Application of Arithmetic Mean & Harmonic Mean in Time, Speed & Distance - Sambit Rath, IIM B
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Sambit Rath is an IIM Bangalore alumnus. A B. Tech pass out from KIIT University Bhubaneswar, he has worked with TCS for 3+ years before joining IIM. He is a sports & music enthusiast. He is set to join HCL Technologies in Sales and Marketing domain after his MBA.
TSD (Time Speed and Distance) has been one of the most important topics for CAT aspirants. Not because more questions tend to appear from this topic, but because of the fact that whatever the amount of questions is, everything is doable.
The only problem is that sometimes the questions are lengthy, and one has to skip it under time constraint. Exams like CAT, IIFT etc. tend to test our time management skills. Hence, some shortcuts always come handy in such exams. Concepts of AM (Arithmetic Mean) and HM (Harmonic Mean) can be used to solve many TSD (Time Speed Distance) problems quite easily & speedily.
Quite often we find that the given speeds (or time taken) are in an Arithmetic Progression (AP). 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 (HP).
This is because speed ‘s’ is inversely proportional to time ‘t’ given the distance ‘d’ is constant. So, distance remaining constant, if speeds are in AP then time taken are in Harmonic Progression and vice versa.
For those of us 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 little trained eyes to identify the above, it will be useful if we can keep a watch for it. See the following data to realize that either time taken or speeds are in an Arithmetic Progression. This will save a lot of time in management entrance tests like CAT & of Course in IIFT i.e. where there is time crunch.
We can go through the following illustrations to get a good grasp on the concept.
A, B, C leave point P, one after the other in the given order, with equal time intervals between their departures. If all three simultaneously meet at Q, given that speed of A and C is 40 kmph and 60 kmph, find speed of B.
If 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 arithmetic progression.
Then, their speeds will be in a Harmonic Progression.
The required speed will be the harmonic mean of 30 and 60 i.e. 2 *30 *60/90 = 40 kmph
Using this technique lots of difficult question can be solved easily. If one has understood the concept of application of arithmetic mean, geometric mean, harmonic mean in Time, Speed & Distance.
1. If I travel at 10 kmph, I reach office at 10 am, if I travel at 15 kmph, I reach office at 10:30 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.
At 10, I reach at 10
At x, I have to reach at 10:15
At 15, I reach at 10:30
Leaving at same time and reaching at 10 am, 10:15 am and 10:30 am suggests that the time travelled are in arithmetic progression. (distance is also constant)
Thus, speeds are in harmonic progression and required speed x is the harmonic mean of 10 & 15
i.e. x=2 *10 *15/25 = 12 kmph
2. 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?
1) 11 km/hr
2) 12 km/hr
3) 13 km/hr
4) 14 km/hr
Ans:12 (using the same concept i.e. reaching times are in arithmetic progression hence Speed must be in harmonic progression)
3. 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]
Ans: Here, it is a little different. We can see that the speeds are in Harmonic Progression i.e 30,40,60 are in HP or 40 is the harmonic mean of 30 40 and 60. Hence, time must be in AP.
Arun starts and time ‘t’, hence, Barun must have started at ‘t-2’ and since the times are in arithmetic progression, so Kiranmala might have started ‘t-4’ i.e. 4 hours after Arun started.
The question which would have taken 3-4 minutes by conventional methods is solved within 30 seconds now.
4. 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?
Ans: We can see the speed here. If usual speed be ‘s’ then question gives time constraints for ‘s+5’ and ‘s-5’. We can see that s, s-5, s+5 are in arithmetic progression & hence, time taken should be in harmonic progression.
We need to calculate speed ‘s’ when the time taken is ‘t’
Given the conditions that
At s+5, he requires t-20 time
& at s-5, he requires t+30 time
Now t is the harmonic mean of (t-20) & (t+30). Hence, t=2(t-20)(t+30)/2t+10= > t=120
At s+5, he requires 100 & at s-5, he requires 150 time
Since ‘s’ is inversely proportional to ‘t’ then, s/s-5 =150/120= > s=25
5. 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)
Ans: let man’s speed be ‘u’, escalator speed be ‘v’. hence, while moving up speed is ‘u+v’, & while moving down speed is ‘u-v’.
u,u-v and u+v are in arithmetic progression.Hence the times should be in harmonic progression. 2*30*90/120=45 seconds.
6. 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?
Ans: Speed upstream = b-r, Speed of still water = b, Speed downstream = b+r ;
Speeds are in arithmetic progression, so time must be in harmonic progression ; 84, t, t-9 are in hp === > solving this we get t = 72; so ans is 72-9 = 63
7. 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?
Ans: 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 arithmetic progression === > time must be in Harmonic Progression == > t = 2*20*30/50 = 24 min ( this is the beauty of aritmetic mean/harmonic mean concept)
8. 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.
Ans: At speed s+5, time is t-10
At speed s-5, time is t+15
Clearly s+5, s-5 and s are in arithmetic progression, hence times must be in harmonic progression, t must be the harmonic mean
(Similar to the 4th question)
t=2(t-10)(t+15)/2t+5 hence, time t=60. Again by using ratio, ‘s’ comes out to be 25.
Some more Explanation
Can you realize that if usual speed is s, then the three speeds are (s – 5), s and (s + 5) i.e. in arithmetic progression? Thus, the time taken are in harmonic progression. The time taken as per the data in question is (t + 15), tand (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 is 5 kmph. Thus speeds are 20 kmph,25 kmph and 30 kmph. Now distance can be found using any combination of speed and time.