Speed, Time and Distance


  1. If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/ hour, it is late by 3 minutes. The correct time to cover the journey is
    1. 58 minutes
    2. 2 hours
    3. 1 hour
    4. 59 minutes

  1. View Hint View Answer Discuss in Forum

    Distance of journey = x km
    Difference of time = 12 – 3 = 9 minutes

    =
    9
    hour =
    3
    hour
    6020

    ∴  
    x
    x
    =
    3
    708020

    ⇒  
    x
    x
    =
    3
    782

    ⇒ 
    8x − 7x
    =
    3
    562

    ⇒ 
    x
    =
    3
    562

    ⇒  x =
    3
    × 56 = 84 km
    2

    ∴   Required correct time
    =
    84
    hours – 12 minutes
    70

    =
    84
    × 60 − 12 minutes
    70

    = 72 – 12 = 60 minutes
    = 1 hour

    Correct Option: C

    Distance of journey = x km
    Difference of time = 12 – 3 = 9 minutes

    =
    9
    hour =
    3
    hour
    6020

    ∴  
    x
    x
    =
    3
    708020

    ⇒  
    x
    x
    =
    3
    782

    ⇒ 
    8x − 7x
    =
    3
    562

    ⇒ 
    x
    =
    3
    562

    ⇒  x =
    3
    × 56 = 84 km
    2

    ∴   Required correct time
    =
    84
    hours – 12 minutes
    70

    =
    84
    × 60 − 12 minutes
    70

    = 72 – 12 = 60 minutes
    = 1 hour


  1. A train travelling at a speed of 55 km/hr travels from place X to place Y in 4 hours. If its speed is increased by 5 km/hr., then the time of journey is reduced by
    1. 25 minutes
    2. 35 minutes
    3. 20 minutes
    4. 30 minutes

  1. View Hint View Answer Discuss in Forum

    Distance between stations X and Y = Speed × Time
    = 55 × 4 = 220 km.
    New speed = 55 + 5 = 60 kmph

    ∴   Required time =
    220
    60

    =
    11
    hours
    3

    = 3 hours 40 minutes.
    ∴   Required answer
    = 4 hours – 3 hours 40 minutes
    = 20 minutes

    Correct Option: C

    Distance between stations X and Y = Speed × Time
    = 55 × 4 = 220 km.
    New speed = 55 + 5 = 60 kmph

    ∴   Required time =
    220
    60

    =
    11
    hours
    3

    = 3 hours 40 minutes.
    ∴   Required answer
    = 4 hours – 3 hours 40 minutes
    = 20 minutes



  1. If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.
    1. 5 km
    2. 4 km
    3. 6 km
    4. 4.5 km

  1. View Hint View Answer Discuss in Forum

    Let the distance of school be x km, then

    x
    x
    =
    20
    3460

    ⇒ 
    x
    =
    1
    ⇒ x =
    12
    = 4 km
    1233


    Second Methdod :
    Here, S1 = 3, t1 = 10
    S2 = 4, t2 = 10
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (3 × 4)(10 + 10)
    4 − 3

    = 12 ×
    20
    = 4 km
    60

    Correct Option: B

    Let the distance of school be x km, then

    x
    x
    =
    20
    3460

    ⇒ 
    x
    =
    1
    ⇒ x =
    12
    = 4 km
    1233


    Second Methdod :
    Here, S1 = 3, t1 = 10
    S2 = 4, t2 = 10
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (3 × 4)(10 + 10)
    4 − 3

    = 12 ×
    20
    = 4 km
    60


  1. Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is
    1. 3 km
    2. 4 km
    3. 3.5 km
    4. 2 km

  1. View Hint View Answer Discuss in Forum

    Let the distance of the office be x km, then

    x
    x
    =
    8
    5660

    ⇒ 
    6x − 5x
    =
    2
    3015

    ⇒  x = 2 × 2 = 4 km

    Second Methdod :
    Here, S1 = 5, t1 = 6
    S2 = 6, t2 = 2
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (6 × 5)(6 + 2)
    6 − 5

    = 30 ×
    8
    = 4 km
    60

    Correct Option: B

    Let the distance of the office be x km, then

    x
    x
    =
    8
    5660

    ⇒ 
    6x − 5x
    =
    2
    3015

    ⇒  x = 2 × 2 = 4 km

    Second Methdod :
    Here, S1 = 5, t1 = 6
    S2 = 6, t2 = 2
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (6 × 5)(6 + 2)
    6 − 5

    = 30 ×
    8
    = 4 km
    60



  1. A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :
    1. 10 minutes
    2. 13 minutes 20 sec
    3. 13 minutes
    4. 11 minutes 20 sec

  1. View Hint View Answer Discuss in Forum

    Speed of train =
    Distance
    Time

    =
    10
    kmph
    12
    60

    =
    10 × 60
    = 50 kmph
    12

    New speed = 45 kmph
    ∴   Required time =
    10
    hour
    45

    =
    2
    × 60 minutes
    9

    =
    40
    minutes
    3

    or 13 minutes 20 seconds

    Correct Option: B

    Speed of train =
    Distance
    Time

    =
    10
    kmph
    12
    60

    =
    10 × 60
    = 50 kmph
    12

    New speed = 45 kmph
    ∴   Required time =
    10
    hour
    45

    =
    2
    × 60 minutes
    9

    =
    40
    minutes
    3

    or 13 minutes 20 seconds