Speed, Time and Distance


  1. If you travel 39 km at a speed of 26 km/h, another 39 km at a speed of 39 km/h and again 39 km at a speed of 52 km/h, what is your average speed for the entire journey?









  1. View Hint View Answer Discuss in Forum

    Required average speed = Total distance covered/Total time taken
    = (3 x 39)/[39/26 + 39/39 +39/52]

    Correct Option: D

    Required average speed = Total distance covered/Total time taken
    = (3 x 39)/[39/26 + 39/39 +39/52]
    = 3 x 13/(1/2 + 1/3 + 1/4)
    = 3 x 13 x 12/13 = 36 km/h


  1. Dalbir Singh, a polieman, is 114 m behind a thief. Dalbir Singh runs 21 m and the thief 15 m in a minute. In what time will Dalbir singh catch the thief?









  1. View Hint View Answer Discuss in Forum

    Relative speed = 21 - 15 = 6 m per minute.
    Relative distance = 114 m

    Correct Option: A

    Relative speed = 21 - 15 = 6 m per minute.
    Relative distance = 114 m
    Time taken to catche = 144/6 min = 19 min



  1. A person travels a certain distance at 3 km/h and reaches 15 min late. If he travels at 4 km/h, he reaches 15 min earlier. The distance he has to travel is. ?









  1. View Hint View Answer Discuss in Forum

    Let the certain distance be d and time t.
    Now, by given condition,
    d/3 = ( t + 15 ) min ( t +15 )/60 h and
    d/4 = (t - 15) min = ( t - 15 )/60 h

    Correct Option: B

    Let the certain distance be d and time t.
    Now, by given condition,
    d/3 = ( t + 15 ) min ( t +15 )/60 h
    ⇒ 20d = t + 15
    ⇒ t = 20d - 15 ......(i)

    And from another condition,
    d/4 = (t - 15) min = ( t - 15 )/60 h
    ⇒ 15d = t - 15
    ⇒ t = 15d + 15 ...(ii)

    From Eqs. (i) and (ii), we get
    20d - 15 = 15d + 15
    ⇒ 5d = 30
    ∴ d = 6 km.


  1. A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 40 min less. If he had moved 2 km/ slower, he would have taken 40 min more. The distance (in
    km) is ?









  1. View Hint View Answer Discuss in Forum

    Let distance and original speed of the man be d km and s km/h.
    Then,
    d/s - d/s + 3 = 2/3 ....(i)
    and d/(s - 2) - d/s = 2/3 ...(ii)

    Correct Option: D

    Let distance and original speed of the man be d km and s km/h.
    Then,
    d/s - d/s + 3 = 2/3
    ⇒ [d(s + 3 - s)]/[s(s + 3)] = 2/3
    ⇒ 9d = 2s(s + 3) ...(i)

    And d/(s - 2) - d/s = 2/3
    ⇒ [d(s - s + 2)]/[s(s - 2)] = 2/3
    ⇒ 3d = s(s - 2) ...(ii)

    From Eqs.(i) and (ii), we get
    3s(s - 2) = 2s(s + 3)
    ⇒ 3s2 - 6s = 2s2 + 6s
    ⇒ s2 = 12s
    ⇒ s = 12

    From Eq.(ii), we get
    3d = 12(12 - 2)
    ∴ d = 40 km



  1. A man riding a bicycle form his house at 10km/h and reaches his office late by 6 min. He increases his speed by 2km/h and reaches 6 min before. How far is the office form his house?









  1. View Hint View Answer Discuss in Forum

    Here, a = 10 km/h
    b = ( 10 + 2 ) = 12 km/h,
    t1 = 6 min = 1/10 h
    and t2= 1/10 h.
    ∴ Required distance = ab( t1 + t2 )/(b - a)

    Correct Option: C

    Here, a = 10 km/h
    b = ( 10 + 2 ) = 12 km/h,
    t1 = 6 min = 1/10 h
    and t2= 1/10 h.
    ∴ Required distance = ab( t1 + t2 )/(b - a)
    = 10 x 12 x ( 1/10 + 1/10 ) /(12 - 10) = (10 x 12 x 1)/(5 x 2) = 12km