Q. 235.0( 4 Votes )

A train takes 2 h

Answer :

Let the speed of the train be x km/hr


The total journey distance, d = 300km….(i)


Case 1: with usual speed


usual speed of the train, s = x km/hr


We know Distance = speed × time


So usual time,



Case 2: with increased speed


new speed of the train, s1 = (x + 5) km/hr


We know Distance = speed × time


So increases time,



Now it is given the train takes 2 hours less after an increase in speed.


So increased time = original time – 2 hours


i.e., t1 = t – 2


Substituting values from equation (ii) and (iii), we get




300x = (x + 5) (300 – 2x)


300x = x(300 – 2x) + 5(300 – 2x)


Canceling the like terms, we get,


300x = 300x – 2x2 + 1500 – 10x


2x2 + 10x – 1500 = 0


Dividing throughout by 2, we get


x2 + 5x – 750 = 0


Now we will be factorizing this by splitting the middle term, we get


x2 + 30x – 25x – 750 = 0


x (x + 30) – 25(x + 30) = 0


(x + 30) (x – 25) = 0


(x + 30) = 0 or (x – 25) = 0


x = – 30 or x = 25


But x is the speed of the train and it cannot be negative, hence the usual speed of the train is 25km/hr.


OR






Canceling the like terms, we get



– ab = x (a + b + x)


– ab = ax + bx + x2


x2 + (a + b) x + ab = 0


Comparing this with standard equation, i.e., Ax2 + Bx + C = 0, we get


A = 1, B = (a + b), C = ab


So, the value of x of a quadratic equation can be found by using the formula,



Now substituting the corresponding values, we get



But we know the expansion of (a + b)2 = a2 + b2 + 2ab, substituting this in above equation, we get




But we know a2 + b2 – 2ab = (a – b)2, substituting this in above equation, we get




Now, this has two possibilities,





⇒ x = – b or x = – a


These are the required values of x.

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