Speed, Time and Distance


  1. A journey takes 4 hours 30 minutes at a speed of 60 km/hr. If the speed is 15 m/s, then the
    journey will take









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    60 kmph =
    60 × 5
    m/sec
    18

    =
    50
    m/sec.
    3

    ∴  Speed ∝
    1
    Time

    ⇒  S1 × T1 = S2 × T2
    ⇒ 
    50
    ×
    9
    = 15 × T2
    32

    ⇒  T2 =
    75
    = 5 hours
    15


    Second Method :
    Here, S1 = 60, t1 = 4
    1
    =
    9
    22

    S2 = 15 ×
    18
    = 54
    5

    S1t1 = S2t2
    60 ×
    9
    = 54 × t2
    2

    t2 =
    270
    = 5 hours
    54

    Correct Option: A

    60 kmph =
    60 × 5
    m/sec
    18

    =
    50
    m/sec.
    3

    ∴  Speed ∝
    1
    Time

    ⇒  S1 × T1 = S2 × T2
    ⇒ 
    50
    ×
    9
    = 15 × T2
    32

    ⇒  T2 =
    75
    = 5 hours
    15


    Second Method :
    Here, S1 = 60, t1 = 4
    1
    =
    9
    22

    S2 = 15 ×
    18
    = 54
    5

    S1t1 = S2t2
    60 ×
    9
    = 54 × t2
    2

    t2 =
    270
    = 5 hours
    54


  1. The distance between 2 places R and S is 42 km. Anita starts from R with a uniform speed of 4 km/h towards S and at the same time Romita starts from S towards R also with some uniform speed. They meet each other after 6 hours. The speed of Romita is











  1. View Hint View Answer Discuss in Forum

    Speed of Romita = x kmph
    (let)
    Distance = Speed × Time
    According to the question,
    4 × 6 + x × 6 = 42
    ⇒  6x = 42 – 24 = 18
    ⇒  x = 18 ÷ 6 = 3 kmph
    Second Method :
    Distance from R to S = S1t1+S2t2
    42 = 4 × 6 + x × 6
    6x = 18 x = 3 km/hr.

    Correct Option: E

    Speed of Romita = x kmph
    (let)
    Distance = Speed × Time
    According to the question,
    4 × 6 + x × 6 = 42
    ⇒  6x = 42 – 24 = 18
    ⇒  x = 18 ÷ 6 = 3 kmph
    Second Method :
    Distance from R to S = S1t1+S2t2
    42 = 4 × 6 + x × 6
    6x = 18 x = 3 km/hr.



  1. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at the rate 4 kmph and partly on bicycle at the rate 9 kmph. The distance travelled on foot is









  1. View Hint View Answer Discuss in Forum

    Distance travelled by farmer on foot = x km (let)
    ∴  Distance covered by cycling = (61–x ) km.

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    61 − x
    = 9
    49

    ⇒ 
    9x + 61 × 4 − 4x
    = 9
    9 × 4

    ⇒  5x + 244 = 9 × 9 × 4 = 324
    ⇒  5x = 324 – 244 = 80
    ⇒  x =
    80
    = 16 km.
    5


    Second Method :
    Here, t = 9, x= 61
    u = 4, v = 9
    Time taken =
    vt − x
    v − u

    =
    9 × 9 − 61
    9 − 4

    =
    20
    = 4 hrs.
    5

    Distance travelled
    = 4 × 4 = 16 km

    Correct Option: A

    Distance travelled by farmer on foot = x km (let)
    ∴  Distance covered by cycling = (61–x ) km.

    Time =
    Distance
    Speed

    According to the question,
    x
    +
    61 − x
    = 9
    49

    ⇒ 
    9x + 61 × 4 − 4x
    = 9
    9 × 4

    ⇒  5x + 244 = 9 × 9 × 4 = 324
    ⇒  5x = 324 – 244 = 80
    ⇒  x =
    80
    = 16 km.
    5


    Second Method :
    Here, t = 9, x= 61
    u = 4, v = 9
    Time taken =
    vt − x
    v − u

    =
    9 × 9 − 61
    9 − 4

    =
    20
    = 4 hrs.
    5

    Distance travelled
    = 4 × 4 = 16 km


  1. A bus moving at 40 km per hour covers a distance in 6 hours 15 minutes. If it travels the same distance at 50 km per hour how long will it take to cover the distance ?









  1. View Hint View Answer Discuss in Forum

    Distance = Speed × TIme

    =
    40 × 6
    1
    km
    4

    =
    40 × 25
    km = 250 km
    4

    New speed = 50 kmph
    ∴  Required time
    =
    Distance
    =
    250
    = 5 hours
    Speed50


    Second Method :
    Here, S1 = 40, t1 = 6
    15
    =
    25
    604

    S2 = 50, t2 = ?
    S1t1 = S2t2
    40 ×
    25
    = 50 × t2
    4

    t2 = 5 hrs.

    Correct Option: D

    Distance = Speed × TIme

    =
    40 × 6
    1
    km
    4

    =
    40 × 25
    km = 250 km
    4

    New speed = 50 kmph
    ∴  Required time
    =
    Distance
    =
    250
    = 5 hours
    Speed50


    Second Method :
    Here, S1 = 40, t1 = 6
    15
    =
    25
    604

    S2 = 50, t2 = ?
    S1t1 = S2t2
    40 ×
    25
    = 50 × t2
    4

    t2 = 5 hrs.



  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. 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