Question

distance between Akola and Bhusawal 168 km an Express train takes 1 hour less than a passenger train to cover the distance find the average speed of each train if the average speed of the express train is more the more by 14 km per hour then speed of the passenger train

Answer 1

Answer:

12.49km/h

Step-by-step explanation:

express speed =14km/h

          distance =168km

         taken time= 12hour

passenger train time = 13hour

       speed of pssnger train=12.92km/h

Answer 2

Answer:

The average speed of passenger train = 42 km/hr

The average speed of express train = 56 km/hr

Step-by-step explanation:

Given,

The distance between Akola and Bhusawal is = 168 km

Consider the average speed of passenger train = (x) km/hr

Consider the average speed of express train = (x+14) km/hr

we know that

$Speed=\frac{Distance}{Time}\\\\\therefore{Time}=\frac{Distance}{Speed}$

The time take by the passenger train to reach the distance = $\frac{168}{x} \text{ hr}$

The time take by the express train to reach the distance = $\frac{168}{x+14} \text{ hr}$

As per problem $\frac{168}{x}-\frac{168}{x+14}=1\\\\\Rightarrow\frac{168(x+14)-168x}{x(x+14)}=1\\\\\Rightarrow\frac{168x+2352-168x}{x(x+14)}=1\\\\\Rightarrow\frac{2352}{x(x+14)}=1\\\\\Rightarrow{x}^{2}+14x=2352\\\\\Rightarrow{x}^{2}+14x-2352=0$ ..................equation 1

For Quadratic Equation $ax^{2}+bx+c=0 \text{ [where x is the variable and a, b and c are known values]}$

the value of $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$b^{2}-4ac \text{ is called Discriminant (D)}$

when Discriminant (D) is positive, we get two Real solutions  for x

when Discriminant (D) is zero we get just ONE real solution (both answers are the same)

when Discriminant (D) is negative we get a pair of Complex solutions

In equation 1 the Discriminant (D) is = $b^{2}-4ac= 14^{2}-4\times1\times(-2352)=196+9408=9604$

As the Discriminant (D) is positive, we get two Real solutions  for x

therefore the value of x is either $x=\frac{-b+\sqrt{D}}{2a}\text{or}\text{ } x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{-b+\sqrt{D}}{2a}=\frac{-14+\sqrt{9604}}{2}=\frac{-14+98}{2}=\frac{84}{2}=42$

or, $x=\frac{-b-\sqrt{D}}{2a}=\frac{-14-\sqrt{9604}}{2}=\frac{-14-98}{2}=\frac{-112}{2}=-56$

As speed is not consider as a negative value so we take the positive value of x=42

therefore,

The average speed of passenger train = 42 km/hr

The average speed of express train = (42+14)=56 km/hr