Speed, Time and Distance


  1. A man covered a distance of 3990 km partly by air, partly by sea and remaining by land. The time spent in air, on sea and on land is in the ratio 1 : 16 : 2 and the ratio of average speed is 20 : 1 : 3 respectively. If total average speed is 42 km per hr, find the distance covered by sea.









  1. View Hint View Answer Discuss in Forum

    Total distance travelled = 3990 km
    Distance = Time × Speed
    Ratio of time spent = 1 : 16 : 2
    Ratio of speed = 20 : 1 : 3
    ∴  Ratio of time × speed
    = 20 × 1 : 16 × 1 : 2 × 3
    = 20 : 16 : 6
    Sum of the ratios
    = 20 + 16 + 6 = 42
    ∴  Distance covered by sea

    =
    3990
    × 16 = 1520 kms
    42

    Correct Option: C

    Total distance travelled = 3990 km
    Distance = Time × Speed
    Ratio of time spent = 1 : 16 : 2
    Ratio of speed = 20 : 1 : 3
    ∴  Ratio of time × speed
    = 20 × 1 : 16 × 1 : 2 × 3
    = 20 : 16 : 6
    Sum of the ratios
    = 20 + 16 + 6 = 42
    ∴  Distance covered by sea

    =
    3990
    × 16 = 1520 kms
    42


  1. A railway engine is proceeding towards A at uniform speed of 30 km/hr. While the engine is 20 kms away from A an insect starting from A flies again and again between A and the engine relentlessly. The speed of insect is 42 km per hr. Find the distance covered by the insect till the engine reaches A.









  1. View Hint View Answer Discuss in Forum

    Relative speed of insect
    = 30 + 42 = 72 km per hr.
    Distance between railway engine and insect = 20 km.
    Engine and insect will meet for the first time after

    =
    20
    hr.
    72

    Distance covered in this period
    =
    20
    × 42 =
    35
    km
    723

    The insect will cover
    35
    km in returning to A.
    3

    The distance covered by engine in this period
    =
    20
    × 30 =
    25
    km
    723

    Since, the insect when reaches A,
    the engine will cover
    25
    to A.
    3

    ∴  Remaining distance between A and engine
    = 20 −
    25
    +
    25
    33

    Again, engine and insect will meet after
    10
    =
    5
    hr.
    3 × 72108

    The distance covered by the insect in this period
    =
    5
    × 42 =
    35
    km
    10818

    tand again the insect will cover
    35
    km in returning.
    18

    ∴ Total distance covered by the insect =
    70
    +
    70
    +.....
    318


    =
    70
    1 +
    1
    +.......∞
    36

    It is a Geometric Progression to infinity with common ratio 1/6.
    =
    70
    1
    31 − (1/6)

    ∴  S =
    a
    1 − r

    =
    70
    ×
    1
    3(5/6)

    =
    70
    ×
    6
    = 28 km
    35

    Correct Option: D

    Relative speed of insect
    = 30 + 42 = 72 km per hr.
    Distance between railway engine and insect = 20 km.
    Engine and insect will meet for the first time after

    =
    20
    hr.
    72

    Distance covered in this period
    =
    20
    × 42 =
    35
    km
    723

    The insect will cover
    35
    km in returning to A.
    3

    The distance covered by engine in this period
    =
    20
    × 30 =
    25
    km
    723

    Since, the insect when reaches A,
    the engine will cover
    25
    to A.
    3

    ∴  Remaining distance between A and engine
    = 20 −
    25
    +
    25
    33

    Again, engine and insect will meet after
    10
    =
    5
    hr.
    3 × 72108

    The distance covered by the insect in this period
    =
    5
    × 42 =
    35
    km
    10818

    tand again the insect will cover
    35
    km in returning.
    18

    ∴ Total distance covered by the insect =
    70
    +
    70
    +.....
    318


    =
    70
    1 +
    1
    +.......∞
    36

    It is a Geometric Progression to infinity with common ratio 1/6.
    =
    70
    1
    31 − (1/6)

    ∴  S =
    a
    1 − r

    =
    70
    ×
    1
    3(5/6)

    =
    70
    ×
    6
    = 28 km
    35



  1. Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 a.m. and 9.51 a.m. respectively at the speed of 25 kmph and 20 kmph respectively for journey towards Y. A train R leaves station Y at 11.30 a.m. at a speed of a 30 kmph. for journey towards X. When will P be at equal distance from Q and R ?









  1. View Hint View Answer Discuss in Forum


    Distance covered by P till 11.30 a.m.
    = (11.30 a.m. – 8 a.m.) × 25km

    = 3
    1
    × 25 = 87.5 km.
    2

    Distance covered by Q till 11.30 a.m.
    = (11.30 – 9.51 am) × 20
    = 1
    39
    hrs. × 20 = 33 km
    60

    So, at 11.30 a.m. the three trains will be at positions shown below :

    P gains 5 km every hour over Q.
    Relative speed of P w.r.t. R = 20 + 30 = 50 km per hr
    Let P be at equal distance from Q and R after t hours.
    ∴  (87.5 – 33) + 5t
    = 132.5 –55t
    or,   54.5 + 5t = 132.5 – 55t
    or,   60 t = 78
    or,   t =
    78
    hrs.
    60

    = 1 hr 18 minutes
    11.30 am + 1 hr. 18 min.
    = 12.48 pm
    At 12.48 pm, P would have covered a distance
    = (12.48 pm – 8 am) × 25
    = 120 km
    Therefore, P will be at equal distance from Q and R at 12.48 pm

