Speed, Time and Distance


  1. If a man walks at the rate of 5 km/hour, he misses a train by 7 minutes. However if he walks at the rate of 6 km/hour, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is









  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.
    Difference of time = 7 + 5 = 12

    minutes =
    1
    hour
    5

    Time =
    Distance
    Speed

    According to the question,
    x
    x
    =
    1
    565

    ⇒ 
    6x − 5x
    =
    1
    305

    ⇒ 
    x
    =
    1
    305

    ⇒  x =
    30
    = 6 km.
    5

    Correct Option: D

    Let the required distance be x km.
    Difference of time = 7 + 5 = 12

    minutes =
    1
    hour
    5

    Time =
    Distance
    Speed

    According to the question,
    x
    x
    =
    1
    565

    ⇒ 
    6x − 5x
    =
    1
    305

    ⇒ 
    x
    =
    1
    305

    ⇒  x =
    30
    = 6 km.
    5


  1. A train passes an electrical pole in 20 seconds and passes a platform 250 m long in 45 seconds. Find the length of the train.









  1. View Hint View Answer Discuss in Forum

    If the length of train be x metre,
    then speed of train

    =
    x
    =
    x + 250
    2045

    ⇒ 
    x
    =
    x + 250
    49

    ⇒  9x = 4x + 1000
    ⇒  9x – 4x = 1000
    ⇒  5x = 1000
    ⇒  x =
    1000
    = 200 metre
    5

    Correct Option: B

    If the length of train be x metre,
    then speed of train

    =
    x
    =
    x + 250
    2045

    ⇒ 
    x
    =
    x + 250
    49

    ⇒  9x = 4x + 1000
    ⇒  9x – 4x = 1000
    ⇒  5x = 1000
    ⇒  x =
    1000
    = 200 metre
    5



  1. Walking at
    6
    th of his usual speed a man
    7
    is 25 minutes late. His usual time to cover this distance is









  1. View Hint View Answer Discuss in Forum

    Time and speed are inversely proportional.

    ∴ 
    7
    × Usual time – Usual time = 25 minutes
    6

    ⇒ 
    7
    − 1= 25 minutes
    6

    ⇒  Usual time ×
    1
    = 25 minutes
    6

    ∴  Usual time = 25 × 6
    = 150 minutes
    = 2 hours 30 minutes
    Second Method :
    Here, A = 6, B = 7,
    t =
    25
    =
    5
    hrs.
    6012

    Usual time =
    A
    × time
    Diff. of A and B

    =
    6
    ×
    5
    =
    5
    hrs.
    (7 − 6)122

    = 2 hours 30 minutes

    Correct Option: A

    Time and speed are inversely proportional.

    ∴ 
    7
    × Usual time – Usual time = 25 minutes
    6

    ⇒ 
    7
    − 1= 25 minutes
    6

    ⇒  Usual time ×
    1
    = 25 minutes
    6

    ∴  Usual time = 25 × 6
    = 150 minutes
    = 2 hours 30 minutes
    Second Method :
    Here, A = 6, B = 7,
    t =
    25
    =
    5
    hrs.
    6012

    Usual time =
    A
    × time
    Diff. of A and B

    =
    6
    ×
    5
    =
    5
    hrs.
    (7 − 6)122

    = 2 hours 30 minutes


  1. Two trains of lengths 150m and 180m respectively are running in opposite directions on parallel tracks. If their speeds be 50 km/hr and 58 km/hr respectively, in what time will they cross each other?









  1. View Hint View Answer Discuss in Forum

    Relative speed = (50 + 58) kmph

    = 108 ×
    5
    m/sec.
    18

    = 30 m/sec
    ∴  Required time =
    Total length of trains
    Relative Speed

    =
    150 + 180
    seconds
    30

    =
    330
    seconds
    30

    = 11 seconds

    Correct Option: D

    Relative speed = (50 + 58) kmph

    = 108 ×
    5
    m/sec.
    18

    = 30 m/sec
    ∴  Required time =
    Total length of trains
    Relative Speed

    =
    150 + 180
    seconds
    30

    =
    330
    seconds
    30

    = 11 seconds



  1. Two trains start at the same time from Aligarh and Delhi and proceed towards each other at the rate of 14 km and 21 km per hour respectively. When they meet, it is found that one train has travelled 70 km more than the other. The distance between two stations is









  1. View Hint View Answer Discuss in Forum

    Let the trains meet each other after t hours.
    Distance = Speed × Time
    According to the question,
    21t – 14t = 70

    ⇒  7t = 70 ⇒ t =
    70
    = 10 hours
    7

    ∴  Required distance
    = 21t + 14t = 35t
    = 35 × 10 = 350 km.
    Second Method :
    Here, a = 21, b = 14, d = 70
    Required distance =
    a + b
    × d
    a − b

    =
    21 + 14
    × 70
    21 − 14

    =
    35
    × 70 = 350 km.
    7

    Correct Option: A

    Let the trains meet each other after t hours.
    Distance = Speed × Time
    According to the question,
    21t – 14t = 70

    ⇒  7t = 70 ⇒ t =
    70
    = 10 hours
    7

    ∴  Required distance
    = 21t + 14t = 35t
    = 35 × 10 = 350 km.
    Second Method :
    Here, a = 21, b = 14, d = 70
    Required distance =
    a + b
    × d
    a − b

    =
    21 + 14
    × 70
    21 − 14

    =
    35
    × 70 = 350 km.
    7