Question

A train travels 360 km with uniform speed.The speed of the train is increased by 5 km/hr,
it takes 48 minutes less to cover the same distance. Find the initial speed of the train.

Answer 1

Initial speed of the train = 45km/h

Step-by-step explanation:

Let the speed of the train be x km/h

and let the time be y.

Speed = Distance / time

In first statement,

Speed = 360/y km/h = x km/h

In the second statement,

Speed = 360/y -0.8 km/h [ 48 mins = 4/5 hrs = 0.8hrs]

= (x + 5) km/h

Rewriting the equation, we have

360/y = x ------> Equation (i)

&

360/ y - 0.8 = x + 5. ---------> Equation (ii)

Substituting equation (i) in (ii), we have

$\frac{360}{y - 0.8} = \frac{360}{y} + 5$

$\frac{360}{y - 0.8}= \frac{360 + 5y}{y}$

By Cross multiplication, we get

360y = (360 + 5y)(y - 0.8)

360y = (360y - 288 + 5y² - 4y)

360y - 360y = 5y² - 4y - 288

=> 5y² - 4y - 288 = 0

Solving the equation

y = [-b + or - √(b² -4×a×c) ] ÷ 2 × a

= [ -(-4) + or - √(-4² - 4×3×-288) ] ÷ 2 × 5

= [ 4 + or - √(16 + 5760) ] ÷ 10

= [ 4 + or - √5776 ] ÷ 10

= [ 4 + or - 76 ] ÷ 10

Therefore,

y = (4 + 76) /10 or (4 - 76) /10

= 80/10 or -72/10

Since time cannot be negative,

We assume time taken = 80/10 hrs = 8 hrs

Therefore, initial speed, x = 360/8 = 45km/h

I Hope You Find It Helpful!