Application of AM & HM in TSD Problems


  • QA/DILR Mentor | Be Legend


    Authored by Nitin Gupta, Founder, Director at AlphaNumeric.

    Quite often in questions we find that the given speeds (or time taken) are in an Arithmetic Progression. And if distance covered at the speeds is constant, then time taken (or speeds) will be inversely proportional i.e. they will be in Harmonic Progression.

    For those who have forgotten, the Arithmetic Mean of a and b is (a + b)/2 and the Harmonic Mean of a and b is 2ab/(a + b)

    Though it requires a little trained eyes to identify the above, it will be useful if you keep a watch for it. See the following data to realise that either time taken or speeds are in an Arithmetic Progression.

    A, B, C leave point P, one after the other in the given order, with equal time intervals between their departure. If all three imultaneously meet at Q, given that speed of A and C is 40 kmph and 60 kmph, find speed of B.

    The time taken by A, B, C over constant distance PQ will be of the type t, t – x and t – 2x i.e. in an AP. Thus, their speeds will be in a Harmonic Progression.
    The required speed will be the HM of 30 and 60 i.e. 2 * 30 * 60/90 = 40 kmph

    If I travel at 10 kmph, I reach office at 10:30 am, if I travel at 15 kmph, I reach office at 10:00 am. At what speed should I travel so that I reach office at 10:15. Assume I leave home at same time and take the same route.

    Leaving at same time and reaching at 10 am, 10:15 am and 10:30 am suggests that the time travelled are in AP. Thus, speeds are in HP and required speed is the HM of 10 & 15
    i.e.2 * 10 * 15/25 = 12 kmph

    Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? [CAT 2006]
    a) 3
    b) 3.5
    c) 4
    d) 4.5
    e) 5

    speed ---30---40---60 (AP)
    wheras time t---(t-2) ----- ???
    since speed is in HP, so time Must be in AP, so ans is t-4 ===> 4hrs

    A man can row UP stream in 84 minute. He can row the same distance in 9 min less than he could row it in still water. How long will he take to row down with the stream?

    upstream = b - r, still water = b, downstream = b + r ;
    so time must be in hp ;
    84, t, t-9 are in hp. solving this u will get t = 72; so ans is 72 - 9 = 63

    A boy is walking along the direction of 2 parallel railway tracks. on one of these tracks, trains are going on 1 direction at equal intervals. on the other track, trains are going in the opp direction at the same equal intervals. the speed of every train is same . In one direction, a train crosses the boy every 20 mins. and in the opp direction the train passes the boy every 30 mins. if the boy stands still beside the tracks, at intervals of how many will two consecutive trains going in the same direction cross him?

    when boy & train are travelling in same direction - relative speed = t-b;
    for opposite direction =t+b;
    when boy is stand stil, train will move at = t ;
    so here t-b, t, t+b are in AP ===> time must be in HP ==> t = 2 * 20 * 30/50 = 24 min

    Everyday I cover the distance from my home and office at a usual speed and take a certain time at the usual speed. When I increase my usual speed by 5 kmph, I take 10 minutes less than usual. If I reduce my usual speed by 5 kmph, I take 15 minutes more than usual. Find the distance from home to office.

    Can you realise that if usual speed is s, then the three speeds are (s – 5), s and (s + 5) i.e. in AP?
    Thus, the time taken are in HP.
    The time taken as per the data in question is (t + 15), t and (t – 10) and hence we have t = 60 min;
    Thus time taken are 75 mins, 60 mins and 50 mins.
    Speeds will be in ratio of 1/75 : 1/60 : 1/50 i.e. 4 : 5 : 6.
    And we know the difference in speeds are 5 kmph. Thus speeds are 20 kmph, 25 kmph and 30 kmph.
    Now distance can be found using any combination of speed and time.

    Some problems for your practice. Use only AM/HM concept while solving them.

    1. Anuva takes 20min less than the usual time to reach her office if her speed increases by 5km/hr , and takes 30 min more than the usual time if her speed decreases by 5 km/hr. What is her usual speed.

    2. A man can walk up a moving “up” escalator in 30 second. The same can walk down this moving “up” escalator in 90 seconds. Assume his walking speed is same Upwards & downwards. How much time he will take to walk up the escalator when escalator is not moving ? (cat 1994)

    3. If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon? [CAT 2004]
      a) 11 km/hr
      b) 12 km/hr
      c) 13 km/hr
      d) 14 km/hr


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