Question

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 per hour more than its original speed its takes 3 hours to complete total journey what is the original averages

Answer 1

Let the original speed of train is x km/h
time taken to cover 63 km with speed x km/h, T₁ = distance/time = 63/x hours

Again, question said , speed of train now (x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h , T₂= 72/(x + 6) km/h

A/C to question ,
T₁ + T₂ = 3 hours
⇒63/x + 72/(x + 6) = 3
⇒ 21/x + 24/(x + 6) = 1
⇒ (21x + 126 + 24x) = x(x + 6)
⇒45x + 126 = x² + 6x
⇒ x² - 39x - 126 = 0
⇒x² - 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

Answer 2

Let the speed of the train is Xkm/h
Time taken to cover 63km with X km/h speed is :
T1= Distance/Speed
T1=63/X hours---------(1)

It's given that train Travels a distance of 72 km at an average speed 6 km per hour more than its original speed.

Time taken to cover 72km with speed (x+6)Km/H

T2=72/x+6 hr------------(2)

But given that total time is 3hrs.
T1+T2=3hrs

63/x+72(/x+6)=3
taking 3 common,

21/x+24/(x+6)=121

(x+6)+24x=x(x+6)

21x+126+24x=x²+6x

x²+6x-45x-126=0

x²-39x-126=0

x²-42x+3x-126=0

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

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

x=42, -3

since speed cannot be negative:
x=42Km/h

∴Original speed of train=42km/h