Speed, Time and Distance


  1. A person, who can walk down a
    hill at the rate of 4
    1
    km/hour
    2
    and up the hill at the rate of 3 km/hour, ascends and comes
    down to his starting point in 5 hours. How far did he ascend ?









  1. View Hint View Answer Discuss in Forum

    Let the required distance be x km.

    ∴ 
    x
    +
    x
    = 5
    9/23

    ⇒   x
    2
    +
    1
    = 5
    93

    ⇒  x
    2 + 3
    = 5
    9

    ⇒ x =
    5 × 9
    = 9 km.
    5

    Correct Option: D

    Let the required distance be x km.

    ∴ 
    x
    +
    x
    = 5
    9/23

    ⇒   x
    2
    +
    1
    = 5
    93

    ⇒  x
    2 + 3
    = 5
    9

    ⇒ x =
    5 × 9
    = 9 km.
    5


  1. A man walks a certain distance and rides back in 4 hours 30 minutes. He could ride both
    ways in 3 hours. The time required by the man to walk both ways is









  1. View Hint View Answer Discuss in Forum

    Suppose, time taken while walking be x hours
    And, time taken on riding be y hours
    ∴  According to question

    x + y = 4
    1
    hours     ...(i)
    2

    Then, 2y = 3 hours
    y = 1
    1
    hours
    2

    From equation (i)
    x = 4
    1
    − 1
    1
    = 3 hours
    22

    Time required to walk both ways = 6 hours

    Correct Option: D

    Suppose, time taken while walking be x hours
    And, time taken on riding be y hours
    ∴  According to question

    x + y = 4
    1
    hours     ...(i)
    2

    Then, 2y = 3 hours
    y = 1
    1
    hours
    2

    From equation (i)
    x = 4
    1
    − 1
    1
    = 3 hours
    22

    Time required to walk both ways = 6 hours



  1. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they next meet at the starting point again ?









  1. View Hint View Answer Discuss in Forum

    Required time = LCM of 252,
    308 and 198 seconds.
    Now, 252 = 2 × 2 × 3 × 3 × 7
    308 = 2 × 2 × 7 × 11
    198 = 2 × 3 × 3 × 11
    ∴  LCM = 2 × 2 × 3 × 3 × 7 × 11
    = 36 × 77 seconds

    =
    36 × 77
    minutes
    60

    =
    231
    = 46 minutes 12 seconds
    5

    Correct Option: A

    Required time = LCM of 252,
    308 and 198 seconds.
    Now, 252 = 2 × 2 × 3 × 3 × 7
    308 = 2 × 2 × 7 × 11
    198 = 2 × 3 × 3 × 11
    ∴  LCM = 2 × 2 × 3 × 3 × 7 × 11
    = 36 × 77 seconds

    =
    36 × 77
    minutes
    60

    =
    231
    = 46 minutes 12 seconds
    5


  1. A man can reach a certain place in 30 hours.
    If he reduces his speed by
    1
    th, he goes
    15
    10 km less in that time. Find his speed per hour.









  1. View Hint View Answer Discuss in Forum

    Let the speed of man be x kmph.

    ∴ 30x − 30x −
    x
    = 10
    15

    ⇒  30x − x +
    x
    = 10
    15

    ⇒ 
    x
    =
    10
    1530

    ⇒   x =
    150
    = 5 kmph
    30

    Correct Option: D

    Let the speed of man be x kmph.

    ∴ 30x − 30x −
    x
    = 10
    15

    ⇒  30x − x +
    x
    = 10
    15

    ⇒ 
    x
    =
    10
    1530

    ⇒   x =
    150
    = 5 kmph
    30



  1. A and B start at the same time with speed of 40 km/hr and 50 km/hr respectively. If in covering the journey A takes 15 minutes longer than B, the total distance of the journey is :









  1. View Hint View Answer Discuss in Forum

    Let the distance be x km

    Time taken by A =
    x
    hrs.
    40

    Time taken by B =
    x
    hrs.
    50

    Now,   
    x
    x
    =
    15
    405060

    5x − 4x
    =
    15
    20060

    ∴  x =
    15
    × 200 = 50km
    60


    Second Method :
    Distance =
    Product of speed
    × Diff. in time
    Diff. of speed

    =
    40 × 50
    ×
    15
    = 50km
    50 − 4060

    Correct Option: C

    Let the distance be x km

    Time taken by A =
    x
    hrs.
    40

    Time taken by B =
    x
    hrs.
    50

    Now,   
    x
    x
    =
    15
    405060

    5x − 4x
    =
    15
    20060

    ∴  x =
    15
    × 200 = 50km
    60


    Second Method :
    Distance =
    Product of speed
    × Diff. in time
    Diff. of speed

    =
    40 × 50
    ×
    15
    = 50km
    50 − 4060