Time, Speed & Distance Primer  Ravi Handa

There are some topics in Quantitative Aptitude, like permutation & combination, where you can easily find out the answer, (it is in the options) but it turns out to be wrong. And then there are some topics in which you read the question, understand it but cannot even begin solving it. You get stuck at the first step and you have no idea about how to even approach the question.
The irritating fact is that you understood the question properly. It happens very frequently with questions on Time Speed & Distance (TSD). I have always been a big advocate of skipping questions which you cannot solve. More often than not, TSD questions should be skipped if you cannot figure out how to start within the first minute. Typically the questions on TSD are based upon certain ideas, which if you are not aware of can make solving the question extremely difficult and time consuming. I do not think that I am even aware of all ideas / types of questions in TSD but there are a few popular ones which have been doing the rounds in the past few years.
I am going to cover some of them in this post and probably revisit some more in the months to come. I am also going to discuss the reasoning behind those ideas. It is very important that you understand the logic behind the formulae before you actually start using them. If you don't, there is a very high probability that you will make a mistake.
To begin with, some of the very basic ideas that you should be aware of are:
Speed = Distance / Time
If Distance is constant, then Speed and Time are inversely proportional to each other
Two bodies moving in the same direction would have the relative speed of S(1)  S(2) and two bodies moving in the opposite direction would have the relative speed of S(1) + S(2)In this particular post, I am going to talk about the motion of two bodies in a straight line starting from opposite ends.
Case 1:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2).
They take times of T(1) & T(2) to reach their destinations.In such a case, the relationship between the times taken and the speeds will be S(1) / S(2) = T(2) / T(1)
The distances that both bodies have travelled are PQ and QP respectively.
PQ = S(1) T(1)
QP = S(2) T(2)Also, PQ = QP so the above equations can be equated to get the desired result.
Let us say that they meet at a point R in between, then
R would divide the distance PQ in the ratio of S(1):S(2)
They started at the same time and they are meeting at point R, so the time taken by both the bodies will be the same.
If we assume that time as T
PR = S(1) T
QR = S(2) TDividing the above two equations would give us the desired result.
Case 2:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). After meeting each other, they take times of T(1) & T(2) to reach their destinations.
In such a case, the relationship between the times taken and the speeds will be S(1)/S(2) = √(T(2)/T(1))
Also, the time taken for the two bodies to to meet will be √(T(1) x T(2))
Let us assume that the two bodies meet at a point R, after time T.
For the first body, PR = S(1)T and RQ = S(1)T(1)
For the second body, QR = S(2)T and RP = S(2)T(2)
We know that PR = RP
=> S(1)T = S(2)T(2)
We know the QR = RQ
=> S(2)T = S(1)T(1)
Dividing the above two equations, we will get
Using the above result, we can obtain the value of T
Case 3:
Two bodies start from opposite ends P & Q to reach their destinations at the same time and move towards each other with speeds S(1) & S(2). Before meeting, they take times of T(1) & T(2) to reach the meeting point.
In such a case, the relationship between the times taken and the speeds will be S(1) / S(2) = √ (T(2) / T(1))
Also, the time taken for the two bodies to reach their destinations after meeting will be √ (T(1) x T(2))
The logic for this would be exactly the same as Case 2. Try working this out on your own.
Case 4:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). They reach the opposite ends and reverse directions. They continue this to and fro motion.
If the distance between the two bodies in the beginning is D, then
Time taken by them to meet for the first time, T = D / S(1) + S(2)
If R is the first meeting point, PR / QR = S(1) / S(2)
The idea that I am going to discuss now is only valid in the case when the greater speed is less than twice of the lesser speed. {If S(1) > S(2), then S(1) < 2S(2)}
After the first meeting, if the bodies continue to move they will reach their respective ends and start the return journey. They will meet again on the return journey and then proceed further. After the first meeting, they would have covered 2D distance. Since the distance has doubled, time taken will also double.
So, the total distance covered by the bodies for their first, second, third ... nth meetings will be:
D, 3D, 5D ... and (2n1)D
So, the total times taken by the bodies for their first, second, third ... nth meetings will be:
D / S(1) + S(2), 3 D / S(1) + S(2), 5 D / S(1) + S(2) ... and (2n1) D / S(1) + S(2)
In case you need to figure out the point at which the bodies meet for the nth time, consider only one of the bodies, say 1.
