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Question

A train, travelling at a uniform speed for 360km, would have taken 48 minutes less to travel the same distance if its speed were 5km/h more. Find the original speed of the train.

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Solution

Step 1: Construct the equation based on given condition.
let the original speed of the train =x km/h.
Then the increased speed of the train
=(x+5)km/h.
and distance =360 km.
According to the question,
360x360x+5=45
(speed=distancetime)

[48minutes=4860hours=45hours]
360(x+5)360xx(x+5)=45
360x+1800360xx(x+5)=45
1800x2+5x=45

1800×54=x2+5x
x2+5x=1800×54=2250
x2+5x2250=0.

Step 2: Find the roots using factorization.
By splitting the middle term, we get :
x2+(50x45x)2250=0
x2+50x45x2250=0
x(x+50)45(x+50)=0
x(x+50)(x+45)=0
x+50=0 or x45=0
x50 because speed can not be negative
So, x45=0
x=45
Hence, the original speed of the train is 45km/h.

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