Question

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 kilometres more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey
A. Original speed = 10 Km/hr, Length of the journey = 90 Km
B. Original speed = 15 Km/hr, Length of the journey = 100 Km
C. Original speed = 25 Km/hr, Length of the journey = 110 Km
D. Original speed = 30 Km/hr, Length of the journey = 120 Km

Answer 1

Hint: First we need to assume the unknown values as variables. Then use the speed equation and the given scenarios to find the equations of the unknown. Solve these equations to get the desired answer.

Complete step-by-step answer:
Given the problem, a train is completing its journey according to two scenarios.
In the first scenario, 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 and it reaches its destination 45 minutes late.
In the second scenario, after covering a distance of 18 km in addition to the one in first case 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 and it reaches its destination 9 minutes earlier than the previous case.
Let us assume the speed of the train to be $x$ km/hr.
Let the total distance to be covered in the journey be $y$ Km.
We know that,
$
  {\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}{\text{ (1)}} \\
   \Rightarrow x = \dfrac{y}{{{\text{time}}}} \\
   \Rightarrow {\text{time}} = \dfrac{y}{x} \\
$
Given the problem in the first scenario, the train covers the first 30 Kms with its original speed.
$ \Rightarrow $Speed for first 30 Kms $ = x$Km/hr
After that for the remaining journey, its speed is reduced to 4/5 of its original speed.
$ \Rightarrow $Speed of the train for the remaining $y - 30$ Kms $ = \dfrac{{4x}}{5}$Km/hr
Using equation (1) in the above, we get
Time taken by train to cover 30 Kms $ = \dfrac{{30}}{x}$ hours.
Similarly, time taken by train to cover remaining $y - 30$ Kms $ = \dfrac{{y - 30}}{{\dfrac{4}{5}x}} = \dfrac{{5\left( {y - 30} \right)}}{{4x}}$ hours.
It is given in the problem that the train reaches its destination 45 minutes late during the above scenario.
$ \Rightarrow $Total time taken for the journey during above scenario = Total time taken in constant average speed + 45 minutes
We know that one minute $ = \dfrac{1}{{60}}$hour.
Therefore, 45 minutes $ = \dfrac{{45}}{{60}}$hour.
Using this result in above (as all the time units are in hours), we get
$ \Rightarrow $$
  \dfrac{{30}}{x} + \dfrac{{5\left( {y - 30} \right)}}{{4x}} = \dfrac{y}{x} + \dfrac{{45}}{{60}} \\
   \Rightarrow \dfrac{{120 + 5y - 150}}{{4x}} = \dfrac{{60y + 45x}}{{60x}} \\
   \Rightarrow 5y - 30 = 4y + 3x \\
   \Rightarrow y - 3x = 30{\text{ (2)}} \\
$
Now considering the second scenario.
The speed remains constant for 18 Kms more than the previous case that is $30 + 18 = 48$Kms.
$ \Rightarrow $Speed for first 48 Kms $ = x$Km/hr
Similarly like in the first scenario, the speed of the train reduces to 4/5 of its original speed in the remaining journey.
$ \Rightarrow $Speed of the train for the remaining $y - 48$ Kms $ = \dfrac{{4x}}{5}$Km/hr
Using equation (1) in the above, we get
Time taken by train to cover 48 Kms $ = \dfrac{{48}}{x}$ hours.
Similarly, time taken by train to cover remaining $y - 48$ Kms $ = \dfrac{{y - 48}}{{\dfrac{4}{5}x}} = \dfrac{{5\left( {y - 48} \right)}}{{4x}}$ hours.
It is given the problem that during the second scenario, the train reaches its destination 9 minutes earlier than the first scenario.
The train was delayed by 45 minutes in the first case.
Hence the train gets delayed by $45 - 9 = 36$minutes during the second scenario.
$ \Rightarrow $Total time taken for the journey during above scenario = Total time taken in constant average speed +36 minutes
We know that one minute $ = \dfrac{1}{{60}}$hour.
Therefore, 36 minutes $ = \dfrac{{36}}{{60}}$hour.
Using this result in above (as all the time units are in hours), we get
$
  \dfrac{{48}}{x} + \dfrac{{5\left( {y - 48} \right)}}{{4x}} = \dfrac{y}{x} + \dfrac{{36}}{{60}} \\
   \Rightarrow \dfrac{{192 + 5y - 240}}{{4x}} = \dfrac{{60y + 36x}}{{60x}} \\
   \Rightarrow 5y - 48 = 4y + \dfrac{{12}}{5}x \\
   \Rightarrow 25y - 240 = 20y + 12x \\
   \Rightarrow - 12x + 5y = 240{\text{ (3)}} \\
$
Hence, we have two linear equations obtained which we need to solve,
\[
  y - 3x = 30{\text{ (2)}} \\
   - 12x + 5y = 240{\text{ (3)}} \\
\]
Multiplying equation (2) by 4 and then subtracting equation (3) form equation (2), we get
$
  4y - 12x + 12x - 5y = 120 - 240 \\
   \Rightarrow - y = - 120 \\
   \Rightarrow y = 120 \\
$
Substituting $y = 120$ in equation (2), we get
\[
   \Rightarrow 120 - 3x = 30 \\
   \Rightarrow x = 30 \\
\]
Hence, the original speed of the train comes out to be $x = 30$Km/hr and the total length of the journey is equal to $y = 120$Km.
Therefore option (D). Original speed = 30 Km/hr, Length of the journey = 120 Km is the correct answer.

Note: In questions like above the text of the problem needs to be read and analysed carefully in order to assume the correct variables and form their equations. In order to find a number of unknown values, we need the same amount of distinct equations involving those unknown variables. Linear equations obtained could be solved by any one of the methods. Speed equation is needed to be kept in mind in problems like the above.