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# A train, travelling at a uniform speed for $ 360 \mathrm{~km} $, would have taken 48 minutes less to travel the same distance if its speed were $ 5 \mathrm{~km} / \mathrm{h} $ more. Find the original speed of the train.

Given:

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more.

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 360 km at original speed$=\frac{360}{x}$ hours

Time taken by the train to travel 360 km when the speed is 5 km/hr more than the original speed$=\frac{360}{x+5}$ hours

$48$ minutes in hours$=\frac{48}{60}$ hours. (since 1 hour = 60 minutes)

According to the question,

$\frac{360}{x}-\frac{360}{x+5}=\frac{48}{60}$

$\frac{360(x+5)-360(x)}{(x)(x+5)}=\frac{12\times4}{12\times5}$

$\frac{360(x+5-x)}{x^2+5x}=\frac{4}{5}$

$5(360)(5)=4(x^2+5x)$ (On cross multiplication)

$25(90)=x^2+5x$

$x^2+5x-2250=0$

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

$x^2+50x-45x-2250=0$

$x(x+50)-45(x+50)=0$

$(x+50)(x-45)=0$

$x+50=0$ or $x-45=0$

$x=-50$ or $x=45$

Speed cannot be negative. Therefore, the value of $x$ is $45$ km/hr.

The original speed of the train is $45$ km/hr.

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