Speed, Time and Distance


  1. A hare sees a dog 100 metres away from her and scuds off in the opposite direction at a speed of 12 km per hr. A minute later the dog perceives her and chases her at a speed of 16 km per hr. How soon will the dog overtake the hare and at what distance from the spot when the hare took flight?









  1. View Hint View Answer Discuss in Forum


    Let the hare at B sees that dog is at A.
    ∴  AB = 100 metres
    Again, let C be the position of the hare when the dog sees her.
    ∴  AB = 100 metres
    Again, let C be the position of the hare when the dog sees her.
    ∴  BC= the distance covered by the hare in 1 minute

    =
    12 × 1000 × 1
    = 200 metres
    60

    ∴  AC = AB + BC
    = 100 + 200 = 300 metres
    Thus, hare has a start of 300 metres.
    Now, the dog gains 16 – 12 = 4 kms
    4000 metres in 1 hour i.e. 60 minutes
    ∴  The distance gained by dog in 1 minute
    =
    4000
    =
    200
    metres
    603

    ∵ 
    200
    metres is covered in 1 minute
    3

    ∴  300 metres is covered in
    300 × 3
    =
    9
    minutes
    2002

    Again the distance walked by hare in
    9
    minutes
    2

    =
    12000
    ×
    9
    = 900 metres
    602

    ∴  Total distance from
    B = 200 + 900 = 1100 metres.

    Correct Option: D


    Let the hare at B sees that dog is at A.
    ∴  AB = 100 metres
    Again, let C be the position of the hare when the dog sees her.
    ∴  AB = 100 metres
    Again, let C be the position of the hare when the dog sees her.
    ∴  BC= the distance covered by the hare in 1 minute

    =
    12 × 1000 × 1
    = 200 metres
    60

    ∴  AC = AB + BC
    = 100 + 200 = 300 metres
    Thus, hare has a start of 300 metres.
    Now, the dog gains 16 – 12 = 4 kms
    4000 metres in 1 hour i.e. 60 minutes
    ∴  The distance gained by dog in 1 minute
    =
    4000
    =
    200
    metres
    603

    ∵ 
    200
    metres is covered in 1 minute
    3

    ∴  300 metres is covered in
    300 × 3
    =
    9
    minutes
    2002

    Again the distance walked by hare in
    9
    minutes
    2

    =
    12000
    ×
    9
    = 900 metres
    602

    ∴  Total distance from
    B = 200 + 900 = 1100 metres.


  1. A monkey climbing up a greased pole ascends 12 metres and slips down 5 metres in alternate minutes. If the pole is 63 metres high, how long will it take him to reach the top?









  1. View Hint View Answer Discuss in Forum

    In 1 minute the monkey climbs 12 metres but then he takes 1 minute to slip down 5
    metres. So, at the end of 2 minutes the net ascending of the monkey is 12 – 5 = 7 metres.
    So, to cover 63 metres the above

    process is repeated
    63
    = 9
    7

    times. Obviously, in 9 such happenings the monkey will slip 8 times, because on 9th time, it
    will climb to the top. Thus, in climbing 8 times and slipping 8 times, he covers 8 × 7 = 56 metres.
    Time taken to cover 56 metres
    =
    56 × 2
    = 16 minutes
    7

    Remaining distance
    = 63 – 56 = 7 metres
    Time taken to ascend 7 metres
    =
    7
    minutes
    12

    ∴  Total time taken = 16 +
    7
    12

    = 16
    7
    minutes.
    12

    Correct Option: C

    In 1 minute the monkey climbs 12 metres but then he takes 1 minute to slip down 5
    metres. So, at the end of 2 minutes the net ascending of the monkey is 12 – 5 = 7 metres.
    So, to cover 63 metres the above

    process is repeated
    63
    = 9
    7

    times. Obviously, in 9 such happenings the monkey will slip 8 times, because on 9th time, it
    will climb to the top. Thus, in climbing 8 times and slipping 8 times, he covers 8 × 7 = 56 metres.
    Time taken to cover 56 metres
    =
    56 × 2
    = 16 minutes
    7

    Remaining distance
    = 63 – 56 = 7 metres
    Time taken to ascend 7 metres
    =
    7
    minutes
    12

    ∴  Total time taken = 16 +
    7
    12

    = 16
    7
    minutes.
    12



  1. Ravi can walk a certain distance in 40 days when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each day?









  1. View Hint View Answer Discuss in Forum

    Working hours per day= 24 – 9 = 15 hrs.
    Total working hours for 40 days
    = 15 × 40 = 600 hrs.
    On doubling the distance, the time required becomes twice but on walking twice as fast, the time required gets halved. Therefore, the two together cancel each other with respect to time required. Increasing rest to twice reduces walking hours per day to
    24 – (2 × 9) = 6 hrs.
    ∴  Total number of days required to cover twice the distance, at twice speed with twice the rest.

