Question

a train travels at a certain speed of a distance of 63km and then travels a distance of 72km at an average speed 6km/h more than its original speed. it takes 3 hours to complete total journey. what is the original average

Answer 1

As per my assumptions and calculations, original average speed is 42km/h

Answer 2

Answer:

Original speed of train = 42 km/h

Step-by-step explanation:

Given : A train travels at a certain speed of a distance of 63 km and then travels a distance of 72 km at an average speed 6 km/h more than its original speed. It takes 3 hours to complete total journey.

To find : What is the original average?

Solution :

Let the original speed of train is x km/h

Relation between distance, speed and time is

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

Time taken to cover 63 km with speed x km/h,

$T_1=\frac{63}{x}$

Now, speed of train is (x + 6) km/h

Time taken to cover 72 km with speed (x + 6) km/h,

$T_2=\frac{72}{x+6}$

According to question,

$T_1+T_2=3$

Substitute the value,

$\frac{63}{x}+\frac{72}{x+6}=3$

$\frac{63(x+6)+72x}{x(x+6)}=3$

$\frac{63x+378+72x}{x^2+6x}=3$

$135x+378=3(x^2+6x)$

$45x+126=x^2+6x$

$x^2-39x-126=0$

$x^2- 42x + 3x - 126 = 0$

$x(x - 42) + 3(x - 42) = 0$

$(x + 3)(x - 42) = 0$

∴ x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative

Hence, original speed of train = 42 km/h