Time and distance is one of the main sources for tricky quantitative questions in the well fame, well designed B-school entrance exams. An aspirant can expect 4 to 6 questions from this topic in CAT and CAT level exams. MAT and CMAT are playing a lot with this topic. For a better performance in this area, aspirants are required to acquire a certain level of awareness and skills which are mentioned below:

- Well awareness on basic concepts (ie. the relationship between the main parameters such as Distance, Speed and Time)
- Effective application of the concepts such as Ratio and Inverse & Direct variations
- Quick and intellectual percentage calculations
- Logical thinking and visualization/imagination skills

In this chapter we will look at problems in the following different dimensions:

- Basic time speed distance relation based on problems
- Basic Problems of trains
- Inverse direct variation problem
- Average Speed

## Basic Concept (Relation between time, speed and distance)

If the distance between 2 cities X and Y is 100km and a car is travelling from X to Y at a constant speed of 50 kmph then it will take 2 hrs to reach Y. Here

- `"Distance = 100km"`
- `"Speed = 50 kmph"`
- `"Time = 2h"`

Let's see how these three parameters are connected with each other.

`"Time" = "Distance"/"Speed"`

`"Speed" = "Distance"/"Time"`

## Standard units of parameters

Distance normally expressed in form of kilometer(km) or meter(m), while speed has the standard unit expressions such as kilometer per hour or kmph(km/h)or meter per second (m/s)and time is in the standard units of hours(hrs) and seconds(s).

### Conversion of units of speed

`1 " kmph" = (1 " km")/(1 " hr") = (1000 " m ")/ (3000 " s ") = 5/18 "m/s"`

SImilarly, `1 "m/s" = 18/5 "kmph"`

`q " m/s" = q xx 18/5 " kmph"`

### Examples

`"Speed " = 18 kmph = 5 m/s`

`"Time required to cover the distance" = "Speed"/"Distance" = 1000/5 = 200 sec = 3 min 20 sec`

**Method I: Using Algebra**

Let the distance he covered by bus is `x km`

`:.` Distance covered by train is `620 - x`

`-> x/60 + (620 - x)/80 = 9`

`-> 4x + 1860 - 3x = 2169`

`-> x = 300km`

**Method II: Using work back/substitution/trial and error**

Note: Work back method always start with a middle quantity as per the ascending order of option values. We can start the substitution with 300 or 320.Speed of bus is 60 kmph and the total time required in an integer value. So most probably the distance covered by bus is an integral multiple of 60(not always!). So here we can start our substitution from 300.

If 300 km is the distance covered by bus then the remaining 320 km covered by train.

`:.` Time taken by bus `= 300/6 = 5hr`

And time taken by train `= 320/80 =4hr`

Total time `= 5 + 4 = 9hr`, hence verified the answer our substitution is correct.

`"Distance covered by bus" = 300 km"`

**Method III: Alligation**

Average speed of entire journey `= 620/9`

The resultant ratio is; `80/9 : 100/9 -> 4:5`, This is the ratio of time distribution for each mode of travel.

`:.` Time duration for the journey by bus `=5 hr`

Distance covered by bus `= 5 xx 60 = 300 km`

## Basic problems on trains

This is the basic train related problems (or the first version of the train problems). In this area we are dealing with the kind of problems in which train crosses a stationary object rather than other moving objects. Most of these types of questions are intended to test the relationship between the basic parameters only and more often the distance covered is the sum of the lengths of the train and that of the covering object. Depends upon the situation, the length (width) of the covering objects may be negligible.

You will get a clear idea about the concept through the following set of concept review questions.

`"Time" = 200/10 = 20 " m/s"`

` = 20 xx 18/5 " kmph" = 72 " kmph"`

`= 2x + x = 3x`

Speed of the train is `135 " kmph" = 135 xx 5/18 "m/s"`

We know that, `"distance" = "speed" xx "time"`

`3x = 135 xx 5/18 xx 10 xx 1/3 = 375m`

`-> "Length of train, 2x" = 250m`