PROBLEMS ON TIME AND DISTANCE WITH SOLUTION

Problem 1 :

A passenger train takes 3 hours less than a slow train for journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the passenger train, find the speed of two trains.

Solution :

Let x be the speed of the of the passenger train.

Speed of the slow train is 10 km/hr less than that of the passenger train.

So x - 10 be the speed of the slow train.

Distance has to be covered = 600 km.

Time = Distance/Speed

Let T1 be the time taken by passenger train.

Let T2 be the time taken by the slow train.

The differences of time taken by both trains are 3 hours.

T= 600/x

T2 = 600/(x - 0)

T2–T1 = 3 hours

[600/(x – 10)] – (600/x) = 3

600[(1/(x - 10) - (1/x)] = 3

x - (x - 10)/(x- 10x) = 1/200

(x - x + 10)/(x- 10x) = 1/200

2000 = x- 10x

x- 10x - 2000 = 0

x– 50x + 40x - 2000 = 0

x(x – 50) + 40(x – 50) = 0

(x + 40)(x – 50) = 0

By solving, we get x = -40 and x = 50.

Therefore the speed of passenger train = 50 km/hr.

Speed of slow train = 40 m/hr.

Problem 2 :

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 T1 be the time taken by the fast train.

Let T2 be the time taken by the slow train.

T1 = 192/x

T2 = 192/(x – 20)

48/60 = 4/5 hours

T1 – T2 = 4/5

[192/(x - 20) - 192/x] = 4/5

192[(x – x + 20)/x(x - 20)] = 4/5

192(20)/x– 20x = 4/5

3840(5) = 4(x– 20x)

19200 = 4x2 – 80x

4800 = x2 – 20x

x– 20x – 4800 = 0

x– 60x + 40x - 4800 = 0

(x – 60)(x + 40) = 0

x = 60 and x = -40

Speed of fast train is 60 km/hr.

Problem 3 :

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

Solution :

Let x be the average speed of the train.

So x – 10 be the decreased speed.

Time = Distance/Speed

T1 and Tbe the time taken by the train to cover the distance with speed of x km/hr and (x - 10) km/hr respectively.

T1 = 300/x

T2 = 300/(x – 10)

T1 – T2 = 1

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

3000/(x2 – 10x) = 1

3000 = x2 – 10 x

x2 – 10x = 3000

x– 10x – 3000 = 0

x– 60x + 50x – 3000 = 0

x(x – 60) + 50(x – 60) = 0

(x + 50)(x – 60) = 0

By solving, we get

x = -50 and x = 60

So speed of the 60 km/hr.

Problem 4 :

The time taken by a train to travel a distance of 250 km was reduced by 5/4 hours when average speed was increased by 10km/hr. Calculate the average speed.

Solution :

Distance to be covered = 250 km.

Let x be the required average speed.

If the average speed was increased by 10 km/hr.

(x + 10) be the increased speed.

Let T1 be the time taken to cover the distance in the average speed of x km/hr.

Let T2 be the time taken to cover the distance in the average speed of (x + 10) km/hr

Time = Distance/Speed

T1 = 250/x

T2 = 250/(x + 10)

T1 – T2 = 5/4

250/x – 250/(x + 10) = 5/4

250[(x + 10 – x)/x(x + 10)] = 5/4

2500/(x2 + 10x) = 5/4

2500(4) = 5(x+ 10x)

10000 = 5x+ 10x

Now we are going to divide the whole equation by 5, so we get

x+ 10x = 2000

x+ 10x – 2000 = 0

x+ 50x - 40x - 2000 = 0

x(x + 50) – 40(x + 50) = 0

(x – 40)(x + 50) = 0

By solving, we get

x = 40 and x = -50

Therefore the required average speed = 40 km/hr.

Increased speed :

= (40 + 10)

= 50 km/hr

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