[Aptitude] TSD: Relative speed, Moving towards eachother

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  1. Relative speed concept
  2. Case  : Two people start together simultaneously
  3. Case: Two people don’t start together

TnW= Time and Work
TSD=Time speed and Distance.
But we solve them using same STD formula, as you’ve seen in my previous articles.

TSD got four main type of problems

  • 1. Trains and Platforms
  • 2. Boats and streams
  • 3. Relative speed
  • 4. Circular track.

Relative speed Problems

In these questions, generally two people start from opposing directions and ride at different speeds. You’ve to find the speed, distance or time when they’ll meet each other.
This type of TSD question routinely appears in various competitive exams.

Relative speed concept

  • Recall the pipes and cisterns concept. If Pipe A and B fill up the tank. When they work together, the combined speed is A+B.
  • Similarly if A and B are riding their cars and coming towards each other, the relative speed increases and becomes A+B
  • If Pipe A fills up the tank and pipe B empties the tank, then combined speed is A-B.
  • Similarly, if A and B are riding their cars and moving away from each other, the relative speed is A-B (subtract higher speed from smaller speed.)

Case  : Two people start together simultaneously

Jethalal starts from Pune to Mumbai and at the same time Roshansingh Sodhi starts from Mumbai to Pune. Their speed is 25kmph and 35 kmph respectively. If the distance between Pune and Mumbai is 120 kms.

  1. When will these two gentlemen meet?
  2. The place where they meet, has a hotel. How far is this hotel from Pune?

If you’ve mastered the Universal STD Table MethodTM , this question Is a walk in the park.
Since they’re moving towards each other,
(Relative Speed)Jetha and Sodhi
=Jetha’s speed + Sodhi’s Speed

Now the usual STD table

Jetha Sodhi Together
Speed 25 35 25+35=60
Time ??
Distance 120

Since they start together, the speed is 60kmph from the beginning.
Run STD on last column
Speed x time = distance
60 x time =120
Time =120/60 =2 hrs.
Ans1: They’ll meet after two hours.

Q2. The place where they meet has a hotel. How far is this hotel from Pune?

From the answer1: we know that Jetha and sodhi will meet after two hours.
Since Jetha is riding from Pune side, in two hours he will cover

Jetha Sodhi Together
Speed 25 35 25+35=60
Time 2 2 2
Distance 50kms 70 120

So applying STD formula on Jetha’s column, you get distance covered by Jetha = 50 kms.
Therefore, the said hotel (meeting place) is 50 kilometers away from Pune.
Similarly, it is 35 x 2 =70 kms from Mumbai.

Alternatively, Since hotel is 50 kms away from Pune and Pune is 120 kms away from Mumbai so
Hotel is 120 minus 50  = 70 kms away from Mumbai.

Case: Two people don’t start together

Jethalal starts from Ahmedabad to Mumbai, at 8 AM with 25kmph speed.
At 10 AM, Sodhi starts from Mumbai to Ahmedabad at 35kmph speed.
The meet at the hotel, how far is this hotel from Ahmedabad, IF the distance between A’bad and Mumbai is 600 kilometers.

Given: Jethalal starts @8AM and Sodhi starts @10 AM.
That means for 2 hours, Jethalal worked alone. Because 10AM minus 8 AM = 2 hours.
Fill up the table.

Jethalal Sodhi 8AM: Jetha starts
Speed 25 35 25
Time 2
distance 600 600 25 x 2 =50

In the last column run STD
Speed x time = distance
25 x 2 =50 kms

You get that from 8AM to 10 PM, Jetha alone covered 25 km x 2 hr =50 kilometers of journey.
So remaining work = total 600 minus 50 = 550 kms.

Since Sodhi also started at 10AM, from now onwards the remaining 550kms will be covered by both of them together. They’re moving towards each other
Relative speed =
=Jetha + Sodhi
=25 + 35
=60 kmph

Make new column

Jethalal Sodhi 8AM: Jetha starts 10AM: Sodhi joins
Speed 25 35 25 25+35=60
Time 2 ??
distance 600 600 25 x 2 =50 600-50=550

Run STD on last column
Speed x time = distance
60 x time = 550
Time =55/6


What’s the total time taken before they meet each other ?
For the first two hours (8 to 10 AM), Jetha rode alone. Then both of them together covered remaining distance in 55/6 hrs. So total time
=2 hours + 55/6 hrs
=67/6 hrs.

