If you’ve mastered the STD table method for Time n Work my previous articles, then Average speed is a piece of cake.

**Absolutely No need to mugup any formula for average speed**.

# Case 1: Distance covered in two phases

Champaklal covers 20 kms at the speed of 15kmph and other 10kms at the speed of 6kmph. What is his average speed throughout this journey?

# Approach #1: Good ol’ STD Table

## Step1: Fill up the Table

Phase 1 | Phase 2 | |

Speed | 15 | 6 |

Time | ?? | ?? |

Distance | 20 | 10 |

We don’t know how much time he took for each phase but we can find it using STD formula

Speed x time = distance

15 x time = 20

Time for phase1=20/15=4/3

Similarly find the time taken for phase 2 and update the table

Phase 1 | Phase 2 | |

Speed | 15 | 6 |

Time | 4/3 | 5/3 |

Distance | 20 | 10 |

## Important note about STD table:

- While solving Time n Work problems, you can only do addition or subtraction in
**Speed and distance cells only**. - But for solving average speed you can do only do addition or subtraction in
**Time and Distance cells only.**

## Step2: Addition in last column

Now make a new column and do the addition of phase 1 and 2.

Phase 1 | Phase 2 | Total=Phase1+2 | |

Speed | 15 | 6 | ?? |

Time | 4/3 | 5/3 | 4/3+5/3=9/3=3 |

Distance | 20 | 10 | 20+10=30 |

Run STD formula for last column

Speed x time = distance

Speed x 3 = 30

Speed =30 / 3

Speed =10kmph

Final answer: Champak’s Average speed = 10kmph

# Approach #2: Alligiation method

Average speed is nothing but a “weightage average” and since journey was finished in two phases, we can run the alligiation method.

IF you don’t know the Alligiation (Wine+Water) method, Click ME

Step1 is same here. Fill up all the cells in STD table just like we did in above method.

Phase 1 | Phase 2 | |

Speed | 15 | 6 |

Time | 4/3 | 5/3 |

Distance | 20 | 10 |

Now run the alligiation method (Wine and Water Mixture) using speed as ‘concentration’ and Time as ‘volume”

Speed (concentration) |
6 | Average speed m (between 6 to 15) |
15 |

Time (Volume) |
5/3 | 4/3 |

Run the visual move

Answer: average speed =10kmph

# Case 2: Distance covered in three phases

Jethalal travels for 30 minutes @50kmph

For the other 20 minutes, he travels @speed of 30kmph

For the 60 minutes, at the speed of 60kmph.

Find his average speed for this journey.

For STD table to work, Everything must be in same unit, you can see here time is given in minutes but speed is given in kmph. Better convert time into hour.

30 minutes = ½ hour

20 minutes=20/60=1/3 hour

60 minutes= obviously 1 hour!

Now we are in business.

## Step1: Fill up all cells of STD table

Phase 1 | Phase 2 | Phase 3 | |

Speed (kmph) | 50 | 30 | 60 |

Time (hr) | ½ | 1/3 | 1 |

Distance |

Run STD formula on phase1

Speed x time = distance

50 x ½ =distance

Distance=25 km

Same procedure for phase 2 and 3.

Update the table.

Phase 1 | Phase 2 | Phase 3 | |

Speed (kmph) | 50 | 30 | 60 |

Time (hr) | ½ | 1/3 | 1 |

Distance | 25 | 10 | 60 |

## Step2: Addition in last column

Make a new column for total of Phase 1+2+3

Remember we can only add distance and time but not speed.

Phase 1 | Phase 2 | Phase 3 | Total | |

Speed (kmph) | 50 | 30 | 60 | ?? |

Time (hr) | ½ | 1/3 | 1 | ½+1/3+1=11/6 |

Distance | 25 | 10 | 60 | 25+10+60=95 |

Run STD fomula on last column

Speed x time = distance

Speed x 11/6=95

Speed =95×6/11=51.81kmph

**Final answer**: Jethalal’s average speed is 51.81kmph

# Alligiation for three phase?

We can also run alligiation method on three phase journey too but it’ll be too lengthy calculation- anyways here is the ‘direction’:

First run alligiation on phase 1 and 2. You get an avg speed(a)

Now run alligiation on avg speed(a) and Phase 3 and you get final average speed.