Boats and Streams


  1. A man 8 km/hr in still water. If the river is running at 2 km/hr, it taken 32 minutes to row to a place and back. How for is the place ?









  1. View Hint View Answer Discuss in Forum

    D/(8 - 2) + D/(8 + 2) = 32/60

    Correct Option: C

    ∵ D/(8 - 2) + D/(8 + 2) = 32/60
    ⇒ D/6 + D/10 = 32/60
    ⇒ 10D + 6D = 32
    ∴ D = 2 km


  1. A boat covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km downstream in 61/2 hours. The velocity of the current is ?









  1. View Hint View Answer Discuss in Forum

    Let the speed upstream be U km/hr and the speed downstream be V km/hr respectively.
    Then 24/U + 36/V = 6 ...(i)
    and 36/U + 24/V = 13/2 ...(ii)

    Correct Option: C

    Let the speed upstream be U km/hr and the speed downstream be V km/hr respectively.
    Then 24/U + 36/V = 6 ...(i)
    and 36/U + 24/V = 13/2 ...(ii)

    Solving these 2 equations we get
    ∴ U = 8 km/hr and V = 12 km/hr
    ∴ Velocity of current = (12 - 8)/2 km/hr = 2 km/hr



  1. A boat goes at 14 kmph along the stream and 8 kmph against the stream. The speed of the boat (in kmph) in still water is :









  1. View Hint View Answer Discuss in Forum

    Given in question , Rate downstream = 14 kmph
    Rate upstream = 8 kmph

    Speed of boat in still water =
    1
    (Rate downstream + Rate upstream)
    2

    Correct Option: B

    Given in question , Rate downstream = 14 kmph
    Rate upstream = 8 kmph

    Speed of boat in still water =
    1
    (Rate downstream + Rate upstream)
    2

    Speed of boat in still water =
    1
    (14 + 8) = 11 kmph
    2


  1. A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 km/hr and 4 km/ hr respectively, the distance of the destination from the starting place is









  1. View Hint View Answer Discuss in Forum

    Let the distance of the destination from the starting point be d km.
    Rate downstream = (10 + 4) kmph = 14 kmph
    Rate upstream = (10 – 4) kmph = 6 kmph
    According to the question,

    d
    +
    d
    = 5
    146

    3d + 7d
    = 5
    42

    ⇒ 10d = 42 × 5
    ⇒ d =
    42 × 5
    = 21 km
    10

    We can find the required answer with the help of given formula :
    Here, p = 10, q = 4, t = 5
    d =
    t(p² - q²)
    2p

    Correct Option: C

    Let the distance of the destination from the starting point be d km.
    Rate downstream = (10 + 4) kmph = 14 kmph
    Rate upstream = (10 – 4) kmph = 6 kmph
    According to the question,

    d
    +
    d
    = 5
    146

    3d + 7d
    = 5
    42

    ⇒ 10d = 42 × 5
    ⇒ d =
    42 × 5
    = 21 km
    10

    We can find the required answer with the help of given formula :
    Here, p = 10, q = 4, t = 5
    d =
    t(p² - q²)
    2p

    d =
    5(10² - 4²)
    2 × 10

    d =
    5(100 - 16)
    2 × 10

    d =
    84
    = 21
    4



  1. In a fixed time, a boy swims double the distance along the current that he swims against the current. If the speed of the current is 3 km/hr, the speed of the boy in still water is









  1. View Hint View Answer Discuss in Forum

    Let the rate of swimming in still water be p kmph
    ∴ Rate down-stream = (p + 3) kmph
    ∴ Rate up-stream = (p – 3) kmph
    According to the question,
    (p + 3) t = 2 (p – 3) × t
    ⇒ p + 3 = 2p – 6
    ⇒ p = 9 kmph
    Using the given formula :
    Here, t1 = 2k , t2 = k
    Speed of Stream = 3 km/hr

    Speed of Boy
    =
    t1 + t2
    Speed of Streamt1 - t2

    Correct Option: B

    Let the rate of swimming in still water be p kmph
    ∴ Rate down-stream = (p + 3) kmph
    ∴ Rate up-stream = (p – 3) kmph
    According to the question,
    (p + 3) t = 2 (p – 3) × t
    ⇒ p + 3 = 2p – 6
    ⇒ p = 9 kmph
    Using the given formula :
    Here, t1 = 2k , t2 = k
    Speed of Stream = 3 km/hr

    Speed of Boy
    =
    t1 + t2
    Speed of Streamt1 - t2

    Speed of Boy
    =
    2k + k
    32k - k

    Speed of Boy = 9 km/hr