**Solving Word Problems Involving Distance Speed and Time :**

In this section, we will learn how to solve word problems involving distance speed and time.

The relationship between distance, speed and time.

Time = Distance / Speed

Speed = Distance / Time

Distance = Time ⋅ Speed

To convert minutes into hour, we should divide the given minutes by 60.

**Example 1 :**

The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr

**Solution :**

Distance between two stations A and B = 192 km

Fast train takes 48 minutes less then the time taken by the slow train.

Let x be the speed taken by the fast train.

Speeds of two trains differ by 20 km/hr

So speed of slow train is x – 20.

Time = Distance/speed

Let t_{1} and t_{2 }be the time taken by the faster train and slower train respectively.

Let be the time taken by the slow train

t_{1 } = 192 / x

t_{2} = 192 / (x – 20)

48 / 60 ==> 4/5 hours

t_{2 - }t_{1} = 4/5

[192 / (x -20) – 192 / x] = 4/5

192[ (x – x + 20)/x (x - 20) ] = 4/5

19200 = 4x² – 80x

Dividing the entire equation by 4, we get

4800 = x^{2} – 20x

x^{2} – 20x – 4800 = 0

(x – 60) (x + 40) = 0

x - 60 = 0 x = 60 |
x + 40 = 0 x = -40 |

So, speed of the faster train = 60 km/hr

Speed of slow train = x – 20

= 60 - 20

= 40 km/hr

Speed of the faster train = 60 km/hr.

Speed of the slow train = 40 km/hr.

**Example 2 :**

A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10 km/hr, the journey would take 1 hour longer. What is the average speed.

**Solution :**

Let x be the average speed of the train

If its speed was decreased by 10 km/hr, the journey would take 1 hour longer.

So, x – 10 be the decreased speed

Time = Distance / Speed

Let t_{1 }and t_{2} time taken to cover the distance in the speed of x and x - 10 km/hr.

t_{1} = 300/x

t_{2} = 300/(x – 10)

t_{1} – t_{2} = 1

( 300/x ) - ( 300/(x - 10) ) = 1

300 [1/(x - 10) – 1/x] = 1

300 [x – x + 10]/[x(x -10)] = 1

3000/(x² – 10x) = 1

3000 = x^{2} – 10x

x^{2} – 10x = 3000

x^{2} – 10x – 3000 = 0

(x + 50)(x – 60) = 0

x + 50 = 0 x = -50 |
x - 60 = 0 x = 60 |

So speed of the 60 km/hr.

After having gone through the stuff given above, we hope that the students would have understood how to solving word problems involving distance speed and time.

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