Speed, Time and Distance


  1. Two bullets were fired at a place at an interval of 12 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 11 minutes 40 seconds. The speed of sound is 330 metres per second. What is the approximate speed of the car?
    1. 34 kmph
    2. 32 kmph
    3. 36 kmph
    4. 38 kmph

  1. View Hint View Answer Discuss in Forum

    If the car were not moving, the person would have heard the two sounds at an interval of 12 minutes. Therefore, the distance travelled by car in 11 minutes 40 seconds is equal to the distance that could have been covered by sound in 12 min – 11 min. 40 seconds = 20 seconds.
    Distance covered by sound in 20 seconds
    = 330 × 20 = 6600 m
    In 11 min 40 seconds
    or 700 seconds the car travels 6600 m.
    In 1 second the car will travel

    6600
    metre =
    66
    metre
    7007

    ∴ 
    66
    metre per second
    7

    =
    66
    ×
    18
    km per hr.
    75

    =
    1188
    km per hr.
    35

    = 33
    33
    km per hr. ≈ 34 kmph
    35

    Correct Option: A

    If the car were not moving, the person would have heard the two sounds at an interval of 12 minutes. Therefore, the distance travelled by car in 11 minutes 40 seconds is equal to the distance that could have been covered by sound in 12 min – 11 min. 40 seconds = 20 seconds.
    Distance covered by sound in 20 seconds
    = 330 × 20 = 6600 m
    In 11 min 40 seconds
    or 700 seconds the car travels 6600 m.
    In 1 second the car will travel

    6600
    metre =
    66
    metre
    7007

    ∴ 
    66
    metre per second
    7

    =
    66
    ×
    18
    km per hr.
    75

    =
    1188
    km per hr.
    35

    = 33
    33
    km per hr. ≈ 34 kmph
    35


  1. A carriage driving in a fog passed a man who was walking at the rate of 6 km per hr. in the same direction. He could see the carriage for 4 minutes and it was visible to him up to a distance of 200 metres. Find the speed of the carriage.
    1. 8.75 kmph
    2. 8.5 kmph
    3. 8 kmph
    4. 9 kmph

  1. View Hint View Answer Discuss in Forum

    The distance covered by man in 4 minutes

    =
    6 × 1000 × 4
    = 400 metres
    60

    The distance covered by carriage in 4 minutes
    = 200 + 400 = 600 metres
    ∴  
    600
    ×
    60
    km per hr.
    41000

    = 9 km per hr.

    Correct Option: D

    The distance covered by man in 4 minutes

    =
    6 × 1000 × 4
    = 400 metres
    60

    The distance covered by carriage in 4 minutes
    = 200 + 400 = 600 metres
    ∴  
    600
    ×
    60
    km per hr.
    41000

    = 9 km per hr.



  1. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km per hr. faster than the second train. If after two hours they are 50 km apart, find the average speed of faster train.
    1. 18 kmph
    2. 15 kmph
    3. 20 kmph
    4. 25 kmph

  1. View Hint View Answer Discuss in Forum

    Let the speed of the second train be x km per hr. Then the speed of the first train is x + 5 km per hr.

    Let O be the position of the railway station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = 2 (x + 5) km.
    Distance travelled by the 2nd train in 2 hours= OB = 2x km.
    By Pythagoras theorem,
    AB2 + OA2 + OB2
    ⇒  502 = [2 (x + 5)]2 + [2x]2
    ⇒  2500 = 4 (x + 5)2 + 4x2
    ⇒  2500 = 4 (x2 + 10x + 25) + 4x2
    ⇒  8x2 + 40x – 2400 = 0
    ⇒  x2 + 5x – 300 = 0
    ⇒  x2 + 20x – 15x – 300 = 0
    ⇒  x (x + 20) – 15 (x + 20) = 0
    ⇒  (x – 15) (x + 20) = 0
    ⇒  x = 15, – 20
    But x cannot be negative
    ∴  x = 15
    ∴  The speed of the second train is 15 km per hr. and the speed of the first train is 20 km per hr.

    Correct Option: C

    Let the speed of the second train be x km per hr. Then the speed of the first train is x + 5 km per hr.