We’ve to find how far is the hotel from A’bad.
Since Jethala is riding from A’bad, whatever distance he covered before meeting Sodhi  = that is our answer. So make a new column

Jethalal Sodhi 8AM: Jetha starts 10AM: Sodhi joins Jetha’s Journey
Speed 25 35 25 60 25
Time 2 55/6 (2+55/6)=67/6
distance 600 600 50 550

Run STD on last column
Speed x time = distance
25 x 67/6 = distance
Distance= approx. 279.16 kms

Final answer: The hotel is 279.16 kms away from Ahmedabad.

Alternatives approach

Jetha and Sodhi travelled together for 55/6
So from 10 AM onwards, Jetha covered 25 x 55/6=229.16 kms
Apart from that, between 8 to 10AM, Jetha already covered 50 kms
So total distance covered by Jetha, before the meeting
=279.16= hotel’s distance from A’bad.

From Mumbai

The hotel is 600-279.16=320.84 kms away from Mumbai
Sodhi travelled for 55/6 hours before Meeting jetha
Distance covered by Sodhi = 35 x 55/6 =320.83=hotel’s distance from Mumbai.
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26 Comments on “[Aptitude] TSD: Relative speed, Moving towards eachother”

  1. Case : Two people start together simultaneously

    since Jetha starts from Pune side, place of hotel will be 70 km from Mumbai

  2. your explaination regarding all aptitude related problems and the economy jargons made it ashtonishingly easily for us, thank u very much for that sir

  3. Time-Distance-Speed, Probability, problems on Pipes used to scare me all the time. But I must admit after reading your articles, I am at ease with all of these. Thank you and best wishes for your work.

    open source for aptitude. :P

  4. the way you have explained is like pulling apart the continuous interlocking knots, u pull it apart and it stand out as a plain thread…..!!!!! i am in love with this STD table it simply rocks and i apply it readily on the types of problems where this method is applicable, no longer mathematics is a hard nut now. ani suggest all non engineering background aspirants and specially medical professionals to follow MRUNAL BHAI’S website and improve your numeracy bit and have an extra edge to beat your own cut offs and the upsc cut off as well. Thanks Mrunal bhai. one expectation if you can clarify the question asked in the previous years are in similar lines with the strategy by giving the list of questions at the end of the examples it would help us practice and become confident of solving such problems.

  5. A hare takes 9 leaps in the same time as a dog takes 4.but the dog’s leap is 7/3 m long while hare’s is only 1 m.how many leaps will the dog have to take before catching hare if hare had a head start of 16 m?

    1. 192.
      Explanation :
      A dog will cover 4*7\3 m distance in the time a hare will cover 9 m distance. Lets consider they will meet after x hour. So distance to be covered by dog = 16 + distance covered by hare.

      28\3x = 9x + 16.

      Solving this , x=48. so leaps taken by dog = 4x = 4*48 = 192.

  6. Discremation is on at all levels at every place trouhgh out the world with the locals and the people migrated people from the other countries and with in the countries from the distant places. It is a part of human nature. SIKHS due to their turban and sabat surat (physical perfactness ) have a major attraction. BUT the sikhs never taken serious steps to make their identity and always remain confused with some what similar identity with muslim world of south east ASia. Sikhs almost 95% never remaind sikhs in the foriegn countries. Who will bother about sikhs for their identity, when they it self not remain sikhs. To establish sikh identity trouhgh out the world the sikhs has to maintain their physical perfactness and purity of mind.waheguru ji ka khalsa ,waheguru ji ke fateh.

  7. a bus running at a speed of 30kmph leaves trivandrum at 10am &another bus running @speed of 40kmph leaves d same place @3pm in d same direction.how many kms from trivandrum will they be togethr??

  8. Two men starting from the same place at the rate of 5kmph and 5.5 kmph. If they walk in same direction, what time it will take them to be 8.5km apart?

  9. The way you make complicated problem look ridiculously easy is quite amazing. Keep up the good work

  10. after how much time will they meet the second time ??questions like these ,how can we solve them ??

  11. ABCD is a square running track with sides 100m each.Alex and Bob start from point A and C with speeds 5m/s and 10m/s respectively towards point B .After they meet they continue running in respective directions.After how many seconds will they meet for the second time from the instant they started running??

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