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Aptitude - Time and Distance ------------------------------------In this section you will find aptitude questions and answers of various difficulty levels on Time and Distance with explanation for various interview, competitive examination and entrance test in an easy to understand way. You can also checkout Tips and Tricks, Videos related to the topic.Use Green Board or space provided for Rough work whenever you need.
Formulae : Speed = Distance / Time Time = Distance / speed Distance = speed * time 1 km/hr = 5/18 m/s 1 m/s = 18/5 Km/hr If the ratio of the speed of A and B is a:b,then the ratio of the time taken by them to cover the same distance is 1a:1b or b:a Suppose a man covers a 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) km/hr
Questions and Answers :
PadhleBeta.net - Aptitude - Time and Distance 1) P walks at 4 km/hr and 4 hours after his start, Q cycles after him at 10 km/hr. How far from the start does Q catch up with P? 1) 26.7 km
3) 27.7 km
2) 25.7 km
4) 27.3 km
Solution : Suppose after x km from the start Q catches up with P. Then, the difference in the time taken by P to cover x km and that taken by Q to cover x km is 4 hours. x/4 - x/10 = 4 or x = 26.7 km.
PadhleBeta.net - Aptitude - Time and Distance 2) Two trains start from P and Q respectively and travel towards each other at a speed of 50 km/hr and 40 km/hr respectively. By the time they meet, the first train has travelled 100 km more than the second. The distance between P and Q is :
1) 800 km
3) 360 km
2) 900 km
4) 540 km
Solution : At the time of meeting, let the distance travelled by P the second train be x km. Then, distance covered by the first train is (x + 100) km x/40 = (x+100)/50 50x = 40x + 4000 xx+x+100) km = 900 km.
PadhleBeta.net - Aptitude - Time and Distance
3) 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 train (in km/hr), supposing that speed travels at 330 meters per second, is:
1) 144
3) 59.4
2) 67.8
4) 89.6
Solution : Let the speed of the train be x m/sec. Then, Distance travelled by the train in 10 min. = Distance travelled by sound in 30 sec. Speed of the train = 16.5 m/sec = [16.5*18/5] km/hr = 59.4 km/hr PadhleBeta.net - Aptitude - Time and Distance 4) P walks around a circular field at the rate of one round per hour while Q runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at:
1) 6.48 am
3) 7.29 a.m
2) 5.30 am
4) 7.42 a.m
Solution : Since P and Q move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two Relative speed of P and Q = (6-1) = 5 rounds per hour. Time taken to complete one round at this speed = 1/5 hr = 12 min.
PadhleBeta.net - Aptitude - Time and Distance 5) Two boys starting from the same place walk at the rate of 5 km/hr and 5.5 km/hr respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?
1) 17 hrs
3) 16 hrs
2) 18 hrs
4) 12 hrs
Solution : to be 0.5 km apart, they take 1 hour. To be 8.5 km apart, they take [1/0.5 * 8.5] hrs = 17 hrs. PadhleBeta.net - Aptitude - Time and Distance 6) Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?
1) 10
3) 18
2) 25
4) 30
Solution : Due to stoppages, it covers 9 km less. Time taken to cover 9 km = [9/54 * 60] min = 10 min PadhleBeta.net - Aptitude - Time and Distance 7) A boy 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.
1) 8 km/hr
3) 5 km/hr
2) 51/2 km/hr
4) 31/2 km/hr
Solution : Let the speed be x km/hr. Then, => after reducing 1/15th part of x provides reduced speed => Reduced Speed = [ (x) - { (1/15) * x } ] => x-(x/15) = (14/15)x hence forth 30x - 30 * (14/15)x = 10 2x = 10 x = 10/2 = 5 km/hr PadhleBeta.net - Aptitude - Time and Distance 8) If a boy walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is :
1) 68 km
3) 100 km
2) 42 km
4) 50 km
Solution : Let the actual distance travelled be x km. Then, x/10 = (x+20)/14 14x = 10x + 200 4x = 200
x = 50 km PadhleBeta.net - Aptitude - Time and Distance 9) A girl rides her bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. Her average speed for the entire trip is approximately.
1) 12.8 km/hr
3) 16.3 km/hr
2) 10.8 km/hr
4) 14.4 km/hr
Solution : Total distance travelled = (10+12) km/hr = 22 km/hr. total time taken = [10/12 + 12/10] hrs = 61/30 hrs Average speed = [22*30/61] km/hr = 10.8 km/hr PadhleBeta.net - Aptitude - Time and Distance 10) Three boys are walking from a place P to another place Q. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach Q by these boys will be :
1) 4 : 2 : 5
3) 20 : 15 : 12
2) 5 : 3 : 4
4) 15 : 20 : 12
Solution : Ratio of speeds = 4 : 3 : 5 Ratio of times taken = 1/4 : 1/3 : 1/5 = 15 : 20 : 12. PadhleBeta.net - Aptitude - Time and Distance
11) Rahul has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/hr) must be :
1) 3 km/hr
3) 18 km/hr
2) 7 km/hr
4) 12 km/hr
Solution : Remaining distance = 3 km and Remaining time = [1/3 * 45] min = 15 min = 1/4 hour. Required speed = (3*4) km/hr = 12 km/hr. PadhleBeta.net - Aptitude - Time and Distance 12) A boy on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is :
1) 71.11 km/hr
3) 43.5 km/hr
2) 82.6 km/hr
4) 77.7 km/hr
Solution : Total time taken = [160/64 + 160/8] hrs = 9/2 hrs. Average speed = [320*2/9] km/hr = 71.11 km/hr.
