Time, Speed & Distance - Part 1 - Vikas Saini



  • Simple basic formulae for Time Speed and distance.
    Time = T, Speed = S, Distance = d
    S = D / T
    D = S * T
    T = D / S

    Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.
    a) 12 min
    b) 25 min
    c) 40 min
    d) 60 min

    Distance shall be same.
    D = ST
    D = 4 x 2.(45 x 100/60)
    D = 4 x 2.75 = 11 km.
    Now if speed is 16.5 then time T = D / S
    T = 11 / 16.5 = 2 / 3 h = 2 x 60 / 3 = 40 min.

    Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
    a) 4
    b) 6
    c) 8
    d) 10

    Due to stoppage, affect in speed = 54 – 45 = 9kmph.
    Per hour stoppage = 9 x 60 / 54 = 10 min.

    Two trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?
    a) 11:9
    b) 7:3
    c) 18:4
    d) 7:5

    Speed is directly proportional to Distance.
    Suppose their Speed are S1 and S2 respectively.
    D1 : D2 = S1 : S2.
    One train has travelled 110 km, therefore another train has covered 200 – 110 = 90 km.
    Speed ratio = D1 : D2 = 110 : 90 = 11:9.

    Sachin can cover a distance in 1hr 24 min by covering 2/3 of the distance at 4 kmph and the rest at 5 kmph. The total distance is?
    a) 5
    b) 6
    c) 7
    d) 8

    Time = D / S
    T = 2D/3 x 4 + D/3 x 5
    1 + (24 x 100/60) = D/6 + D/15
    1.4 = 21 D/90
    D = 6 km.

    A train covers a distance in 50 min, if it runs at a speed of 48 kmph on an average. The speed at which the train must run to reduce the time of journey to 40 min will be.
    a) 90
    b) 75
    c) 60
    d) 45

    Distance D = S T
    D = 48 x 50/60 = 40 km.
    Now time = 40 min.
    Speed = D / T = 40 / (40/60) = 60 km.

    A passenger train takes two hours less for a journey of 300 km if its speed is increased by 5 km/hr from its normal speed. The normal speed is:
    a) 35 km/hr
    b) 50 km/hr
    c) 25 km/hr
    d) 30 km/hr

    300 = S x T.
    300 = (S+5)(T-2)
    ST = ST – 2S + 5T – 10.
    2S = 5T – 10.
    2ST = 5T^2 – 10T.
    600 = 5T^2 – 10T.
    T^2 – 2T + 1 = 121.
    T – 1 = 11.
    T = 12.
    S = 25.

    If Ashu goes his office at speed of 12 kmph then gets late by 20 minutes. If he goes speed of 15 kmph then he still reaches late by 10 minutes. How much distance is office from his home ?

    D = ST
    12 (T + 20/60) = 15 (T + 10/60)
    12T + 4 = 15T + 15/6.
    3T = 1.5
    T = 0.5 hrs.
    D = 10 km.

    Two stations AA and BB are 110 km apart on a straight line. One train starts from AA at 7 am and travel towards BB at 20 km/hr speed. Another train starts from BB at 8 am and travel towards AA at 25 km/hr speed. At what time will they meet?
    a) 9 am
    b) 10 am
    c) 11 am
    d) None of these

    One train covers 20 km till 8 AM.
    Now left distance = 110 – 20 = 90.
    Now relative speed of both trains = 25+20 = 45 kmph.
    Time when they will meet = 90 / 45 = 2 hour.
    Time = 8+2 = 10 AM.

    The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?
    a) 950 ft
    b) 1450 ft
    c) 1200 ft
    d) 800 ft

    Let total distance travelled by cart = D.
    D/40 – D/48 = 5.
    D = 1200 ft.


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