    Correct Option: A


    Distance covered by P till 11.30 a.m.
    = (11.30 a.m. – 8 a.m.) × 25km

    = 3
    1
    × 25 = 87.5 km.
    2

    Distance covered by Q till 11.30 a.m.
    = (11.30 – 9.51 am) × 20
    = 1
    39
    hrs. × 20 = 33 km
    60

    So, at 11.30 a.m. the three trains will be at positions shown below :

    P gains 5 km every hour over Q.
    Relative speed of P w.r.t. R = 20 + 30 = 50 km per hr
    Let P be at equal distance from Q and R after t hours.
    ∴  (87.5 – 33) + 5t
    = 132.5 –55t
    or,   54.5 + 5t = 132.5 – 55t
    or,   60 t = 78
    or,   t =
    78
    hrs.
    60

    = 1 hr 18 minutes
    11.30 am + 1 hr. 18 min.
    = 12.48 pm
    At 12.48 pm, P would have covered a distance
    = (12.48 pm – 8 am) × 25
    = 120 km
    Therefore, P will be at equal distance from Q and R at 12.48 pm


  1. A person travels a certain distance on a bicycle at a certain speed. Had he moved 3 km/hour faster, he would have taken 40 minutes less. Had he moved 2km/hour slower, he would have taken 40 minutes more. Find the distance.









  1. View Hint View Answer Discuss in Forum

    Let the original speed of the person be x km/hr. and the distance be y km.
    Case : I

    y
    y
    = 40 minutes
    xx + 3

    or,  
    40
    hr.
    60

    or,  
    y
    y
    =
    40
    =
    2
    xx + 3603

    or,   y
    1
    1
    =
    1
    x(x + 3)3

    or,   y
    x + 3 − x
    =
    2
    x (x + 3)3

    or,  
    3y
    =
    2
    x (x + 3)3

    or,   2 x (x + 3) = 9y       ...(i)
    Case : II
    =
    y
    y
    =
    40
    x − 2x60

    or,   y
    1
    +
    1
    +
    2
    x − 2x3

    or,   y
    x − x + 2
    =
    2
    x (x − 2)3

    or,  
    2y
    =
    2
    x (x − 2)3

    or, x (x – 2) = 3y       ... (ii)
    On dividing equation (i) by (ii) we have,
    2x(x + 3)
    =
    9y
    x (x − 2)3y

    or,  
    2(x + 3)
    = 3
    (x − 2)

    or,   2x + 6 = 3x – 6
    or,   3x – 2x = 6 + 6 = 12
    or,   x = 12 km/hr.
    ∴  Original speed of the person
    = 12 km/hr.
    Putting the value of x in equation (ii)
    12 (12 – 2) = 3y
    or,   3y = 12 × 10
    or,   y =
    12 × 10
    = 40
    3

    ∴  The required distance = 40km.

    Correct Option: B

    Let the original speed of the person be x km/hr. and the distance be y km.
    Case : I

    y
    y
    = 40 minutes
    xx + 3

    or,  
    40
    hr.
    60

    or,  
    y
    y
    =
    40
    =
    2
    xx + 3603

    or,   y
    1
    1
    =
    1
    x(x + 3)3

    or,   y
    x + 3 − x
    =
    2
    x (x + 3)3

    or,  
    3y
    =
    2
    x (x + 3)3

    or,   2 x (x + 3) = 9y       ...(i)
    Case : II
    =
    y
    y
    =
    40
    x − 2x60

    or,   y
    1
    +
    1
    +
    2
    x − 2x3

    or,   y
    x − x + 2
    =
    2
    x (x − 2)3

    or,  
    2y
    =
    2
    x (x − 2)3

    or, x (x – 2) = 3y       ... (ii)
    On dividing equation (i) by (ii) we have,
    2x(x + 3)
    =
    9y
    x (x − 2)3y

    or,  
    2(x + 3)
    = 3
    (x − 2)

    or,   2x + 6 = 3x – 6
    or,   3x – 2x = 6 + 6 = 12
    or,   x = 12 km/hr.
    ∴  Original speed of the person
    = 12 km/hr.
    Putting the value of x in equation (ii)
    12 (12 – 2) = 3y
    or,   3y = 12 × 10
    or,   y =
    12 × 10
    = 40
    3

    ∴  The required distance = 40km.



  1. A steamer goes downstream from one port to another in 4 hours. It covers the same distance upstream in 5 hours. If the speed of the stream be 2 km/hr, find the distance between the two ports.









  1. View Hint View Answer Discuss in Forum

    Let the speed of steamer in still water = x kmph
    ∴  Rate downstream = (x + 2) kmph
    Rate upstream = (x – 2) kmph
    Obviously, distance covered downstream and upstream are equal
    ⇒  4 (x + 2) = 5 (x – 2)
    ⇒  4x + 8 = 5x – 10
    ⇒  5x – 4x = 10 + 8 ⇒ x = 18
    ∴  Rate downstream
    = 18 + 2 = 20 kmph
    Therefore, the required distance
    = Speed downstream × Time
    = 20 × 4 = 80 km.

    Correct Option: C

    Let the speed of steamer in still water = x kmph
    ∴  Rate downstream = (x + 2) kmph
    Rate upstream = (x – 2) kmph
    Obviously, distance covered downstream and upstream are equal
    ⇒  4 (x + 2) = 5 (x – 2)
    ⇒  4x + 8 = 5x – 10
    ⇒  5x – 4x = 10 + 8 ⇒ x = 18
    ∴  Rate downstream
    = 18 + 2 = 20 kmph
    Therefore, the required distance
    = Speed downstream × Time
    = 20 × 4 = 80 km.