PadhleBeta.net - Aptitude - Time and Distance 13) Sound is said to travel in air at about 1100 feet per second. A person hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the person from the wood chopper?
1) 2420 ft
3) 2460 ft
2) 2500 ft
4) 25.30 ft
Solution : Distance = [1100 * 11/5] feet = 2420 feet.
PadhleBeta.net - Aptitude - Time and Distance 14) Rita goes to her school from her house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If she takes 5 hours in going and coming. the distance between her house and school is : 1) 5.5 km
3) 6.2 km
2) 6 km
4) 7 km
Solution : Average speed = [2*3*2 / 3+2] km/hr = 12/5 km/hr. Distance travelled = [12/5 * 5] km = 12 km. Distance between house and school = [12/2] km = 6 km. PadhleBeta.net - Aptitude - Time and Distance 15) Amit travels equal distances with speeds of 3 km/hr, 4 km/hr and 5 km/hr and takes a total time of 47 minutes. The total distance (in km) is : 1) 30 km
3) 12 km
2) 5 km
4) 3 km
Solution : Let the total distance be 3x km.
Then, x/3 + x/4 + x/5 = 47/60 47x/60 = 47/60 x = 1. total distance = (3*1) km = 3 km. PadhleBeta.net - Aptitude - Time and Distance 16) A man takes 5 hours 45minutes to walk to a place and ride back. He would have gained 2 hours byriding both ways. Calculate the time that man would take to walk both ways is
1) 7 hrs 45 min.
3) 3 hrs 15 min.
2) 9 hrs
4) 11 hrs 45 min.
Solution : Let the distance be x km. Then,(time taken to walk x km) + (Time taken to ride x km) = 23/4 hrs. (Time towalk 2x km) + (Time to ride 2x km) = 23/2 hrs. But, time taken to ride 2x km = 15/4 hrs. Time taken to walk 2x km [23/2 - 15/4] hrs = 31/4hrs = 7 hrs 45 min.
PadhleBeta.net - Aptitude - Time and Distance 17) A and B walk around acircular track. Starting at 9 a.m. from the same point in the oppositedirections. A and B walk at a speed of 2 rounds per hour and 3 rounds per hourrespectively. How many times shall they cross each other before 9.30 a.m.?
1) 9
3) 7
2) 8
4) 11
Solution : Relative speed = (2 + 3) = 5 rounds per hour. So, they cross each other 5 times in an hour and 2 times in half an hour. Hence, they cross each other 7 times before 9.30 a.m.
PadhleBeta.net - Aptitude - Time and Distance 18) A thief initially at a distance of 200m away from policeman starts running and the policemanstarts to chase him. The thief and the policeman run at the rate of 10 km and 11 km
per hour respectively.Calculate the distance betweenthem after 6 minutes.
1) 9m
3) 0m
2) 6m
4) 11m
Solution : Relative speed of the policeman = (11-10) km / hr = 1 km/ hr. Distance covered in 6 minutes = [1/60 * 6] km =1/10 km = 100 m. Distance between the thief and policeman = (100-100) m =0 m.
PadhleBeta.net - Aptitude - Time and Distance 19) How many minutes does Gaurav take to cover a distance of 400 m , if he runs at a speed of 20 km/hr?
1) 1 1/4
3) 1 1/8
2) 1 1/5
4) 1 1/3
Solution : Gaurav's speed = 20 km/hr = 20 * (5/18) m/sec = 50/9 m/sec Therefore , time taken to cover a distance of 400 m = 400 * (50/9) sec = 72 seconds = 1 12/60 = 1 1/5 minutes
PadhleBeta.net - Aptitude - Time and Distance 20) A cyclist covers a distance of 750 meters in 2 min 30 sec. What is the speed in km/hr of the cyclist?
1) 20 km/hr
3) 25 km/hr
2) 15 km/hr
4) 18 km/hr
Solution : Speed of cyclist = (750/150) m/sec = 5 m/sec = 5 * 18/5 km/hr =18 km/hr
PadhleBeta.net - Aptitude - Time and Distance 21) A dog takes 4 leaps for every 5 leaps of a hare. 3 leaps of the dog are equal to 4 leaps of the hare. Compare their speeds.
