Speed, Time and Distance


  1. When Alisha goes by car at 50 kmph, she reaches her office 5 minutes late. But when she takes her motorbike, she reaches 3 minutes early. If her office is 25 kms away, what is the approximate average speed at which she rides her motorbike ?
    1. 68 kmph
    2. 62 kmph
    3. 58 kmph
    4. 52 kmph

  1. View Hint View Answer Discuss in Forum

    Difference of time = 5 + 3 = 8 minutes

    =
    8
    hour =
    2
    hour
    6015

    If the speed of motorbike be x kmph, then
    25
    25
    =
    2
    50x15

    ⇒ 
    25
    =
    1
    2
    x215

    ⇒ 
    25
    =
    15 − 4
    =
    11
    x3030

    ⇒ 11x = 25 × 30
    ⇒ x =
    25 × 30
    11

    =
    750
    = 68.18 kmph
    11

    ≈ 68 kmph

    Correct Option: A

    Difference of time = 5 + 3 = 8 minutes

    =
    8
    hour =
    2
    hour
    6015

    If the speed of motorbike be x kmph, then
    25
    25
    =
    2
    50x15

    ⇒ 
    25
    =
    1
    2
    x215

    ⇒ 
    25
    =
    15 − 4
    =
    11
    x3030

    ⇒ 11x = 25 × 30
    ⇒ x =
    25 × 30
    11

    =
    750
    = 68.18 kmph
    11

    ≈ 68 kmph


  1. A man covers a total distance of 100 km on bicycle. For the first 2 hours, the speed was 20 km/hr and for the rest of the journey, it came down to 10 km/hr. The average speed will be
    1. 12
      1
      km/hr
      2
    2. 13 km/hr
    3. 15
      1
      km/hr
      8
    4. 20 km/hr

  1. View Hint View Answer Discuss in Forum

    Distance covered in first 2 hours
    = 2 × 20 = 40 km.
    Remaining distance
    = 100 – 40 = 60 km.
    Time taken in covering 60 km at 10 kmph

    =
    60
    = 6 hours
    10

    ∴  Required average speed =
    Total distance
    Total Time

    =
    100
    kmph
    2 + 6

    =
    100
    kmph
    8

    =
    25
    kmph
    2

    = 12
    1
    kmph
    2

    Correct Option: A

    Distance covered in first 2 hours
    = 2 × 20 = 40 km.
    Remaining distance
    = 100 – 40 = 60 km.
    Time taken in covering 60 km at 10 kmph

    =
    60
    = 6 hours
    10

    ∴  Required average speed =
    Total distance
    Total Time

    =
    100
    kmph
    2 + 6

    =
    100
    kmph
    8

    =
    25
    kmph
    2

    = 12
    1
    kmph
    2



  1. A train travels 500 m in first minute. In the next 4 minutes, it travels in each minute 125 m more than that in the previous minute. The average speed per hour of the train during those 5 minutes will be
    1. 30 km/hr
    2. 45 km/hr
    3. 50 km/hr
    4. 55 km/hr

  1. View Hint View Answer Discuss in Forum

    Total distance covered by train in 5 minutes
    = (500 + 625 + 750 + 875 + 1000) metre = 3750 metre
    = 3.75 km.

    Time = 5 minutes =
    5
    hour
    60

    =
    1
    hour
    12

    Speed of train =
    Distance
    Time

    =
    3.75
    kmph
    (1/12)

    = (3.75 × 12) kmph
    = 45 kmph

    Correct Option: B

    Total distance covered by train in 5 minutes
    = (500 + 625 + 750 + 875 + 1000) metre = 3750 metre
    = 3.75 km.

    Time = 5 minutes =
    5
    hour
    60

    =
    1
    hour
    12

    Speed of train =
    Distance
    Time

    =
    3.75
    kmph
    (1/12)

    = (3.75 × 12) kmph
    = 45 kmph


  1. A train runs at an average speed of 75 km/hr. If the distance to be covered is 1050 kms, how long will the train take to cover it ?
    1. 13 hrs
    2. 12 hrs
    3. 15 hrs
    4. 14 hrs

  1. View Hint View Answer Discuss in Forum

    Time =
    Distance
    Speed

    =
    1050
    = 14 hours
    75

    Correct Option: D

    Time =
    Distance
    Speed

    =
    1050
    = 14 hours
    75



  1. A man walks from his house at an average speed of 5 km per hour and reaches his office 6 minutes late. If he walks at an average speed of 6 km/h he reaches 2 minutes early. The distance of the office from his house is
    1. 6 km
    2. 9 km
    3. 12 km
    4. 4 km

  1. View Hint View Answer Discuss in Forum

    Required distance of office from house = x km. (let)

    Time =
    Distance
    Speed

    ∴  According to the question,
    x
    x
    =
    6 + 2
    =
    2
    566015

    ⇒ 
    6x − 5x
    30

    =
    2
    15

    ⇒ 
    x
    =
    2
    3015

    ⇒ x =
    2
    × 30 = 4 km.
    15


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

    =
    (5 × 6)(6 + 2)
    6 − 5

    = 30 ×
    8
    = 4 km.
    60

    Correct Option: D

    Required distance of office from house = x km. (let)

    Time =
    Distance
    Speed

    ∴  According to the question,
    x
    x
    =
    6 + 2
    =
    2
    566015

    ⇒ 
    6x − 5x
    30

    =
    2
    15

    ⇒ 
    x
    =
    2
    3015

    ⇒ x =
    2
    × 30 = 4 km.
    15


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

    =
    (5 × 6)(6 + 2)
    6 − 5

    = 30 ×
    8
    = 4 km.
    60