Speed, Time and Distance


  1. A train met with an accident 3 hours after starting, which detains it for one hour, after which it proceeds at 75% of its original speed. It arrives at the destination 4 hours late. Had the accident taken place 150 km further along the railway line,
    the train would have arrived only 3
    1
    hours late.
    2
    Find the length of the trip and the original speed of the train.









  1. View Hint View Answer Discuss in Forum

    Let A be the starting point, B the terminus. C and D are points where accidents take place.

    ∴ 0.75 =
    3
    4

    By travelling a 3/4 of its original speed, the train would take 4/3 of its usual time i.e., 1/3 more of the usual time.
    ∴  1/3 of the usual time taken to travel the distance CB.
    = 4 – 1 = 3 hrs.     ...(i)
    and 1/3 of the usual time taken to travel the distance
    DB = 3
    1
    − 1 = 2
    1
    hrs.   ...(ii)
    22

    Subtracting equation (ii) from (i)
    we can write,
    1/3 of the usual time taken to travel the distance
    CD = 3 − 2
    1
    =
    1
    hr.
    22

    ∴  Usual time taken to travel
    CD (150 km) =
    1
    2
    1
    3

    =
    3
    hr.
    2

    Usual speed of the train
    =
    3
    = 9 hrs.
    1
    3

    Total time = 3 + 9 = 12 hrs.
    ∴  Length of the trip = 12 × 100 = 1200 km.

    Correct Option: B

    Let A be the starting point, B the terminus. C and D are points where accidents take place.

    ∴ 0.75 =
    3
    4

    By travelling a 3/4 of its original speed, the train would take 4/3 of its usual time i.e., 1/3 more of the usual time.
    ∴  1/3 of the usual time taken to travel the distance CB.
    = 4 – 1 = 3 hrs.     ...(i)
    and 1/3 of the usual time taken to travel the distance
    DB = 3
    1
    − 1 = 2
    1
    hrs.   ...(ii)
    22

    Subtracting equation (ii) from (i)
    we can write,
    1/3 of the usual time taken to travel the distance
    CD = 3 − 2
    1
    =
    1
    hr.
    22

    ∴  Usual time taken to travel
    CD (150 km) =
    1
    2
    1
    3

    =
    3
    hr.
    2

    Usual speed of the train
    =
    3
    = 9 hrs.
    1
    3

    Total time = 3 + 9 = 12 hrs.
    ∴  Length of the trip = 12 × 100 = 1200 km.


  1. A train meets with an accident after travelling 30 kms, after which it moves
    with
    4
    th of its original speed and arrives
    5
    at the destination 45 minutes late. Had the accident happened 18 kms further on, it would have been 9 minutes before. Find the distance of journey and original speed of the train.









  1. View Hint View Answer Discuss in Forum

    Let the original speed be x and distance be y
    Case I.
    Time taken by train to travel

    30 km =
    30
    x

    Time taken by train after accident =
    y − 30
    4/5x

    Total time taken =
    30
    +
    y − 30
    x4/5x

    Case II :
    Time taken by train to travel
    48 km =
    48
    x

    Time taken by train after accident =
    y − 48
    4/5x

    Total time taken =
    48
    +
    y − 48
    x4/5x

    30
    +
    y − 30
    48
    +
    y − 48
    x4/5xx4/5x

    =
    9
    [∵ Difference between time is 9 minutes]
    60

    y − 30
    y − 48
    +
    30
    48
    4/5x4/5xxx

    =
    9
    60

    y − y − 30 + 48
    +
    (− 18)
    =
    9
    4/5xx60

    5(18)
    18
    =
    9
    4x4x60

    ⇒ 
    90 − 72
    =
    9
    4x60

    x =
    18 × 60
    = 30
    4 × 9

    Hence, original speed = 30 kmph
    Also,
    30
    +
    y − 30
    =
    y
    +
    45
    x4/5xx60

    [Original time + 45 minute = New time]
    ⇒  3x – y = –30
    ⇒  3(30) – y = –30
    ⇒  i.e. Distance = 120 km

    Correct Option: D

    Let the original speed be x and distance be y
    Case I.
    Time taken by train to travel

    30 km =
    30
    x

    Time taken by train after accident =
    y − 30
    4/5x

    Total time taken =
    30
    +
    y − 30
    x4/5x

    Case II :
    Time taken by train to travel
    48 km =
    48
    x

    Time taken by train after accident =
    y − 48
    4/5x

    Total time taken =
    48
    +
    y − 48
    x4/5x

    30
    +
    y − 30
    48
    +
    y − 48
    x4/5xx4/5x

    =
    9
    [∵ Difference between time is 9 minutes]
    60

    y − 30
    y − 48
    +
    30
    48
    4/5x4/5xxx

    =
    9
    60

    y − y − 30 + 48
    +
    (− 18)
    =
    9
    4/5xx60

    5(18)
    18
    =
    9
    4x4x60

    ⇒ 
    90 − 72
    =
    9
    4x60

    x =
    18 × 60
    = 30
    4 × 9

    Hence, original speed = 30 kmph
    Also,
    30
    +
    y − 30
    =
    y
    +
    45
    x4/5xx60

