Speed, Time and Distance


  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. 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. 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. Ravi has a roadmap with a scale of 1.5 cm for 18 km. He drives on that road for 72 km. What would be his distance covered in that map?









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    ∵  18 km ≡ 1.5 cm

    ∴  1 km ≡
    1.5
    cm
    18

    Correct Option: B

    ∵  18 km ≡ 1.5 cm

    ∴  1 km ≡
    1.5
    cm
    18



  1. Ram arrives at a Bank 15 minutes earlier than scheduled time if he drives his car at 42 km/hr. If he drives car at 35 km/hr he arrives 5 minutes late. The distance of the Bank from his starting point is









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    Let the required distance be x km.
    Difference of time = 15 + 5 = 20 minutes

    =
    1
    hour
    3

    According to the question,
    x
    x
    =
    1
    6x − 5x
    =
    1
    354232103

    ⇒ 
    x
    =
    1
    2103

    ⇒   x =
    210
    = 70 km.
    3

    Correct Option: A

    Let the required distance be x km.
    Difference of time = 15 + 5 = 20 minutes

    =
    1
    hour
    3

    According to the question,
    x
    x
    =
    1
    6x − 5x
    =
    1
    354232103

    ⇒ 
    x
    =
    1
    2103

    ⇒   x =
    210
    = 70 km.
    3