Speed, Time and Distance


  1. Two places A and B are 80 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet each other in 8 hours. If they move in opposite directions towards each other, they meet in 1 hour 20 minutes. Determine the speed of the faster car.









  1. View Hint View Answer Discuss in Forum

    Case : I
    When the cars are moving in the same direction.

    Let A and B be two places and C be the place of meeting.
    Let the speed of car starting from A be x kmph, and that of car starting from B be y kmph.
    Relative speed = (x – y) kmph
    According to the question.
    (x – y) × 8 = 80
    ⇒  x – y = 10     ...(i)
    Case : II
    When the cars are moving in the opposite directions and they meet at point C.

    Relative speed = (x + y) kmph
    Time taken = 1 hour 20 minutes

    = 1 +
    1
    =
    4
    hours
    33

    ∴  (x + y) ×
    4
    = 80
    3

    ⇒  x + y =
    80 × 3
    4

    ⇒  x + y = 60     ...(ii)
    Adding equations (i) and (ii),
    2x = 70
    ⇒  x = 35
    From equation (ii),
    x + y = 60
    ⇒  35 + y = 60
    ⇒  y = 60 – 35 = 25
    ∴  Speed of the faster car = 35 kmph

    Correct Option: C

    Case : I
    When the cars are moving in the same direction.

    Let A and B be two places and C be the place of meeting.
    Let the speed of car starting from A be x kmph, and that of car starting from B be y kmph.
    Relative speed = (x – y) kmph
    According to the question.
    (x – y) × 8 = 80
    ⇒  x – y = 10     ...(i)
    Case : II
    When the cars are moving in the opposite directions and they meet at point C.

    Relative speed = (x + y) kmph
    Time taken = 1 hour 20 minutes

    = 1 +
    1
    =
    4
    hours
    33

    ∴  (x + y) ×
    4
    = 80
    3

    ⇒  x + y =
    80 × 3
    4

    ⇒  x + y = 60     ...(ii)
    Adding equations (i) and (ii),
    2x = 70
    ⇒  x = 35
    From equation (ii),
    x + y = 60
    ⇒  35 + y = 60
    ⇒  y = 60 – 35 = 25
    ∴  Speed of the faster car = 35 kmph


  1. In a 200 metre race, A beats B by 20 metres; while in a 100 metres race, B beats C by 5
    metres. Assuming that the speed of A, B and C remain the same in various races, by how many metres will A beat C in one kilometre race ?









  1. View Hint View Answer Discuss in Forum

    According to the question,
    when A covers the distance of 200 metres, B covers only 200– 20 = 180 metres
    Again, in 100 metre race, B beats C by 5 metres.
    Hence, if B runs 100 metres, C runs 100–5 = 95 metres
    ∵  If B runs 100 m, C runs = 95 m
    ∴  If B runs 180 m, C runs

    =
    95 × 180
    = 171 m
    100

    ∴  A : B : C = 200 : 180 : 171
    Hence, A will beat C by
    = 200–171 = 29 m in 200 m race.
    i.e., 29 × 5 = 145 m in 1 km race.

    Correct Option: C

    According to the question,
    when A covers the distance of 200 metres, B covers only 200– 20 = 180 metres
    Again, in 100 metre race, B beats C by 5 metres.
    Hence, if B runs 100 metres, C runs 100–5 = 95 metres
    ∵  If B runs 100 m, C runs = 95 m
    ∴  If B runs 180 m, C runs

    =
    95 × 180
    = 171 m
    100

    ∴  A : B : C = 200 : 180 : 171
    Hence, A will beat C by
    = 200–171 = 29 m in 200 m race.
    i.e., 29 × 5 = 145 m in 1 km race.



  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.


  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. 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