Speed, Time and Distance


  1. A car can finish a certain journey in 10 hours at the speed of 42 kmph. In order to cover the same distance in 7 hours, the speed of the car (km/h) must be increased by :
    1. 12
    2. 15
    3. 18
    4. 24

  1. View Hint View Answer Discuss in Forum

    Distance covered by car = 42 × 10 = 420 km.
    New time = 7 hours

    ∴  Required speed =
    420
    = 60 kmph.
    7

    ∴  Required increase
    = (60 – 42) kmph
    = 18 kmph

    Correct Option: C

    Distance covered by car = 42 × 10 = 420 km.
    New time = 7 hours

    ∴  Required speed =
    420
    = 60 kmph.
    7

    ∴  Required increase
    = (60 – 42) kmph
    = 18 kmph


  1. A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.
    1. 1028
      6
      km.
      7
    2. 189 km.
    3. 378 km.
    4. 270 km.

  1. View Hint View Answer Discuss in Forum

    Let the distance covered be 2x km.

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    x
    = 5
    15
    =
    1
    6045604

    ⇒ 
    3x + 4x
    =
    21
    1804

    ⇒ 7x =
    21
    × 180
    4

    ⇒  x =
    21 × 180
    = 135 km.
    4 × 7

    ∴  Length of total journey
    = 2 × 135 = 270 km.

    Correct Option: D

    Let the distance covered be 2x km.

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    x
    = 5
    15
    =
    1
    6045604

    ⇒ 
    3x + 4x
    =
    21
    1804

    ⇒ 7x =
    21
    × 180
    4

    ⇒  x =
    21 × 180
    = 135 km.
    4 × 7

    ∴  Length of total journey
    = 2 × 135 = 270 km.



  1. A student goes to school at the
    rate of 2
    1
    km/hr and reaches 6 minutes late.
    2
    If he travels at the speed of 3 km/hr. he is 10 minutes early. What is the distance to the school?
    1. 4 km
    2. 3
      1
      km
      2
    3. 1 km
    4. 3
      1
      km
      4

  1. View Hint View Answer Discuss in Forum

    Let the distance of school be x km.
    Difference of time = 6 + 10

    = 16 minutes =
    16
    hour
    60

    =
    4
    hour
    15

    Time =
    Distance
    Speed

    ∴ 
    x
    x
    =
    4
    5/2315

    ⇒ 
    2x
    x
    =
    4
    5315

    ⇒ 
    6x − 5x
    4
    1515

    ⇒  x = 4 km.
    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
    60

    =
    16
    = 4 km
    4

    Correct Option: A

    Let the distance of school be x km.
    Difference of time = 6 + 10

    = 16 minutes =
    16
    hour
    60

    =
    4
    hour
    15

    Time =
    Distance
    Speed

    ∴ 
    x
    x
    =
    4
    5/2315

    ⇒ 
    2x
    x
    =
    4
    5315

    ⇒ 
    6x − 5x
    4
    1515

    ⇒  x = 4 km.
    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
    60

    =
    16
    = 4 km
    4


  1. A man starts from a place P and reaches the place Q in 7 hours.
    He travels
    1
    th of the distance at 10 km/hour
    4
    and the remaining distance at 12 km/hour. The distance between P and Q is
    1. 72 km
    2. 90 km
    3. 80 km
    4. 70 km

  1. View Hint View Answer Discuss in Forum

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    3x
    = 7
    1012

    ⇒ 
    x
    +
    x
    = 7
    104

    ⇒ 
    2x + 5x
    = 7
    20

    ⇒  7x = 7 × 20
    ∴  x =
    7 × 20
    = 20 km.
    7

    ∴  PQ = 4x = 4 × 20 = 80 km.

    Correct Option: C

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    3x
    = 7
    1012

    ⇒ 
    x
    +
    x
    = 7
    104

    ⇒ 
    2x + 5x
    = 7
    20

    ⇒  7x = 7 × 20
    ∴  x =
    7 × 20
    = 20 km.
    7

    ∴  PQ = 4x = 4 × 20 = 80 km.



  1. A student starting from his house walks at a
    speed of 2
    1
    km/hour and reaches his
    2
    school 6 minutes late. Next day starting at
    the same time he increases his speed by 1 km/hour and reaches 6 minutes early. The distance between the school and his house is
    1. 4 km
    2. 3
      1
      km
      2
    3. 1
      3
      km
      4
    4. 6 km

  1. View Hint View Answer Discuss in Forum

    Distance between school and house = x km (let)

    Time =
    Distance
    Speed

    According to the question,
    x
    x
    =
    6 + 6
    =
    1
    5/27/2605

    (Difference of time = 6 + 6 =12 minutes)
    ⇒ 
    2x
    2x
    =
    1
    575

    ⇒ 
    14x − 10x
    =
    1
    4x
    =
    1
    355355

    ⇒ 4x =
    35
    = 7
    5

    ⇒  x =
    7
    = 1
    3
    km.
    44


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

    S2 = 3
    1
    , t2 = 6
    2

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

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

    =
    35
    ×
    12
    460

    =
    7
    = 1
    3
    km
    44

    Correct Option: C

    Distance between school and house = x km (let)

    Time =
    Distance
    Speed

    According to the question,
    x
    x
    =
    6 + 6
    =
    1
    5/27/2605

    (Difference of time = 6 + 6 =12 minutes)
    ⇒ 
    2x
    2x
    =
    1
    575

    ⇒ 
    14x − 10x
    =
    1
    4x
    =
    1
    355355

    ⇒ 4x =
    35
    = 7
    5

    ⇒  x =
    7
    = 1
    3
    km.
    44


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

    S2 = 3
    1
    , t2 = 6
    2

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

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

    =
    35
    ×
    12
    460

    =
    7
    = 1
    3
    km
    44