Represent the following situations in the form of quadratic equations: A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Given:

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

To do:

We have to find the original speed of the train.

Solution:

Let the original speed of the train be $x$ km/hr.

This implies,

Time taken by the train to travel 480 km at original speed$=\frac{480}{x}$ hours

Time taken by the train to travel 480 km when the speed is 8 km/hr less than the original speed$=\frac{480}{x-8}$ hours

According to the question,

$\frac{480}{x-8}-\frac{480}{x}=3$

$480(\frac{(x)-(x-8)}{(x)(x-8)})=3$

$480(\frac{8}{x^2-8x})=3$

$(480)(8)=3(x^2-8x)$ (On cross multiplication)

$160(8)=x^2-8x$

$x^2-8x-1280=0$

Solving for $x$ by factorization method, we get,

$x^2-40x+32x-1280=0$

$x(x-40)+32(x-40)=0$

$(x-40)(x+32)=0$

$x-40=0$ or $x+32=0$

$x=40$ or $x=-32$

Speed cannot be negative. Therefore, the value of $x$ is $40$ km/hr.

The original speed of the train is $40$ km/hr.

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