    [Original time + 45 minute = New time]
    ⇒  3x – y = –30
    ⇒  3(30) – y = –30
    ⇒  i.e. Distance = 120 km



  1. Ram starts his journey from Bombay to Pune and simultaneously Mohan starts from Pune to Bombay. After crossing each other they finish their remaining
    journey in 6
    1
    and 4 hours respectively.
    4
    What is Mohan’s speed if Ram’s speed is 20 km per hr. ?









  1. View Hint View Answer Discuss in Forum


    Suppose that Ram and Mohan meet at A. Let Ram’s speed be x km per hr. and Mohan’s speed be y km per hr.

    Then AP =
    25
    x km
    x

    and AB = 4y km.
    Now, time taken by Ram in going from
    B to A =
    4y
    x

    and the time taken by Mohan in going from
    P to A =
    25x
    .
    4y

    Obviously time taken is equal
    ∴ 
    4y
    =
    25x
    x4y

    ⇒  16y2 = 25x2
    ⇒ 
    y2
    =
    25
    x216

    ⇒ 
    y
    =
    5
    x4

    ⇒  y =
    5
    x
    4

    Here, x = 20 km per hr.
    ∴  y = Mohan’s speed
    =
    5
    × 20 = 25 km per hr.
    4

    Correct Option: C


    Suppose that Ram and Mohan meet at A. Let Ram’s speed be x km per hr. and Mohan’s speed be y km per hr.

    Then AP =
    25
    x km
    x

    and AB = 4y km.
    Now, time taken by Ram in going from
    B to A =
    4y
    x

    and the time taken by Mohan in going from
    P to A =
    25x
    .
    4y

    Obviously time taken is equal
    ∴ 
    4y
    =
    25x
    x4y

    ⇒  16y2 = 25x2
    ⇒ 
    y2
    =
    25
    x216

    ⇒ 
    y
    =
    5
    x4

    ⇒  y =
    5
    x
    4

    Here, x = 20 km per hr.
    ∴  y = Mohan’s speed
    =
    5
    × 20 = 25 km per hr.
    4


  1. A car driving in the morning fog passes a man walking at 4 km per hr. in the same direction. The man can see the car for 3 minutes and visibility is upto a distance of 130 metres. Find the speed of the car.









  1. View Hint View Answer Discuss in Forum

    Distance covered by man in 3 minutes

    =
    4 × 1000
    m
    × 3 minutes
    60minutes

    =200 metres
    Total distance covered by the car in 3 min.
    = (200 + 130) m = 330 metres
    ∴   Speed of the car =
    330
    m per min.
    3

    = 110 m per minutes
    =
    110
    1000
    1
    60

    =
    33
    km per hr.
    5

    or 6.6 kmph

    Correct Option: B

    Distance covered by man in 3 minutes

    =
    4 × 1000
    m
    × 3 minutes
    60minutes

    =200 metres
    Total distance covered by the car in 3 min.
    = (200 + 130) m = 330 metres
    ∴   Speed of the car =
    330
    m per min.
    3

    = 110 m per minutes
    =
    110
    1000
    1
    60

    =
    33
    km per hr.
    5

    or 6.6 kmph



  1. A and B start simultaneously at 5 km per hr. and 4 km per hr. from P and Q, 180 kms apart, towards Q and P respectively. They cross each other at M and after reaching Q and P turn back immediately and meet again at N. Find the distance MN.









  1. View Hint View Answer Discuss in Forum


    When A and B cross each other at M for the first time, they have together covered the whole distance PQ = 180 km.
    When they meet again at N, they have together covered total distance equal to 3 times of PQ = 3 × 180 = 540 km.

    PM =
    5
    × 180 = 100 km
    5 + 4

    [Distance covered by each will be in the ratio of their speeds)
    QP + PN =
    4
    × 540 = 240 km
    5 + 4

    or PN = 240 – QP = 240 – 180 = 60 km.
    Then, MN = PM – PN
    = 100 – 60 = 40 km.

    Correct Option: B


    When A and B cross each other at M for the first time, they have together covered the whole distance PQ = 180 km.
    When they meet again at N, they have together covered total distance equal to 3 times of PQ = 3 × 180 = 540 km.

    PM =
    5
    × 180 = 100 km
    5 + 4

    [Distance covered by each will be in the ratio of their speeds)
    QP + PN =
    4
    × 540 = 240 km
    5 + 4

    or PN = 240 – QP = 240 – 180 = 60 km.
    Then, MN = PM – PN
    = 100 – 60 = 40 km.