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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