Speed, Time and Distance


  1. A motor-boat goes 2 km upstream in a stream flowing at 3 km per hr. and then returns
    downstream to the starting point in 30 minutes. Find the speed of the motor-boat in still water.
    1. 9.5 kmph
    2. 8.5 kmph
    3. 9 kmph
    4. 8 kmph

  1. View Hint View Answer Discuss in Forum

    Let the speed of the motorboat in still water be Z km per hr.
    Downstream speed= (Z + 3) km per hr.
    Upstream speed = (Z – 3) km per hr.
    Total journey time

    = 30 minutes =
    30
    hr. =
    1
    hour
    602

    We can write,
    2
    +
    2
    =
    1
    Z − 3Z + 32

    or,   2
    (Z + 3) + (Z − 3)
    =
    1
    (Z − 3) + (Z − 3)2

    or,  
    2Z
    =
    1
    Z2 − 94

    or,   Z2 – 9 = 8Z
    or,   Z2 – 8Z – 9 = 0
    or,   Z2 + Z – 9Z – 9 = 0
    or,   Z(Z + 1) – 9 (Z + 1) = 0
    or,   (Z + 1) (Z – 9) = 0
    ∴  Z = – 1 or 9.
    Since speed can’t be negative
    Therefore, the speed of the motor-boat in still water = 9 km per hr.

    Correct Option: C

    Let the speed of the motorboat in still water be Z km per hr.
    Downstream speed= (Z + 3) km per hr.
    Upstream speed = (Z – 3) km per hr.
    Total journey time

    = 30 minutes =
    30
    hr. =
    1
    hour
    602

    We can write,
    2
    +
    2
    =
    1
    Z − 3Z + 32

    or,   2
    (Z + 3) + (Z − 3)
    =
    1
    (Z − 3) + (Z − 3)2

    or,  
    2Z
    =
    1
    Z2 − 94

    or,   Z2 – 9 = 8Z
    or,   Z2 – 8Z – 9 = 0
    or,   Z2 + Z – 9Z – 9 = 0
    or,   Z(Z + 1) – 9 (Z + 1) = 0
    or,   (Z + 1) (Z – 9) = 0
    ∴  Z = – 1 or 9.
    Since speed can’t be negative
    Therefore, the speed of the motor-boat in still water = 9 km per hr.


  1. A goods train 158 metres long and travelling at the speed of 32 km per hr. leaves Delhi at 6 am. Another mail train 130 metres long and travelling at the average speed of 80 km per hr. leaves Delhi at 12 noon and follows the goods train. At what time will the mail train completely cross the goods train?
    1. 4 hours
    2. 4 hours 21.6 sec.
    3. 5 hours 21.6 sec.
    4. None of these

  1. View Hint View Answer Discuss in Forum

    The goods train leaves Delhi at 6 am and mail train at 12 noon, hence after 6 hours
    The distance covered by the goods train in 6 hours at 32 km per hr. = 32 ´ 6 = 192 kms
    The relative velocity of mail train with respect to goods train = 80 – 32 = 48 km per hr.
    To completely cross the goods train, the mail train will have to cover a distance
    = 192 km + 158m + 130m
    = 192km + 0.158 km + 0.130 km
    = 192.288 km more
    Since, the mail train goes 48 kms more in 1 hour.
    ∴  The mail train goes 192.288 kms more in

    =
    192288
    ×
    1
    =
    2003
    100048500

    = 4 hours 21.6 sec.

    Correct Option: B

    The goods train leaves Delhi at 6 am and mail train at 12 noon, hence after 6 hours
    The distance covered by the goods train in 6 hours at 32 km per hr. = 32 ´ 6 = 192 kms
    The relative velocity of mail train with respect to goods train = 80 – 32 = 48 km per hr.
    To completely cross the goods train, the mail train will have to cover a distance
    = 192 km + 158m + 130m
    = 192km + 0.158 km + 0.130 km
    = 192.288 km more
    Since, the mail train goes 48 kms more in 1 hour.
    ∴  The mail train goes 192.288 kms more in

    =
    192288
    ×
    1
    =
    2003
    100048500

    = 4 hours 21.6 sec.



  1. A man standing on a 170 metre long platform watches that a train takes
    7
    1
    seconds to pass him and 21 seconds to cross
    2
    the platform. Find the speed of train.
    1. 12
      16
      m/sec.
      27
    2. 12.5 m/sec.
    3. 12
      13
      m/sec.
      27
    4. None of these

  1. View Hint View Answer Discuss in Forum

    Let the length of the train be x m and its speed y m/sec.
    Distance covered in crossing the platform = 170 + x metres
    and time taken = 21 seconds

    ∴  Speed y =
    170 + x
        ...(i)
    21

    Distance covered to cross the man = x metres
    and time taken = 7
    and time taken = 7
    1
    =
    15
    seconds
    22

    ∴  Speed y =
    x
    =
    2x
        ...(ii)
    15/215

    From equations (i) and (ii),
    170 + x
    =
    2x
    2115

    ⇒  2550 + 15x = 42x
    ⇒  42x – 15x = 2550
    ⇒  27x = 2550
    ⇒  x =
    2550
    = 94
    4
    metres
    279

    From equation (ii),
    y =
    2 × 2550
    15 × 27

    =
    340
    = 12
    16
    m per sec.
    2727

    Hence, speed = 12
    16
    m per sec.
    27

    Correct Option: A

    Let the length of the train be x m and its speed y m/sec.
    Distance covered in crossing the platform = 170 + x metres
    and time taken = 21 seconds

    ∴  Speed y =
    170 + x
        ...(i)
    21

    Distance covered to cross the man = x metres
    and time taken = 7
    and time taken = 7
    1
    =
    15
    seconds
    22

