Speed, Time and Distance


  1. In a kilometre race, A beats B by 30 seconds and B beats C by 15 seconds. If A beats C by 180 metres, the time taken by A to run 1 kilometre is









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    A beats B by 30 seconds and B beats C by 15 seconds.
    Clearly, A beats C by 45 seconds.
    Also, A beats C by 180 metres.
    Hence, C covers 180 metres in 45 seconds.

    ∴  Speed of C =
    180
    = 4 m/sec
    45

    ∴  Time taken by C to cover 1000 m
    =
    1000
    = 250 sec.
    4

    ∴  Time taken by A to cover 1000 m
    = 250–45 = 205 sec.

    Correct Option: B

    A beats B by 30 seconds and B beats C by 15 seconds.
    Clearly, A beats C by 45 seconds.
    Also, A beats C by 180 metres.
    Hence, C covers 180 metres in 45 seconds.

    ∴  Speed of C =
    180
    = 4 m/sec
    45

    ∴  Time taken by C to cover 1000 m
    =
    1000
    = 250 sec.
    4

    ∴  Time taken by A to cover 1000 m
    = 250–45 = 205 sec.


  1. Two cars start at the same time from one point and move along two roads at right angles to each other. Their speeds are 36 km/ hour and 48 km/hour respectively. After 15 seconds the distance between them will be









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    Let O be the starting point. The car running at 36 kmph is moving along OB and that at 48 kmph moving along OA. Also let they reach at B and A respectively after 15 seconds.

    ∴   OA = 48 ×
    5
    × 15 = 200 m
    18

    ∴   and OB = 36 ×
    5
    × 15 = 150 m
    18

    ∴  Required distance = AB
    = √(200)2 + (150)2
    W (By Pythagoras theorem)
    = √40000 + 22500
    = √62500
    = 250 m

    Correct Option: D


    Let O be the starting point. The car running at 36 kmph is moving along OB and that at 48 kmph moving along OA. Also let they reach at B and A respectively after 15 seconds.

    ∴   OA = 48 ×
    5
    × 15 = 200 m
    18

    ∴   and OB = 36 ×
    5
    × 15 = 150 m
    18

    ∴  Required distance = AB
    = √(200)2 + (150)2
    W (By Pythagoras theorem)
    = √40000 + 22500
    = √62500
    = 250 m



  1. A and B run a kilometre and A wins by 25 sec. A and C run a kilometre and A wins by 275 m. When B and C run the same distance, B wins by 30 sec. The time taken by A to run a kilometre is









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    If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) seconds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
    If B covers 1 km in (x + 25) seconds, then C covers 1 km in (x + 55) seconds.
    Thus in x seconds, C covers the distance of 725 m.

    ∴  
    x
    × 1000 = x + 55
    725

    ⇒  x = 145
    ∴   A covers the distance of 1 km in 2 minutes 25 seconds.

    Correct Option: A

    If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) seconds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
    If B covers 1 km in (x + 25) seconds, then C covers 1 km in (x + 55) seconds.
    Thus in x seconds, C covers the distance of 725 m.

    ∴  
    x
    × 1000 = x + 55
    725

    ⇒  x = 145
    ∴   A covers the distance of 1 km in 2 minutes 25 seconds.


  1. A car completes a journey in 10 hours. If it covers half of the journey at 40 kmph and the remaining half at 60 kmph, the distance covered by car is









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    Suppose distance be x km.

    x
    +
    x
    = 10
    2 × 402 × 60

    ⇒  
    x
    +
    x
    = 10
    80120

    ⇒ 
    3x + 2x
    = 10
    240

    ⇒ 
    5x
    = 10
    240

    x = 480 km

    Correct Option: B

    Suppose distance be x km.

    x
    +
    x
    = 10
    2 × 402 × 60

    ⇒  
    x
    +
    x
    = 10
    80120

    ⇒ 
    3x + 2x
    = 10
    240

    ⇒ 
    5x
    = 10
    240

    x = 480 km



  1. A walks at a uniform rate of 4 km an hour; and 4 hours after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A ?









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    Distance covered by A in 4 hours = 4 × 4 = 16 km
    Relative speed of B with respect to A = 10 – 4 = 6 km/hr
    ∴  Time taken to catch A

    =
    16
    =
    8
    hours
    63

    ∴ Required distance =
    8
    × 10 =
    80
    33

    = 26.67 km. ≈ 26.7 km

    Correct Option: D

    Distance covered by A in 4 hours = 4 × 4 = 16 km
    Relative speed of B with respect to A = 10 – 4 = 6 km/hr
    ∴  Time taken to catch A

    =
    16
    =
    8
    hours
    63

    ∴ Required distance =
    8
    × 10 =
    80
    33

    = 26.67 km. ≈ 26.7 km