Speed, Time and Distance


  1. The speed of a car is 54 km/hr. What is its speed in m/sec?
    1. 15 m/sec
    2. 19-44 m/sec
    3. 194.4 m/sec
    4. 150 m/sec

  1. View Hint View Answer Discuss in Forum

    1 kmph =
    5
    m/sec
    18

    ∴  54 kmph =
    5
    × 54
    18

    = 15 m/sec.

    Correct Option: A

    1 kmph =
    5
    m/sec
    18

    ∴  54 kmph =
    5
    × 54
    18

    = 15 m/sec.


  1. A car goes 20 metres in a second. Find its speed in km/hr.
    1. 18
    2. 72
    3. 36
    4. 20

  1. View Hint View Answer Discuss in Forum

    1 m/sec =
    18
    kmph
    5

    ∴  20 m/sec =
    20 × 18
    = 72 kmph
    5

    Correct Option: B

    1 m/sec =
    18
    kmph
    5

    ∴  20 m/sec =
    20 × 18
    = 72 kmph
    5



  1. A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/
    hour and 60 km/hour respectively. Its average speed over this distance is
    1. 30 km/hour
    2. 20 km/hour
    3. 60 km/hour
    4. 40 km/hour

  1. View Hint View Answer Discuss in Forum

    Total distance = 7 × 4 = 28 km.

    Total time
    =
    7
    +
    7
    +
    7
    +
    7
    hours
    10203060

    =
    42 + 21 + 14 + 7
    hours
    60

    =
    84
    hours =
    7
    hours
    605

    ∴  Average speed =
    Total distance
    Total time

    =
    28
    kmph
    7/5

    =
    28 × 5
    = 20 kmph
    7

    Correct Option: B

    Total distance = 7 × 4 = 28 km.

    Total time
    =
    7
    +
    7
    +
    7
    +
    7
    hours
    10203060

    =
    42 + 21 + 14 + 7
    hours
    60

    =
    84
    hours =
    7
    hours
    605

    ∴  Average speed =
    Total distance
    Total time

    =
    28
    kmph
    7/5

    =
    28 × 5
    = 20 kmph
    7


  1. A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is
    1. 540 km
    2. 420 km
    3. 600 km
    4. 620 km

  1. View Hint View Answer Discuss in Forum

    Fixed distance = x km and certain speed = y kmph (let).
    Case I,

    x
    =
    x
    – 1
    y + 10y

    ⇒ 
    x
    + 1 =
    x
      --- (i)
    y + 10y


    Case II,
    x
    =
    x
    − 1 −
    3
    y + 20y4

    =
    x
    4 + 3
    y4

    ⇒ 
    x
    +
    7
    =
    x
      --- (ii)
    y + 204y

    From equations (i) and (ii),
    x
    + 1 =
    x
    +
    7
    y + 10y + 204

    ⇒ 
    x
    x
    =
    7
    − 1
    y + 10y + 204

    ⇒ x
    y + 20 − y − 10
    (y + 10)(y + 20)

    =
    7 − 4
    =
    3
    44

    ⇒ 3 (y + 10) (y + 20) = 40 x
    ⇒ 
    3 (y + 10) (y + 20)
    = x   ---(iii)
    40

    From equation (i),
    3 (y + 10) (y + 20)
    + 1
    40(y + 10)

    =
    3 (y + 10) (y + 20)
    40y

    ⇒  3 (y +20) + 40
    =
    3 (y + 10) (y + 20)
    y

    ⇒  3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
    ⇒  3y2 + 100y = 3y2 + 90y + 600
    ⇒  10y = 600 ⇒ y = 60
    Again from equation (i),
    x
    + 1 =
    x
    y + 10y

    ⇒ 
    x
    + 1 =
    x
    60 + 1060

    ⇒ 
    x
    + 1 =
    x
    7060

    ⇒ 
    x + 70
    =
    x
    7060

    ⇒  6x + 420 = 7x
    ⇒  7x – 6x = 420
    ⇒  x = 420 km.

    Correct Option: B

    Fixed distance = x km and certain speed = y kmph (let).
    Case I,

    x
    =
    x
    – 1
    y + 10y

    ⇒ 
    x
    + 1 =
    x
      --- (i)
    y + 10y


    Case II,
    x
    =
    x
    − 1 −
    3
    y + 20y4

    =
    x
    4 + 3
    y4

    ⇒ 
    x
    +
    7
    =
    x
      --- (ii)
    y + 204y

    From equations (i) and (ii),
    x
    + 1 =
    x
    +
    7
    y + 10y + 204

    ⇒ 
    x
    x
    =
    7
    − 1
    y + 10y + 204

    ⇒ x
    y + 20 − y − 10
    (y + 10)(y + 20)

    =
    7 − 4
    =
    3
    44

    ⇒ 3 (y + 10) (y + 20) = 40 x
    ⇒ 
    3 (y + 10) (y + 20)
    = x   ---(iii)
    40

    From equation (i),
    3 (y + 10) (y + 20)
    + 1
    40(y + 10)

    =
    3 (y + 10) (y + 20)
    40y

    ⇒  3 (y +20) + 40
    =
    3 (y + 10) (y + 20)
    y

    ⇒  3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
    ⇒  3y2 + 100y = 3y2 + 90y + 600
    ⇒  10y = 600 ⇒ y = 60
    Again from equation (i),
    x
    + 1 =
    x
    y + 10y

    ⇒ 
    x
    + 1 =
    x
    60 + 1060

    ⇒ 
    x
    + 1 =
    x
    7060

    ⇒ 
    x + 70
    =
    x
    7060

    ⇒  6x + 420 = 7x
    ⇒  7x – 6x = 420
    ⇒  x = 420 km.



  1. Walking
    6
    th of his usual speed, a man is 12 minutes
    7
    late. The usual time taken by him to cover that distance is
    1. 1 hour
    2. 1 hour 12 minutes
    3. 1 hour 15 minutes
    4. 1 hour 20 minutes

  1. View Hint View Answer Discuss in Forum

    Time and speed are inversely proportional.

    ∴  Usual time ×
    7
    – usual time = 12 minutes
    6

    ⇒  Usual time ×
    1
    = 12 minutes
    6

    ∴  Usual time = 72 minutes
    = 1 hour 12 minutes
    Second Method :
    t =
    12
    =
    1
    hrs.
    605

    Usual time =
    A
    × time
    Diff. of A and B

    =
    6
    ×
    1
    = 1
    1
    hrs.
    (7 − 6)55

    = 1 hrs. 12 minutes

    Correct Option: B

    Time and speed are inversely proportional.

    ∴  Usual time ×
    7
    – usual time = 12 minutes
    6

    ⇒  Usual time ×
    1
    = 12 minutes
    6

    ∴  Usual time = 72 minutes
    = 1 hour 12 minutes
    Second Method :
    t =
    12
    =
    1
    hrs.
    605

    Usual time =
    A
    × time
    Diff. of A and B

    =
    6
    ×
    1
    = 1
    1
    hrs.
    (7 − 6)55

    = 1 hrs. 12 minutes