    ∴  Speed y =
    x
    =
    2x
        ...(ii)
    15/215

    From equations (i) and (ii),
    170 + x
    =
    2x
    2115

    ⇒  2550 + 15x = 42x
    ⇒  42x – 15x = 2550
    ⇒  27x = 2550
    ⇒  x =
    2550
    = 94
    4
    metres
    279

    From equation (ii),
    y =
    2 × 2550
    15 × 27

    =
    340
    = 12
    16
    m per sec.
    2727

    Hence, speed = 12
    16
    m per sec.
    27


  1. Two trains of which one is 50 metres longer than the other are running in opposite directions and cross each other in 10 seconds. If they be running in the same direction then faster train would have passed the other train in 1 minute 30 seconds. The speed of faster train is 90 km per hr. Find the speed of other train.
    1. 25 m/sec.
    2. 20 m/sec.
    3. 30 m/sec.
    4. 35 m/sec.

  1. View Hint View Answer Discuss in Forum

    Let the length of trains be x m and (x + 50)m and the speed of other train be y m per sec.
    The speed of the first train = 90 km per hr.

    = 90 ×
    5
    = 25 m per sec.
    18

    Case : I Opposite direction,
    Their relative speed = (y + 25)m per sec.
    Distance covered = x + x + 50 = 2x + 50 metres
    ∴ Time taken =
    2x + 50
    = 10
    y + 25

    ⇒  2x + 50 = 10y + 250     ...(i)
    Case II. Direction is Same
    Their relative speed = (25 – y) m per sec.
    Distance covered = x + x + 50 = 2x + 50m
    ∴ Time taken =
    2x + 50
    = 90
    25 − y

    ⇒  2x + 50 = 90 (25 – y)     ...(ii)
    From equations (i) and (ii)
    10y + 250 = 2250 – 90y
    ⇒  10y + 90y = 2250 – 250
    ⇒  y =
    2000
    = 20
    100

    Putting y = 20 in equation (i), we have
    2x + 50= 10 × 20 + 250 = 450
    ⇒  2x = 450 – 50 = 400
    ⇒  x =
    400
    = 200
    2

    ∴  x + 50 = 200 + 50 = 250 metres.
    Hence,
    The length of the 1st train = 200 metres.
    The length of the 2nd train = 250 metres.
    The speed of the 2nd train = 20 m per sec.

    Correct Option: B

    Let the length of trains be x m and (x + 50)m and the speed of other train be y m per sec.
    The speed of the first train = 90 km per hr.

    = 90 ×
    5
    = 25 m per sec.
    18

    Case : I Opposite direction,
    Their relative speed = (y + 25)m per sec.
    Distance covered = x + x + 50 = 2x + 50 metres
    ∴ Time taken =
    2x + 50
    = 10
    y + 25

    ⇒  2x + 50 = 10y + 250     ...(i)
    Case II. Direction is Same
    Their relative speed = (25 – y) m per sec.
    Distance covered = x + x + 50 = 2x + 50m
    ∴ Time taken =
    2x + 50
    = 90
    25 − y

    ⇒  2x + 50 = 90 (25 – y)     ...(ii)
    From equations (i) and (ii)
    10y + 250 = 2250 – 90y
    ⇒  10y + 90y = 2250 – 250
    ⇒  y =
    2000
    = 20
    100

    Putting y = 20 in equation (i), we have
    2x + 50= 10 × 20 + 250 = 450
    ⇒  2x = 450 – 50 = 400
    ⇒  x =
    400
    = 200
    2

    ∴  x + 50 = 200 + 50 = 250 metres.
    Hence,
    The length of the 1st train = 200 metres.
    The length of the 2nd train = 250 metres.
    The speed of the 2nd train = 20 m per sec.



  1. A train takes 18 seconds to pass completely through a station 162 metres long and 15 seconds to pass completely through another station 120 metres long. Find the speed of train in km per hr.
    1. 50.4 kmph
    2. 52 kmph
    3. 55 kmph
    4. 60 kmph

  1. View Hint View Answer Discuss in Forum

    Let the length of the train be x metres
    Then, in 18 sec. the train travels (x + 162) metres     ...(i)
    and in 15 sec. the train travels (x + 120) metres
    ∴  In (18 – 15) = 3 sec. the train travels (x + 162)
    – (x + 120) = 42m.

    ∴  In 1 sec the train travels =
    42
    = 14 metres.     ...(ii)
    3

    ∴  In 18 sec. the train travels = 14 × 18 = 252 metres ...(iii)
    From equations (i) and (iii)
    ∴  x + 162 = 252
    ⇒  x = 252 – 162 = 90
    ∴  Length of the train = 90 metres
    Also, from equation (ii) we see
    that in 1hr. the train travels
    = 14 × 60 × 60 metres
    =
    14 × 60 × 60
    km = 50.4 km
    1000

    ∴  The speed of the train
    = 50.4 km per hr.

    Correct Option: A

    Let the length of the train be x metres
    Then, in 18 sec. the train travels (x + 162) metres     ...(i)
    and in 15 sec. the train travels (x + 120) metres
    ∴  In (18 – 15) = 3 sec. the train travels (x + 162)
    – (x + 120) = 42m.

    ∴  In 1 sec the train travels =
    42
    = 14 metres.     ...(ii)
    3

    ∴  In 18 sec. the train travels = 14 × 18 = 252 metres ...(iii)
    From equations (i) and (iii)
    ∴  x + 162 = 252
    ⇒  x = 252 – 162 = 90
    ∴  Length of the train = 90 metres
    Also, from equation (ii) we see
    that in 1hr. the train travels
    = 14 × 60 × 60 metres
    =
    14 × 60 × 60
    km = 50.4 km
    1000

    ∴  The speed of the train
    = 50.4 km per hr.