Speed, Time and Distance


  1. A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is
    1. 60 km
    2. 64 km
    3. 80 km
    4. 55 km

  1. View Hint View Answer Discuss in Forum

    Let the distance between stations be x km, then speed of train

    =
    2x
    = (4x/3) kmph
    45
    60

    ∴  
    x
    = (48/60)
    4x
    − 5
    3

    ⇒ 
    3x
    =
    4
    4x − 155

    ⇒  16x – 60 = 15x
    ⇒  x = 60 km

    Correct Option: A

    Let the distance between stations be x km, then speed of train

    =
    2x
    = (4x/3) kmph
    45
    60

    ∴  
    x
    = (48/60)
    4x
    − 5
    3

    ⇒ 
    3x
    =
    4
    4x − 155

    ⇒  16x – 60 = 15x
    ⇒  x = 60 km


  1. When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is
    1. 6 km
    2. 7 km
    3. 12 km
    4. 16 km

  1. View Hint View Answer Discuss in Forum

    Let the distance be x km.

    ∴  
    x
    x
    =
    12
    101260

    ⇒ 
    6x − 5x
    =
    1
    605

    ⇒  x =
    1
    × 60 = 12 km
    5


    Second Methdod :
    Here, S1 = 10, t1 = 6
    S2 = 12, t2 = 6
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (10 × 12)(6 + 6)
    12 − 10

    =
    120 × 12
    2

    = 60 ×
    12
    km = 12 km
    60

    Correct Option: C

    Let the distance be x km.

    ∴  
    x
    x
    =
    12
    101260

    ⇒ 
    6x − 5x
    =
    1
    605

    ⇒  x =
    1
    × 60 = 12 km
    5


    Second Methdod :
    Here, S1 = 10, t1 = 6
    S2 = 12, t2 = 6
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (10 × 12)(6 + 6)
    12 − 10

    =
    120 × 12
    2

    = 60 ×
    12
    km = 12 km
    60



  1. A student goes to school at the rate of
    2
    1
    km/h and reaches 6 minutes late.
    2
    If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is
    1. 5
    2. 4
    3. 3
    4. 1

  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.

    x
    x
    =
    16
    5/2360

    ⇒ 
    2x
    x
    =
    4
    5315

    ⇒ 
    6x − 5x
    =
    4
    1515

    ⇒  x = 4 km.

    Second Methdod :
    Here, S1 = 2
    1
    , t1 = 6
    2

    S2 = 3, t2 = 10
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    5
    × 3(6 + 10)
    2
    3 −
    5
    2

    = 15 ×
    16
    km = 4 km
    60

    Correct Option: B

    Let the required distance be x km.

    x
    x
    =
    16
    5/2360

    ⇒ 
    2x
    x
    =
    4
    5315

    ⇒ 
    6x − 5x
    =
    4
    1515

    ⇒  x = 4 km.

    Second Methdod :
    Here, S1 = 2
    1
    , t1 = 6
    2

    S2 = 3, t2 = 10
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    5
    × 3(6 + 10)
    2
    3 −
    5
    2

    = 15 ×
    16
    km = 4 km
    60


  1. Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?
    1. 5 km
    2. 8 km
    3. 3 km
    4. 2 km

  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.

    Then,  
    x
    x
    =
    24
    3560

    ⇒ 
    5x − 3x
    =
    2
    155

    ⇒  
    2x
    = 2
    3

    ⇒  2x = 2 × 3 ⇒ x = 3 km

    Second Methdod :
    Here, S1 = 3, t1 = 9
    S2 = 5, t2 = 15
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (3 × 5)(9 + 15)
    5 − 3

    =
    15
    ×
    24
    260

    = 3 km

    Correct Option: C

    Let the required distance be x km.

    Then,  
    x
    x
    =
    24
    3560

    ⇒ 
    5x − 3x
    =
    2
    155

    ⇒  
    2x
    = 2
    3

    ⇒  2x = 2 × 3 ⇒ x = 3 km

    Second Methdod :
    Here, S1 = 3, t1 = 9
    S2 = 5, t2 = 15
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (3 × 5)(9 + 15)
    5 − 3

    =
    15
    ×
    24
    260

    = 3 km



  1. Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is
    1. 20 km
    2. 21 km
    3. 22 km
    4. 24 km

  1. View Hint View Answer Discuss in Forum

    Let the distance of office be x km.

    ∴  
    x
    x
    =
    11
    243060

    ⇒ 
    5x − 4x
    =
    11
    12060

    ⇒ 
    x
    =
    11
    12060

    ⇒  x =
    11
    × 120 = 22 km.
    60


    Second Methdod :
    Here, S1 = 24, t1 = 5
    S2 = 30, t2 = 6
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    24 × 30(5 + 6)
    30 − 24

    =
    720 × 11
    = 22 km
    6 × 60

    Correct Option: C

    Let the distance of office be x km.

    ∴  
    x
    x
    =
    11
    243060

    ⇒ 
    5x − 4x
    =
    11
    12060

    ⇒ 
    x
    =
    11
    12060

    ⇒  x =
    11
    × 120 = 22 km.
    60


    Second Methdod :
    Here, S1 = 24, t1 = 5
    S2 = 30, t2 = 6
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    24 × 30(5 + 6)
    30 − 24

    =
    720 × 11
    = 22 km
    6 × 60