Question

a train traveling a distance of 1200km at a constant speed.when driver of the train learnt that he is getting late ,he increased the speed by 5km per hour .now the journey took 8 hours less and reached in time .find the original speed of the train

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Answer:

Let the Original Speed be x km/hr and New Speed be (x + 20) km/hr.

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Original\:Time - New\:Time=5\:hours\\\\\\:\implies\sf \dfrac{Distance}{Original\:Speed} - \dfrac{Distance}{New\:Speed}=5\\\\\\:\implies\sf \dfrac{1200}{x} - \dfrac{1200}{x + 20} = 5\\\\\\:\implies\sf 1200\bigg\lgroup\dfrac{1}{x} - \dfrac{1}{x + 20}\bigg\rgroup = 5\\\\\\:\implies\sf \dfrac{x + 20 - x}{x(x + 20)} = \dfrac{5}{1200}\\\\\\:\implies\sf \dfrac{20}{x^2 + 20x} = \dfrac{5}{1200}\\\\\\:\implies\sf \dfrac{4}{x^2 + 20x} = \dfrac{1}{1200}\\\\\\:\implies\sf 4 \times 1200= x^2 + 20x\\\\\\:\implies\sf x^2 + 20x - 4800 = 0\\\\\\:\implies\sf x^2 + (80 - 60)x - 4800 = 0\\\\\\:\implies\sf x^2 + 80x - 60x - 4800 = 0\\\\\\:\implies\sf x(x + 80) - 60(x + 80) = 0\\\\\\:\implies\sf (x - 60)(x + 80) = 0\\\\\\:\implies\sf x = 60 \quad or \quad x = - \:80$

⠀

$\therefore\:\underline{\textsf{Ignoring Negative, Original Speed is \textbf{60 km/hr}}}.$