A train takes 2 hours less for a journey of 300km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train.

OR

Solve for x:

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 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)

Cancelling 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

Cancelling 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.