After covering a distance of $30$ km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to $\frac{4}{5}$ of its original speed. Consequently, the train reaches its destination late by $45$ minutes. Had it happened after covering $18$ kilometres more, the train would have reached $9$ minutes earlier. Find the speed of the train and the distance of journey.

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Solution

Let the original speed of the train be $x$ km/hr and the length of the journey be $y$ km. Then,
Time taken $= \frac{y}{x}$ hrs

Case I : When the defect in the engine occurs after covering a distance of $30$ km.
We have,
Speed for first $30$ km $= x$ km/hr
and, Speed for the remaining $(y - 30)$ km $= \frac{4}{5} x$ km/hr

$∴$ Time taken to cover $30$ km $= \frac{30}{x}$ hrs
Time taken to cover $(y - 30)$ km $= \frac{y - 30}{\frac{4 x}{5}}$ hrs $= \frac{5}{4 x} (y - 30)$ hrs
According to the given condition, we have,
$\frac{30}{x} + \frac{5}{4 x} (y - 30) = \frac{y}{x} + \frac{45}{60}$
$\frac{30}{x} + \frac{5 y - 150}{4 x} = \frac{y}{x} + \frac{3}{4}$
$\Rightarrow \frac{120 + 5 y - 150}{4 x} = \frac{4 y + 3 x}{4 x}$
$\Rightarrow 5 y - 30 = 4 y + 3 x$
$\Rightarrow 3 x - y + 30 = 0$ ...... (i)

Case II : When the defect in the engine occurs after covering a distance of $48$ km
Speed for first $48$ km $= x$ km/hr
Speed for the remaining $(y - 48)$ km $= \frac{4 x}{5}$ km/hr
$∴$ Time taken to cover $48$ km $= \frac{48}{x}$ hrs
Time taken to cover $(y - 48)$ km $= ⎛ ⎜ ⎜ ⎜ ⎝ \frac{y - 48}{\frac{4 x}{5}} ⎞ ⎟ ⎟ ⎟ ⎠ = {\frac{5 (y - 48)}{4 x}}$ hr
According to the given condition, the train now reaches $9$ minutes earlier i.e., $36$ minutes later.
$\frac{48}{x} + \frac{5 (y - 48)}{4 x} = \frac{y}{x} + \frac{36}{60}$
$\Rightarrow \frac{48}{x} + \frac{5 y - 240}{4 x} = \frac{y}{x} + \frac{3}{5}$
$\Rightarrow \frac{192 + 5 y - 240}{4 x} = \frac{5 y + 3 x}{5 x}$
$\Rightarrow \frac{5 y - 48}{4} = \frac{5 y + 3 x}{5}$
$\Rightarrow 25 y - 240 = 20 y + 12 x$
$\Rightarrow 12 x - 5 y + 240 = 0$ (ii)
Thus, we have the following system of simultaneous equations:
$3 x - y + 30 = 0$
$12 x - 5 y + 240 = 0$
By using cross-multiplication, we have
$\frac{x}{- 240 + 150} = \frac{- y}{720 - 360} = \frac{1}{- 15 + 12}$
$\Rightarrow \frac{x}{- 90} = \frac{- y}{360} = \frac{1}{- 3}$
$\Rightarrow x = \frac{- 90}{- 3} = 30$ and $y = \frac{- 360}{- 3} = 120$
Hence, the original speed of the train is $30$ km/hr and the length of the journey is $120$ km.

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