After covering a distance of 30Km with a uniform speed, there got some defect in train engine and therefore its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination 45 minutes late. If it had happened after covering 18Km of distance, the train would have reached 9 minutes earlier>Find the speed of the train and the distance of the journey.

This is my question. .

now after letting the original speed of train be $x Km/Hr$ and the time taken be y Hr, threfore distance = $xy$


speed = $x-4x/5$ Km/hr

time = $y + 45/60$

$xy=60xy/300 +45x/300 => +300xy-60xy = 45x => 240xy = 45x$ .............[i]

similarly in CASE II we get the equation:

$240xy = -9x$.........[ii]

but after solving these two equations the answer is coming to 0 which is wrong please tell me the correct solution.


(fast please)


You should have $$\frac{30}{x}+\frac{xy-30}{\frac{4}{5}x} = y+\frac{45}{60}$$ and (if I read the question correctly, with $18$km being in addition to the $30$km) $$\frac{48}{x}+\frac{xy-48}{\frac{4}{5}x} = y+\frac{36}{60}.$$ Multiply through by say $20x$ and simplify to get two simultaneous equations you can easily solve for $x$ and then $y$.


You can use simple equations to solve this:

Let's call the rate of the train $x$.

Now, during those $30\ \text{km}$, the train's rate remains the same.

After this, the train will go $\frac{4x}{5}$ regardless of anything.

However, if the train covered $18$ more km, it would've reached $9$ minutes earlier.

So the difference in time lies between that 18 km stretch.

We can now say the train with the early wreck had a time of $\frac{18}{4x/5} = \frac{45}{2x}$ and the train with the late wreck had a time of $\frac{18}{x}$.

Since these difference in these times is only $9$ minutes:

$$\frac{45}{2x} - \frac{18}{x} = 0.15$$

Solving we get that $x = 30$ km/hr

Use this to find your distance.


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