Speed, Time and Distance


  1. A man is walking at a speed of 10 kmph. After every km, he takes a rest for 5 minutes. How much time will he take to cover a distance of 5 km?
    1. 60 minutes
    2. 50 minutes
    3. 40 minutes
    4. 70 minutes

  1. View Hint View Answer Discuss in Forum

    Time taken in covering 5 Km =
    5
    =
    1
    hour
    102

    = 30 minutes
    That person will take rest for four times.
    ∴  Required time
    = (30 + 4 × 5) minutes
    = 50 minutes

    Correct Option: B

    Time taken in covering 5 Km =
    5
    =
    1
    hour
    102

    = 30 minutes
    That person will take rest for four times.
    ∴  Required time
    = (30 + 4 × 5) minutes
    = 50 minutes


  1. A car driver leaves Bangalore at 8.30 A.M. and expects to reach a place 300 km from Bangalore at 12.30 P.M. At 10.30 he finds that he has covered only 40% of the distance. By how much he has to increase the speed of the car in order to keep up his schedule?
    1. 45 km/hr
    2. 40 km/hr
    3. 35 km/hr
    4. 30 km/hr

  1. View Hint View Answer Discuss in Forum

    Distance covered by car in 2 hours

    =
    300 × 40
    = 120 km
    100

    Remaining distance = 300 – 120 = 180 km
    Remaining time = 4 – 2 = 2 hours
    ∴  Required speed =
    180
    = 90 kmph
    2
    Original speed of car =
    120
    = 60 kmph
    2

    ∴  Required increase in speed
    = 90 – 60 = 30 kmph

    Correct Option: D

    Distance covered by car in 2 hours

    =
    300 × 40
    = 120 km
    100

    Remaining distance = 300 – 120 = 180 km
    Remaining time = 4 – 2 = 2 hours
    ∴  Required speed =
    180
    = 90 kmph
    2
    Original speed of car =
    120
    = 60 kmph
    2

    ∴  Required increase in speed
    = 90 – 60 = 30 kmph



  1. A man travelled a distance of 80 km in 7 hrs partly on foot at the rate of 8 km per hour and partly on bicycle at 16km per hour. The distance travelled on the foot is
    1. 32 km
    2. 48 km
    3. 36 km
    4. 44 km

  1. View Hint View Answer Discuss in Forum

    Journey on foot = x km
    Journey on cycle = (80 – x) km

    ∴ 
    x
    =
    80 − x
    = 7
    816

    ⇒ 
    2x + 80 − x
    = 7
    16

    ⇒  x + 80 = 16 × 7 = 112
    ⇒  x = 112 - 80 = 32 km.

    Second Method :
    Here, x = 80, t = 7
    u = 8, v = 16
    Time =
    vt − x
    v − u

    =
    16 × 7 − 80
    16 − 8

    =
    112 − 80
    8

    =
    32
    = 4 hrs
    8

    Distance travelled
    = 4 × 8 = 32 kms

    Correct Option: A

    Journey on foot = x km
    Journey on cycle = (80 – x) km

    ∴ 
    x
    =
    80 − x
    = 7
    816

    ⇒ 
    2x + 80 − x
    = 7
    16

    ⇒  x + 80 = 16 × 7 = 112
    ⇒  x = 112 - 80 = 32 km.

    Second Method :
    Here, x = 80, t = 7
    u = 8, v = 16
    Time =
    vt − x
    v − u

    =
    16 × 7 − 80
    16 − 8

    =
    112 − 80
    8

    =
    32
    = 4 hrs
    8

    Distance travelled
    = 4 × 8 = 32 kms


  1. Sarita and Julie start walking from the same place in the opposite directions. If Julie walks at a
    speed of 2
    1
    km/hr and Sarita at a speed
    2
    of 2 km/hr, in how much time will they be 18 km apart ?
    1. 4.0 hrs
    2. 4.5 hrs
    3. 5.0 hrs
    4. 4.8 hrs

  1. View Hint View Answer Discuss in Forum

    =
    5
    + 2kmph =
    9
    kmph
    22
    Time =
    Distance
    =
    18
    Relative speed(9/2)

    =
    18 × 2
    = 4 hours
    9

    Correct Option: A

    =
    5
    + 2kmph =
    9
    kmph
    22
    Time =
    Distance
    =
    18
    Relative speed(9/2)

    =
    18 × 2
    = 4 hours
    9



  1. You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is
    1. 4
    2. 5
    3. 10
    4. 2

  1. View Hint View Answer Discuss in Forum

    If the required distance be = x km, then

    x
    x
    =
    10 + 5
    4560

    ⇒ 
    5x − 4x
    =
    1
    204

    ⇒ 
    x
    =
    1
    204

    ⇒ x =
    1
    × 20 = 5 km.
    4


    Second Method :
    Here, S1 = 4, t1 = 5
    S2 = 5, t2 = 10
    Distance =
    (S1 × S2)(t1 + t2)
    S1 − S2

    =
    (4 × 5)(5 + 10)
    5 − 4

    = 20 ×
    15
    = 5 kms
    60

    Correct Option: B

    If the required distance be = x km, then

    x
    x
    =
    10 + 5
    4560

    ⇒ 
    5x − 4x
    =
    1
    204

    ⇒ 
    x
    =
    1
    204

    ⇒ x =
    1
    × 20 = 5 km.
    4


    Second Method :
    Here, S1 = 4, t1 = 5
    S2 = 5, t2 = 10
    Distance =
    (S1 × S2)(t1 + t2)
    S1 − S2

    =
    (4 × 5)(5 + 10)
    5 − 4

    = 20 ×
    15
    = 5 kms
    60