Distance covered by 1 till the nth meeting = [S(1)/(S(1)+S(2))] * (2n1)D
The remainder of the above when divided by 2D will give you the exact location of the nth meeting point.
For example, if the distance covered by 1 till the nth meeting comes out as 700 meters and D = 150 meters, the nth meeting point can be obtained by the remainder of 700/300 which is 100. So, the nth meeting point will be 100 m from P.
I know you would hate me for saying this, but this is not the end of TSD. This is not even the end of motion of two bodies in a straight line. The destination is still far away and hopefully we will reach there in time. We will now see some ideas that will help to solve TSD problems without forming too many equations.
Solving TimeSpeedDistance problems without using equations
I guess my first fascination with problems of Time, Speed and Distance began when I first watch a film called Henna. An important portion of the plot, if you can call it that, had Rishi Kapoor floating from India to Pakistan in a river without drowning. I remember arguing with my friends that if he could float that long he could swim back to India as well. My friends nullified the argument by saying,
Speed_River > Speed_Rishi Kapoor
I know that the reference is a little dated for most readers, but Zeba Bhaktiyar made me look beyond reason. In this post we will discuss some of the ideas that have helped me solve TSD problems without forming too many equations.
Funda 1: Average Speed
We know that the average speed during a journey is given by (Total Distance Covered) / (Total Time Taken); but there are a few special cases which might help in solving questions,
If the distance covered is constant (d1 = d2 = d3 = dn) in each part of the journey, then the average speed is the Harmonic Mean of the values.
Speed_Avg = n / (1/s1 + 1/s2 + 1/s3 ... + 1/sn)
If the time taken is constant (t1 = t2 = t3 = tn) in each part of the journey then the average speed is Arithmetic Mean of the values.
Speed_Avg = (s1 + s2 + s3 ... +sn)/n
Funda 2: Using Progressions (Arithmetic & Harmonic)
In many questions, you will come across a situation when a person is going from point A to point B at various speeds and taking various times. We know that if distance is constant, speed and time are inversely proportional to each other. But this information can also be used to deduce the following two facts,
If the various speeds which are mentioned are in AP, then the corresponding times taken will be in HP.
If the various speeds which are mentioned are in HP, then the corresponding times taken will be in AP.
Let us use these ideas to solve couple of quant questions.
Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour 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?
(Some useless information: Arun Barun Kiranmala is a 1968 Bangladeshi film. Now you can guess what inspires the CAT question setters)
Solution: As you can see that the speeds are in HP, so we can say that the times taken will be in AP. Time difference between Arun and Barun is 2 hours, so the time difference between Barun and Kiranbala will also be 2 hours.
Hence, Kiranbala started 4 hours after Arun.Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swim the course with the river flow?
Solution: Let us say Speed of Rishi Kapoor in still water is RK and Speed of the river is R. Hence, Rishi Kapoors speeds against the river flow, in still water and with the river flow are,
RK  R, RK and RK + R.
As you can see, they are in AP.
Hence, the corresponding times taken will be in HP.
Let us say that the time taken to row down with the stream is t, then 84, t+9 and t are in HP. So,
t + 9 = (2 * 84 * t) / (84 + t)
t^2 + 93t + 756 = 168t
t^2  75t + 756 = 0
t = 63 or 12Funda 3: Special Case
Let us say that two bodies a & b start at the same time from two points P & Q towards each other and meet at a point R in between. After meeting at R, a takes ta time to reach its destination (Q) and b takes tb time to reach its destination (P). Then,
Sa / Sb = √(tb / ta)
Also, the time taken by a & b to meet (i.e. to reach point R from P & Q respectively) is given by,
t = √(ta * tb)Note: The same formulae will be valid if two bodies a & b start at different times from two points P & Q towards each other. They meet at a point R in between after travelling for ta and tb time respectively. After meeting, they take the same amount of time (t) to reach their respective destinations (Q & P).
I hope that these ideas will help you reduce the number of equations that you form while solving TSD problems if not completely eliminate them.