Question

A train was travelling a distance of 1200 km at a constant speed. When driver of the train
learnt that he was running late, he increased the speed by 20 km/h. The journey took 5 hours
less and reached the destination in time. Find the original speed of the train. ​

Answer 1

Answer:-

Given:

Distance travelled by a train = 1200 Km.

If he had increased the speed by 20 km/h , it would reach the destination 5 hrs earlier.

Let the actual Speed of the train be x km/h and let the time taken to travel 1200 km by the train be y hrs.

→ New speed of the train = (x + 20) km/h

→ Time taken by the train to reach the destination = (y - 5) hrs

We know that,

Distance = speed * time.

Hence,

→ x * y = 1200

→ y = 1200/x

(or)

→ (x + 20)(y - 5) = 1200

→ xy - 5x + 20y - 100 = 1200

Substitute y value here.

→ x(1200/x) - 5x + 20(1200/x) = 1200 + 100

→ 1200 - 5x + 24000/x = 1300

→ ( - 5x² + 24000)/x = 1300 - 1200

→ 24000 - 5x² = 100x

→ 5x² + 100x - 24000 = 0

→ 5(x² + 20x - 4800) = 0

→ x² + 80x - 60x - 4800 = 0

→ x(x + 80) - 6(x + 80) = 0

→ (x - 60)(x + 80) = 0

• x - 60 = 0

→ x = 60

• x + 80 = 0

→ x = - 80

Speed cannot be negative, So positive value is taken i.e., x = 60 km/h

Therefore, the original speed of the train is 60 km/h.

Answer 2

Answer:

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

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

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

⠀

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