## Simple, quick and intelligent solution of time distance problems

Simple, quick and intelligent solutions of time distance problems in a few steps are explained using two selected time and distance MCQ problems.

**The basic time and distance concept**

The core concept in all time and distance problems is governed by the classic relation between the three entities, time, distance and speed.

$Distance = Time\times{Speed}$

This is the most basic relationship between these three elements.

Conceptually this means, for a moving object **distance is directly proportional to either speed or time**, the other remaining constant. Furthermore, if distance is held constant, this relationship also means **time and speed will be inversely proportional to each other.**

We can believe this relationship from our every day experience also. If our train moves at a higher speed we will reach our destination in lesser time, isn't it? That is inverse relationship between the speed and time, where the distance is same in both the journeys at different speeds.

Instead, if we move faster for the same duration we will cover a much larger distance. That is direct proportionality between distance and speed. Here time duration is held constant in both cases of travel.

### The derived and advanced time and distance concept

The derived concept is the concept of * relative speed* when two or more than two objects move in the same or opposite direction. One object moving faster than another but in the

**same direction**will pass the second object at a

**relative speed which is the difference between the two speeds**.

But if the **two objects move in opposite directions**, their **speed relative to each other will be the sum of the two speeds.**

All time and distance sums at school level and that appear in competitive tests are based on these two sets of concepts.

We will take two problems on this topic area of time and distance just to highlight how problems can be solved in just a few simple steps by applying relevant problem solving techniques and strategies.

To contrast this simplicity we will present the conventional approach that is mostly followed in many occasions that we are aware of.

### Problem 1.

A motorboat covers 25km upstream and 39km downstream by travelling at same speed for 8 hours. On another occasion with the same speed it covers 35km upstream and 52km downstream in 11 hours. What is the speed of the stream?

- 5 km/hr
- 3 km/hr
- 4 km/hr
- 2 km/hr

**Simple approach - Solution 1:**

**In this solution we will try to solve the problem using as few steps as possible.**

**Problem analysis:**

Each of the two periods of 8 hours and 11 hours consists of **two component time periods**, **one upstream** at a speed that is the **difference in boat speed and stream speed**, and the other **downstream at a speed that is a sum of boat speed and stream speed**.

Let us call the upstream speed of the boat as $SU$ and downstream speed $SD$.

This is the first stage of use of **Abstraction technique.** Conventionally, the variables are straightaway assumed to be boat speed $B$ and stream speed $S$ and the equations are expressed in terms of these variables right from the start.

But we recognize the great **principle of representative**,

Do not deal with the masses till it is absolutely necessary. Instead deal with their representatives.

Politicians and officers who need to meet large number of people always meet with the representatives. In this principle lies practical wisdom about how things should done. If you meet large number of people you won't be able to home in to any single issue and solve it. Most likely your meeting will be a fiasco. Just imagine if you have to do this daily. The much more practical system that evolved is - the masses select a small group of representatives who presents the issues on behalf of the masses.

We identify the **wisdom in this approach** and also recognize that this **act of creating a representative is nothing but an abstraction** of the large number people to a few representative, making them equivalent for this purpose. **In our case instead of prople, it is simply variables.**

**Please note:** this is not * equivalence class mapping, *as

*the characteristics of each member of the masses do not come into the picture at all.*

We rephrase the abstract **principle of representative** and at all stages, * as far as possible* (at the end stage we will have to expand the representative abstracted variables),

**deal with abstracted variables**, not variables in their actual form. In other words

*by applying Abstraction technique and with the help of principle of representative*, we deal with only the $SU$ and $SD$, the upstream and downstream speeds, but not their component speeds as expressed in the relationships,

Upstream speed, $SU = B - S$ and,

Downstream speed, $SD = B + S$.

The actual **act of using the representative variables** is though a well-known mathematical **technique of substitution**. To be able to use our problem solving resource in any domain we would prefer to call this instead the **principle of representative**, which is the core process and more abstract.

In **Substitution technique**,

A complex expression is substituted by a simpler expression for making further derivation simpler.

Finally we observe that in the process of using these representative variables, we transform the equations to a simpler form and in fact use another well known technique originated from mathematics, the **technique of solving a simpler problem.**

Our observation on this last aspect is** - technique of solving a simpler problem may use substitution or any other relevant technique. **It is not necessary that this technique will use only substitution.

**Solution process:**

With this simple assumption we can form two expressions for two occasions:

$\displaystyle\frac{25}{SU} + \displaystyle\frac{39}{SD} = 8$, and

$\displaystyle\frac{35}{SU} + \displaystyle\frac{52}{SD} = 11$.

We know that we would be able to evaluate both $SU$ and $SD$ from these two linear equations.

But again, instead of evaluating $SU$ or $SD$ directly **we will treat their inverses as the target variables**. This is a much simpler form of abstraction and substitution.

We select the second expression (it could have been the first expression as well) and transform it to,

$\displaystyle\frac{25}{SU} + \displaystyle\frac{52}{SD}\times{\frac{25}{35}} = 11\times{\displaystyle\frac{25}{35}}$.

