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.
T1 = 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)/(x2 - 10x) = 1/200
(x - x + 10)/(x2 - 10x) = 1/200
2000 = x2 - 10x
x2 - 10x - 2000 = 0
x2 – 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)/x2 – 20x = 4/5
3840(5) = 4(x2 – 20x)
19200 = 4x2 – 80x
4800 = x2 – 20x
x2 – 20x – 4800 = 0
x2 – 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 T2 be 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
x2 – 10x – 3000 = 0
x2 – 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(x2 + 10x)
10000 = 5x2 + 10x
Now we are going to divide the whole equation by 5, so we get
x2 + 10x = 2000
x2 + 10x – 2000 = 0
x2 + 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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