    =
    600
    = 100 days
    6

    Correct Option: B

    Working hours per day= 24 – 9 = 15 hrs.
    Total working hours for 40 days
    = 15 × 40 = 600 hrs.
    On doubling the distance, the time required becomes twice but on walking twice as fast, the time required gets halved. Therefore, the two together cancel each other with respect to time required. Increasing rest to twice reduces walking hours per day to
    24 – (2 × 9) = 6 hrs.
    ∴  Total number of days required to cover twice the distance, at twice speed with twice the rest.

    =
    600
    = 100 days
    6


  1. On increasing the speed of a train at the rate of 10 km per hr, 30 minutes is saved in a journey of 100 kms. Find the initial speed of train.









  1. View Hint View Answer Discuss in Forum

    Let the original speed be x km/hr
    then, increased speed
    = (x + 10) km/hr
    According to question,

    100
    100
    =
    30
    xx + 1060

    [ ∵  Original time − New time
    = 30 minute or 
    30
    hr   ]
    60

    ⇒  100
    1
    1
    =
    1
    xx + 102

    ⇒ 
    x + 10 − x
    =
    1
    x(x + 10)200

    ⇒  10 × 200 = x (x + 10)
    ⇒  x2 + 10x – 2000 = 0
    ⇒  x2 + 50x – 40x – 2000 = 0
    ⇒  x (x + 50) – 40 (x + 50) = 0
    ⇒  x = – 50, 40
    Speed can’t be negative.
    Hence, Original speed = 40 kmph

    Correct Option: A

    Let the original speed be x km/hr
    then, increased speed
    = (x + 10) km/hr
    According to question,

    100
    100
    =
    30
    xx + 1060

    [ ∵  Original time − New time
    = 30 minute or 
    30
    hr   ]
    60

    ⇒  100
    1
    1
    =
    1
    xx + 102

    ⇒ 
    x + 10 − x
    =
    1
    x(x + 10)200

    ⇒  10 × 200 = x (x + 10)
    ⇒  x2 + 10x – 2000 = 0
    ⇒  x2 + 50x – 40x – 2000 = 0
    ⇒  x (x + 50) – 40 (x + 50) = 0
    ⇒  x = – 50, 40
    Speed can’t be negative.
    Hence, Original speed = 40 kmph



  1. A man travels 400 kms in 4 hours partly by air and partly by train. If he had travelled all the way by air,
    he would have saved
    4
    of the time he was in train
    5
    and would have arrived his destination 2 hours early. Find the distance he travelled by train.









  1. View Hint View Answer Discuss in Forum

    Obviously, 4/5 of total time in train = 2 hours

    ∴  Total time in train =
    5
    × 2 =
    5
    hours
    42

    Total time to cover 400 km is 4 hours
    ∴  Time spent in travelling by air
    = 4 −
    5
    =
    8 − 5
    =
    3
    hours
    222

    If 400 kms is travelled by air, then time taken = 2 hours
    ∴  In 2 hours, distance covered by air = 400 kms
    In 3/2 hours distance covered =
    400
    ×
    3
    = 300 kms
    22

    Distance covered by the train
    = 400 – 300 = 100 kms.

    Correct Option: D

    Obviously, 4/5 of total time in train = 2 hours

    ∴  Total time in train =
    5
    × 2 =
    5
    hours
    42

    Total time to cover 400 km is 4 hours
    ∴  Time spent in travelling by air
    = 4 −
    5
    =
    8 − 5
    =
    3
    hours
    222

    If 400 kms is travelled by air, then time taken = 2 hours
    ∴  In 2 hours, distance covered by air = 400 kms
    In 3/2 hours distance covered =
    400
    ×
    3
    = 300 kms
    22

    Distance covered by the train
    = 400 – 300 = 100 kms.