    Let O be the position of the railway station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = 2 (x + 5) km.
    Distance travelled by the 2nd train in 2 hours= OB = 2x km.
    By Pythagoras theorem,
    AB2 + OA2 + OB2
    ⇒  502 = [2 (x + 5)]2 + [2x]2
    ⇒  2500 = 4 (x + 5)2 + 4x2
    ⇒  2500 = 4 (x2 + 10x + 25) + 4x2
    ⇒  8x2 + 40x – 2400 = 0
    ⇒  x2 + 5x – 300 = 0
    ⇒  x2 + 20x – 15x – 300 = 0
    ⇒  x (x + 20) – 15 (x + 20) = 0
    ⇒  (x – 15) (x + 20) = 0
    ⇒  x = 15, – 20
    But x cannot be negative
    ∴  x = 15
    ∴  The speed of the second train is 15 km per hr. and the speed of the first train is 20 km per hr.


  1. In a flight of 600 kms, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km per hr. and the time of flight increased by 30 minutes. Find the duration of flight.
    1. 1.2 hours
    2. 1 hour
    3. 1.5 hours
    4. 2 hours

  1. View Hint View Answer Discuss in Forum

    Let the original speed be x kmph
    then, new speed = (x – 200) kmph
    According to question,
    Time taken with new speed – time taken with original speed

    = 30 min. i.e
    1
    hr.
    2

    ∴ 
    600
    600
    =
    1
    x − 200x2

    ⇒  600
    1
    1
    =
    1
    x − 200x2

    ⇒  
    x − x + 200
    =
    1
    x(x − 200)1200

    ⇒  24000 = x (x – 200)
    ⇒  x2 – 200x – 24000 = 0
    ⇒  x2 – 600x + 400x – 24000 = 0
    ⇒  x (x – 600) + 400 (x – 600) = 0
    ⇒  (x – 600) (x + 400) = 0
    ⇒  x = 600, – 400
    Speed cannot be negative
    Hence, original speed = 600 kmph and duration of flight
    =
    600
    hr. = 1 hr.
    600

    Correct Option: B

    Let the original speed be x kmph
    then, new speed = (x – 200) kmph
    According to question,
    Time taken with new speed – time taken with original speed

    = 30 min. i.e
    1
    hr.
    2

    ∴ 
    600
    600
    =
    1
    x − 200x2

    ⇒  600
    1
    1
    =
    1
    x − 200x2

    ⇒  
    x − x + 200
    =
    1
    x(x − 200)1200

    ⇒  24000 = x (x – 200)
    ⇒  x2 – 200x – 24000 = 0
    ⇒  x2 – 600x + 400x – 24000 = 0
    ⇒  x (x – 600) + 400 (x – 600) = 0
    ⇒  (x – 600) (x + 400) = 0
    ⇒  x = 600, – 400
    Speed cannot be negative
    Hence, original speed = 600 kmph and duration of flight
    =
    600
    hr. = 1 hr.
    600



  1. A hare, pursued by a grey hound is 50 of her own leaps before him. While the hare takes 4 leaps, the grey hound takes 3 leaps. In one leap, the hare goes 1.75 metres and the grey hound 2.75 metres. In how many leaps, will the grey hound overtake the hare?
    1. 210 leaps
    2. 220 leaps
    3. 230 leaps
    4. 250 leaps

  1. View Hint View Answer Discuss in Forum

    Grey hound and hare make 3 leaps and 4 leaps respectively. This happens at the same time.
    The hare goes 1.75 metres in 1 leap.
    ∴  Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres
    The grey hound goes 2.75 metres in one leap.
    ∴  Distance covered by it in 3
    leaps
    = 3 × 2.75 = 8.25 metres
    Distance gained by grey hound in 3 leaps = (8.25 − 7)
    = 1.25 metres
    Distance covered by hare in 50 leaps = 50 × 1.75 metres
    = 87.5 metres
    Now, 1.25 metres is gained by grey hound in 3 leaps
    ∴  87.5 metres is gained in

    3
    × 87.5 = 210 leaps.
    1.25

    Correct Option: A

    Grey hound and hare make 3 leaps and 4 leaps respectively. This happens at the same time.
    The hare goes 1.75 metres in 1 leap.
    ∴  Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres
    The grey hound goes 2.75 metres in one leap.
    ∴  Distance covered by it in 3
    leaps
    = 3 × 2.75 = 8.25 metres
    Distance gained by grey hound in 3 leaps = (8.25 − 7)
    = 1.25 metres
    Distance covered by hare in 50 leaps = 50 × 1.75 metres
    = 87.5 metres
    Now, 1.25 metres is gained by grey hound in 3 leaps
    ∴  87.5 metres is gained in

    3
    × 87.5 = 210 leaps.
    1.25