- Relative speed concept
- Case : Two people start together simultaneously
- 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.

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

If you’ve mastered the Universal STD Table Method^{TM }, 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

=25+35

=60kmph

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

## Important:

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

=229.16+50

=279.16= hotel’s distance from A’bad.

## From Mumbai

The hotel is 600-279.16=320.84 kms away from Mumbai

Alternatively,

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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