1) 16 :17
3) 8 :9
2) 16 :15
4) 15 :16
Solution : Let the distance covered in one leap of dog be x and that covered in 1 leap of hare be y. Then , from problem statement , 3x = 4y => x = 4/3 y => 4x = 16/3 y Therefore, Ratio of speeds of dog and hare = Ratio of the distances covered by them in same time => 4x : 5y = (16/3)y : 5y = (16/3) : 5 = 16 :15
PadhleBeta.net - Aptitude - Time and Distance 22) While covering a distance of 24 km, the man noticed that after walking for 1 hour and 40 minutes , he covered 5/7 part of remaining distance. Calculate his speed in meters per second.
1) 5/3 m/sec
3) 7/3 m/sec
2) 4/3 m/sec
4) 11/3 m/sec
Solution : Let the speed be x km/hr Then distance covered in 1 hour 40 minutes, i.e. 1 2/3 hours = 5x/3 km. Remaining distance = (24 - 5x/3) km Therefore, 5x/3 = (5/7) * (24-5x/3)
5x/3 = 5/7 {(72-5x)/3} 7x = 72 -5x 12x = 72 x=6 So, speed = 6 km/hr =>6 * 5/18 = 5/3 m/sec
PadhleBeta.net - Aptitude - Time and Distance 23) Devesh can cover a certain distance in 1 hr. 24 minutes by covering two-third of the distance at 4 km/hour and the rest at 5 km/hr. Calculate total distance.
1) 8 km
3) 7 km
2) 6 km
4) 7/5 km
Solution : Let the total distance be x , then [{(2/3) x }/4 ] + [{(1/3) x }/5 ] = 7/5 => x/6 + x/15 = 7/5 => 7x = 42 => x = 6 Therefore , total distance = 6 km
PadhleBeta.net - Aptitude - Time and Distance 24) A train when moves at an average speed of 40 kmph, reaches its destination on time. When its average speedbecomes 35 kmph, then itreaches its destination 15 minutes late. Find the length of journey.
1) 60km
3) 70km
2) 30km
4) 50km
Solution : Difference between timings = 15 min = 1/4 hr. Let the length of journey be x km. Then, x/35 - x/40 = 1/4 8x - 7x = 70 x = 70 km PadhleBeta.net - Aptitude - Time and Distance 25) In covering a distance of 30 km,Kamlesh takes 2 hours more than Pankaj. If Kamlesh doubles his speed, then he wouldtake 1 hour less than Pankaj. Then what is Kamlesh's speed?
1) 11 kmph
3) 5kmph
2) 9 kmph
4) 6 kmph
Solution : Let Kamlesh’s speed be x km/hr. Then, 30/x - 30/2x = 3 6x = 30 x = 5 km/hr PadhleBeta.net - Aptitude - Time and Distance 26) Walking 6/7thof her usual speed, a woman is 12 minutes too late. Then what is the usual timetaken by her to cover that distance?
1) 2hr12min
3) 1hr20min
2) 1hr12min
4) 1hr10min
Solution : current speed = 6/7 of usual speed.
New time = 7/6 of usual time [7/6 of usual time] - (usual time) = 1/5 hr. 1/6 of usual time = 1/5 hr usual time = 6/5 hr = 1 hr 12 min.
PadhleBeta.net - Aptitude - Time and Distance 27) if 120 km is done by train and the rest bycar, thenIt takes 8 hours for a 600kmjourney. If 200kmis done by train and the rest by car, then ittakes 20 minutes more. Then what would be the ratio of the speed of the train tothat of the car?
1) 3:4
3) 5:4
2) 4:3
4) 3:5
Solution : Let speed of the train bex km/hr and of the car be y km/hr. Then, 120/x + 480/y = 8 => 1/x + 4/y = 1/5 -- (i) And, 200/x + 400/y = 25/3 or 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. PadhleBeta.net - Aptitude - Time and Distance 28) if 120 km is done by train and the rest bycar, thenIt takes 8 hours for a 600kmjourney. If 200kmis done by train and the rest by car, then ittakes 20 minutes more. Then what would be the ratio of the speed of the train tothat of the car?
1) 3:4
3) 5:4
2) 4:3
4) 3:5
Solution : Let speed of the train bex km/hr and of the car be y km/hr. Then, 120/x + 480/y = 8 => 1/x + 4/y = 1/5 -- (i) And, 200/x + 400/y = 25/3 or 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. PadhleBeta.net - Aptitude - Time and Distance 29) A man traveleda distance of 61 km in 9 hours. Hetraveled on foot at 4 km/hr and on bicycle at 9 km/hr. Then what is the distancetraveledon foot?
1) 57 km
3) 17 km
2) 9 km
4) 16 km
Solution : Let the distancetraveledon foot be x km. then, distancetraveledon bicycle = (61 - x)km. So, x/4 + (61-x)/9 = 9 9x + 4 (61 - x) = 9 * 36 5x = 80 x = 16 km.
PadhleBeta.net - Aptitude - Time and Distance 30) A lady performs 3/5 of the total journey by train , 17/20 by bus and theremaining 6.5 km on foot. Calculate total distance covered.
1) 130km
3) 170 km
2) 125km
4) 65 km
Solution : Let the total journey be x km. Then 3x/5 + 7x/20 + 6.5 = x 12x + 7x + 20 * 6.5 = 20x x = 130 km.
