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. 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. 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. 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. 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. 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