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 4/5 of its original speed. Consequently, the train reached its destination late by 45 minutes, Had it happened after covering 18 kilometers more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey

Original speed = 10 km/hr, Length of the journey = 90km

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Original speed = 15 km/hr, Length of the journey = 100km

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Original speed = 25 km/hr, Length of the journey = 110km

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Original speed = 30 km/hr, Length of the journey = 120km

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Solution

The correct option is D Original speed = 30 km/hr, Length of the journey = 120km
Let the speed of the train $= x$ km\hr
Length of the journey $= y$ km.
Time taken $= \frac{x}{y}$
According to the first condition.
Speed for the first $30$ km = $x$ km/hr
speed for the remaining $(y - 30)$ km = $\frac{4}{5}$ km/hr
$∴$ time taken to cover $30$ km = $\frac{30}{x}$
time taken to cover $(y - 30) = \frac{y - 30}{\frac{4 x}{5}} = \frac{5}{4 x} (y - 30)$
According to the problem
$\Rightarrow \frac{30}{x} + \frac{5}{4 x} (y - 30) = \frac{y}{x} + \frac{45}{60}$

$\Rightarrow \frac{30}{x} + \frac{5 y - 150}{4 x} = \frac{y}{x} + \frac{3}{4}$

$\Rightarrow 120 + 5 y - 150 = 4 y + 3 x$
$\Rightarrow y - 3 x = 30 \Leftrightarrow 3 x - y = - 30$ ....(1)

According to the second condition.
Speed for the first $48$ km = $x$ km/hr
speed for the remaining $(y - 48)$ km = $\frac{4}{5}$ km/hr
$∴$ time taken to cover $48$ km = $\frac{48}{x}$
time taken to cover $(y - 48) = \frac{y - 48}{4 x / 5} = \frac{5}{4 x} (y - 48)$
According to the problem
$\Rightarrow \frac{48}{x} + \frac{5}{4 x} (y - 48) = \frac{y}{x} + \frac{36}{60}$

$\Rightarrow \frac{30}{x} + \frac{5 y - 150}{4 x} = \frac{y}{x} + \frac{3}{5}$

$\Rightarrow \frac{192 + 5 y - 240}{4 x} = \frac{5 y + 3 x}{5 x}$

$\Rightarrow \frac{5 x - 48}{4 x} = \frac{5 y + 3 x}{5 x}$

$\Rightarrow 25 y - 240 = 20 y + 12 x$
$\Rightarrow - 12 x + 5 y = 240 \Leftrightarrow 12 x - 5 y = - 240$ ........(2)

multiply eq 1 by 5
$\Rightarrow 15 x - 5 y = - 150$ ........(3)

Substract eq2 and eq3
$\Rightarrow 3 x = 90 \Rightarrow x = 30$

put $x = 30$ in eq1
$\Rightarrow 3 \times 30 - y = - 30$
$\Rightarrow y = 120$

Let the speed of the train $= 30$ km/hr
Length of the journey $= 120$ km.

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The train was moving with uniform speed after covering a distance of 30km some defect develop with the engine of the train and for this reason its speed reduced to 4 by 5 of its original speed. The train reaches is destination late by 45 minutes in case the defects had happened after covering 18 km more the train would have reached just 36 minute late find the speed of the train at the start and the distance of Germany

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