51 A train after travelling 60 km meets with an accident and then proceeds at 3 of its former speed and arrives at its destination 40 minutes 4 late. If the accident could have occurred 20 km ahead, it would have reached the destination 10 minute earlier. Find the speed of the train.. Original time taken for B to D = 40 3 => 120 Minutes If accident takes place after 25 km then original time for C to D = (40 10) 3 => 90 minutes Time taken for travelling 25 km = 30 minutes Speed of Train = 25 30 60 = 50 kmph (answer) 52 The Speeds of three Cars are in the ratio 3 : 4 : 5. The time taken by each of them to travel the same distance is in the ratio is

Speed = Distance [Formula] Time 1 For the same distance, Speed Required ratio = 1 3 1 4 1 5 = 1 3 60 1 4 60 1 5 60 = 20 : 15 : 12 (answer) Time 53 The auto rickshaw fare consists of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is 85/- and for a journey of 15 km, the charge paid is 120/-. The fare for a journey of 25 km will be.. Auto rickshaw charge for distance covered = x Fixed charge = y According to charge 10x + y = 85.. (1) 15x + y = 120. (2) From equations (1) and (2) x = 7, y = 15 Hence fare for journey of 25 km = (25x + y) = 25 7 + 15= 190/- (answer) 54 A Railway passenger counts the telegraph poles on the rail road as the passes them. The telegraph poles are at a distance of 50 m. What will be his count in 4 hours if the speed of the train is 45 kmph Distance travelled by train in 4 hours = 45 4 = 180 km No. of telegraph poles = 180 1000 50 = 3600 (answer)

55 An aeroplane first flew with a speed of 440 kmph and covered a certain distance. It still had to cover 770 km less than what it had already covered but it flew with a speed of 660 kmph. The average speed for the entire flight was 500 kmph. Find the total distance covered.. Total Distance Average Speed = Total Time 500 = 2x 770 x 440 + x 770 660 x = 1760 Total distance covered = 2x 770 = 2 1760 770 = 2750 km (answer)