Speed, Time and Distance


  1. A car can cover a certain distance in 4
    1
    hours.
    2
    If the speed is increased by 5 km/hour, it would take
    1
    hour less to cover the same distance.
    2

    Find the slower speed of the car.
    1. 50 km/hour
    2. 40 km/hour
    3. 45 km/hour
    4. 60 km/hour

  1. View Hint View Answer Discuss in Forum

    Let the initial speed of the car be x kmph and the distance be y km.

    Then, y =
    9
    x     ...(i)
    2

    and, y = 4 (x + 5)     ...(ii)
    ∴  
    9x
    = 4 (x + 5)
    2

    ⇒   9x = 8x + 40
    ⇒   x = 40 kmph

    Correct Option: B

    Let the initial speed of the car be x kmph and the distance be y km.

    Then, y =
    9
    x     ...(i)
    2

    and, y = 4 (x + 5)     ...(ii)
    ∴  
    9x
    = 4 (x + 5)
    2

    ⇒   9x = 8x + 40
    ⇒   x = 40 kmph


  1. A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is
    1. 9 km
    2. 5 km
    3. 4 km
    4. 6 km

  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.
    According to the question,

    x
    x
    =
    18
    4560

    ⇒ 
    5x − 4x
    3
    2010

    ⇒  x =
    3
    × 20 = 6 km
    10


    Second Methdod :
    Here, S1 = 4, t1 = 9
    S2 = 5, t2 = 9
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (4 × 5)(9 + 9)
    5 − 4

    = 20 ×
    18
    = 6 km
    60

    Correct Option: D

    Let the required distance be x km.
    According to the question,

    x
    x
    =
    18
    4560

    ⇒ 
    5x − 4x
    3
    2010

    ⇒  x =
    3
    × 20 = 6 km
    10


    Second Methdod :
    Here, S1 = 4, t1 = 9
    S2 = 5, t2 = 9
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (4 × 5)(9 + 9)
    5 − 4

    = 20 ×
    18
    = 6 km
    60



  1. A student walks from his house at
    a speed of 2
    1
    km per hour and reaches his school
    2
    6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?
    1. 5
      km
      4
    2. 7
      km
      4
    3. 9
      km
      4
    4. 11
      km
      4

  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.
    Difference of time

    = 6 + 6 = 12 minutes =
    1
    hr.
    5

    According to the question,
    ⇒ 
    x
    x
    =
    1
    2x
    2x
    =
    1
    5/27/25575

    ⇒ 
    14x − 10x
    1
    355

    ⇒ 
    4x
    =
    1
    ⇒ x =
    35
    =
    7
    km.
    355204


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

    Here, S2 = 3
    1
    , t2 = 6
    2

    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    5
    ×
    7
    × (6 + 6)
    22
    5
    7
    22

    =
    35
    ×
    12
    =
    7
    km
    4604

    Correct Option: B

    Let the required distance be x km.
    Difference of time

    = 6 + 6 = 12 minutes =
    1
    hr.
    5

    According to the question,
    ⇒ 
    x
    x
    =
    1
    2x
    2x
    =
    1
    5/27/25575

    ⇒ 
    14x − 10x
    1
    355

    ⇒ 
    4x
    =
    1
    ⇒ x =
    35
    =
    7
    km.
    355204


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

    Here, S2 = 3
    1
    , t2 = 6
    2

    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    5
    ×
    7
    × (6 + 6)
    22
    5
    7
    22

    =
    35
    ×
    12
    =
    7
    km
    4604


  1. If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is
    1. 13
    2. 15
    3. 19
    4. 21

  1. View Hint View Answer Discuss in Forum

    If the distance be x km, then

    x
    x
    =
    6
    405060

    ⇒ 
    x
    x
    = 1
    45

    ⇒  x = 20 km.
    ∴   Required time =
    20
    hour – 11 minutes
    40

    =
    1
    × 60 − 11 minutes
    2

    = 19 minutes

    Correct Option: C

    If the distance be x km, then

    x
    x
    =
    6
    405060

    ⇒ 
    x
    x
    = 1
    45

    ⇒  x = 20 km.
    ∴   Required time =
    20
    hour – 11 minutes
    40

    =
    1
    × 60 − 11 minutes
    2

    = 19 minutes



  1. A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?
    1. 5
      km
      8
    2. 8 km
    3. 5 km
    4. 10 km

  1. View Hint View Answer Discuss in Forum

    Let x km. be the required distance.
    Difference in time = 2.5 + 5 = 7.5 minutes

    =
    7.5
    hrs. =
    1
    hrs.
    608

    Now, 
    x
    x
    =
    1
    8108

    ⇒ 
    5x − 4x
    =
    1
    408

    ⇒  x =
    40
    = 5 km.
    8


    Second Methdod :
    Here, S1 = 8, t1 = 2.5
    S2 = 10, t2 = 5
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (8 × 10)(2.5 + 5)
    10 − 8

    = 40 ×
    7.5
    = 5 km
    60

    Correct Option: C

    Let x km. be the required distance.
    Difference in time = 2.5 + 5 = 7.5 minutes

    =
    7.5
    hrs. =
    1
    hrs.
    608

    Now, 
    x
    x
    =
    1
    8108

    ⇒ 
    5x − 4x
    =
    1
    408

    ⇒  x =
    40
    = 5 km.
    8


    Second Methdod :
    Here, S1 = 8, t1 = 2.5
    S2 = 10, t2 = 5
    Distance =
    (S1 ×S2)(t1 + t2)
    S2 −S1

    =
    (8 × 10)(2.5 + 5)
    10 − 8

    = 40 ×
    7.5
    = 5 km
    60