Subtracting this equation from the first equation we get,

$\displaystyle\frac{1}{SD}\left(39 - \displaystyle\frac{5\times{52}}{7}\right) = 8 - \displaystyle\frac{55}{7}$

Or, $\displaystyle\frac{13}{7SD} = \displaystyle\frac{1}{7}$.

Or, $SD = 13$, and so from any of two original equations, we get,

$SU = 5$.

Now only we explore the boat speed and the stream speed. As per our oringinal definition, upstream speed $SU$ is the difference of the two speeds whereas downstream speed $SD$ is the sum of the two speeds.

Summing up the two relative speeds we get, $SU + SD = 13 + 5 = 18$ to be equivalent to twice the higher speed of the boat $B$. So boat speed is 9 km/hr and the stream speed $S$ is 4km/hr.

**Answer:** Option c: 4 km/hr.

#### Key concepts used:

The use of ** basic relation between distance, time and speed **and the

**in the form of**

*principle of representative*

*technique of substitution.*#### Conventional approach - Solution 2

In this conventional approach no substitution is used so that the initial expression will be,

$\displaystyle\frac{25}{B - S} + \displaystyle\frac{39}{B + S} = 8$, and

$\displaystyle\frac{35}{B - S} + \displaystyle\frac{52}{B + S} = 11$.

We decide to eliminate $B + S$ and multiply first equation by 4 and the second by 3 to get,

$\displaystyle\frac{100}{B - S} + \displaystyle\frac{156}{B + S} = 32$, and

$\displaystyle\frac{105}{B - S} + \displaystyle\frac{156}{B + S} = 33$.

Now we subtract the first result from the second, giving,

$\displaystyle\frac{105}{B - S} - \displaystyle\frac{100}{B - S} = 1$,

Or, $\displaystyle\frac{5}{B - S}=1$

Or, $B - S = 5$.

From any of the original equations using the value of $B - S$ we get,

$B + S = 13$.

Finally then, $B = 9$ km/hr and $S = 4$ km/hr.

Not much extra deduction but notice the sense of discomfort in dealing with larger expressions in inverse form.

We would just say, choose whichever approach suits your style.

The second problem we solve now is again a problem on time and distance highlighting a different aspect.

### Problem 2.

If a train runs at 40km/hr, it reaches its destination 11 minutes late and if it runs at a speed of 50km/hr, it reaches the destination only 5 minutes late. Find the right time by which the train was to reach its destination.

- 20 minutes
- 19 minutes
- 21 minutes
- 18 minutes

**Simple approach - Solution 1:**

Let us assume $T$ to be the right time in minutes by which the train is to reach its destination and $D$ be the distance of travel. So by the two statements we have,

(T + 11 mins) $\times{}$ 40km/hr = D, and

(T + 5 mins) $\times{} $ 50km/hr = D.

We **take the** **ratio of two equations thus eliminating the variable $D$** which is constant here.

$\displaystyle\frac{(T + 11)\times{4}}{(T + 5)\times{5}} = 1$

Or, $(T + 11)\times{4} = (T + 5)\times{5}$

Or, $T = 19$ minutes.

**Answer:** Option b: 19 minutes.

#### Key concepts used:

At the very outset we choose our unknown variable to be the right time $T$ using our **goal oriented approach**. Always choose your unknown variable as the unknown quantity, value of which you have to finally find.

Next using the **basic concept use technique** we express the problem statements in their simplest form.

Now recognizing this as a case of constant distance $D$, we use the **ratio technique** to eliminate $D$ and find $T$ with ease.

Finally we reach the solution in a few simple steps.

We will take up now the conventional approach.

#### Conventional approach - Solution 2

First the speeds are converted from per hour form to per minute form.

$40km/hr = \displaystyle\frac{2}{3}km/minute$, and similarly,

$50km/hr = \displaystyle\frac{5}{6}km/minute$.

Let us now assume as before, the distance as $D$ km and the right travel time as $T$ minutes.

Thus we have,

$\displaystyle\frac{D}{\frac{2}{3}} = T + 11$, and

$\displaystyle\frac{D}{\frac{5}{6}} = T + 5$.

Subtracting the second equation from the first we get,

$\displaystyle\frac{3D}{2} - \displaystyle\frac{6D}{5} = 6$,

Or, $\displaystyle\frac{5\times{3D} - 2\times{6D}}{10}= 6$

Or, $\displaystyle\frac{3D}{10} = 6$

Or, $3D = 60$

Or $D = 20$ km.

With this value of $D$ from the second original equation we get,

$20 = \displaystyle\frac{5}{6}(T + 5)$

Or, $120 = 5T + 25$,

Or, $5T = 95$,

Or $T = 19$ minutes.

Now you *compare the two solutions carefully to understand where and why you need to apply relevant problem solving strategies and techniques for reaching the solution in a few simple steps.*

We hope these two case examples will not only show you how to use the concepts relevant to this topic area of Time and Distance problems, but **more importantly will provide you with insights into the process of elegant problem solving in a few simple steps.**

### Tutorials and Quick solutions in a few steps on Speed time distance and related topics

**Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers**

**How to solve Time and Distance problems in a few simple steps 1**

**How to solve Time and Distance problems in a few simple steps 2**

**How to solve a GATE level long Work Time problem analytically in a few steps 1**

**How to solve arithmetic boundary condition problems in a few simple steps**

**How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns**