Question

A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.​

Answer 1

$\huge{\sf{\red{Answer-}}}$

Speed of the train = 40 km/hr

$\rule{200}2$

Explanation-

Let us assume that the speed of train be x km/hr.

A train travels a distance of 480 km at a uniform speed.

$\small{\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:{\underline{As\:per\:given\:condition}}}}$

• Distance travelled by train = 480 km
• Assumed speed is x km/hr

We know that -

$\boxed{\sf{Time\:=\:\dfrac{Distance}{Speed}}}$

$\implies\:\sf{T\:=\:\dfrac{480}{x}}$ hr.

If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance.

$\small{\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:{\underline{As\:per\:given\:condition}}}}$

• Distance = 480 km
• New speed = (x - 8) km/hr

Let's denote the new time by T'

$\implies\:\sf{T'\:=\:\dfrac{480}{(x-8)}}$ hr.

ACCORDING TO QUESTION

$\huge{\boxed{T'-T\:=\:3}}$

Substitute the known values from above

$\implies\:\sf{\dfrac{480}{(x-8)}-\dfrac{480}{x}\:=\:3}$

Take 480 as common

$\implies\:\sf{480 \bigg [\dfrac{1}{(x-8)}-\dfrac{1}{x} \bigg]\:=\:3}$

Solve the denominator

$\implies\:\sf{480\bigg[\dfrac{x+8-x}{x(x-8)}\bigg]\:=\:3}$

$\implies\:\sf{\dfrac{480(8)}{x(x-8)}\:=\:3}$

Cross-multiply them

$\implies\:\sf{3840\:=\:3x^2-24x}$

$\implies\:\sf{x^2-8x-1280\:=\:0}$

By Quadratic Formula

In above equation, a = 1, b = -8 and c = -1280

x = $\sf{\dfrac{ - b \pm \sqrt{ {(b)}^{2} - 4ac} }{2a}}$

= $\sf{\dfrac{ - ( - 8) \pm \sqrt{ {( - 8)}^{2} - 4(1)( - 1280)} }{2(1)}}$

= $\sf{\dfrac{ 8 \pm \sqrt{ 64 + 5120} }{2}}$

= $\sf{\dfrac{ 8 \pm \sqrt{ 5184} }{2}}$

= $\sf{\dfrac{8\pm 72}{2}}$

= $\sf{\dfrac{8 + 72}{2}}$, $\sf{\dfrac{8- 72}{2}}$

(Negative one neglected, as speed can't be negative)

= $\sf{\dfrac{80}{2}\:=\:40}$

Therefore,

The speed of the train is 40 km/hr.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the usual speed of the train be x km/h.

Reduced speed = (x - 8) km/h

Total distance = 480 km

Time taken at usual speed = 480x h

Time taken at reduced speed = 480x - 8 h

According to Question,

⇒ 480x - 8 = 480x + 3

⇒ 480x - 8 - 480x = 3

⇒480x - 480x + 3840x × x - 8 = 3

⇒ 3840x² - 8x = 3

⇒ x² - 8x = 1280

⇒ x² - 8x - 1280 = 0

⇒ x² - 40x + 32x - 1280 = 0

⇒ x - 40 + 32x - 40 = 0

⇒ x - 40x + 32 = 0

⇒ x - 40 = 0 or x + 32 = 0

⇒ x = 40 or x = - 32 (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.