Question

A train met with an accident 60 km away from Anantpur station. It completed the remaining journey at$\dfrac{{{5^{{\text{th}}}}}}{6}$ of the previous speed and reached Barmula station 1 hour 12 minutes late. Had the accident taken place 60 km further, it would have been only 1 hour late.
- What is the normal speed of the train?
- What is the distance between Anantpur and Barmula?
$
  {\text{A}}{\text{. 60 km/hr, 420 km}} \\
  {\text{B}}{\text{. 80 km/hr, 800 km}} \\
  {\text{C}}{\text{. 8 km/hr, 80 km}} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint: We consider the normal speed to be a variable ‘x’ and the distance between the places be ‘y’. Using the relative distance and the speed given in each case we establish a relation between speed and the actual distance using the formula of speed.

Complete step by step answer:
Given Data,
First case: Accident at 60 km away and the speed after accident is$\dfrac{{{5^{{\text{th}}}}}}{6}$of the original and reached 1hour 12 minutes late.
Second case: Accident at 6o km further, i.e. 120 km away and reached 1 hour late.
Let the normal speed of the train be ‘x’ and the original distance between anantpur and barmula is ‘y’.
We know the speed of a body is given by,
${\text{speed = }}\dfrac{{{\text{distance travelled}}}}{{{\text{time taken}}}}$.
Using this formula, the time taken by the train to reach barmula from anantpur is$\dfrac{{\text{y}}}{{\text{x}}}$.
First Case –
Accident at 60 km away and the speed after accident is$\dfrac{{{5^{{\text{th}}}}}}{6}$of the original and reached 1hour 12 minutes late.
Time taken till the accident is given by,${\text{t = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}} = \dfrac{{60}}{{\text{x}}}$
Accident happens 60 km away from anantpur station, so the remaining distance = y – 60.
Also given the speed of train after accident is$\dfrac{{{5^{{\text{th}}}}}}{6}$of the actual i.e.$\left( {\dfrac{5}{6} \times {\text{x}}} \right)$
Therefore the time taken to cover the remaining distance = $\dfrac{{{\text{remaining distance}}}}{{{\text{speed after accident}}}}$
Time taken to cover the remaining = ${\text{6 }} \times {\text{ }}\dfrac{{\left( {{\text{y - 60}}} \right)}}{{{\text{5x}}}}$
Given, delay is 1 hour 12 minutes.
We know one hour = 60 minutes.
Therefore delay = 1.2 hour
Now According to the question, we have
The sum of times taken before and after accident = time taken to cover the distance normally + delay
$
   \Rightarrow \dfrac{{60}}{{\text{x}}} + 6 \times \dfrac{{\left( {{\text{y - 60}}} \right)}}{{{\text{5x}}}} = \dfrac{{\text{y}}}{{\text{x}}} + 1.2 \\
   \Rightarrow \dfrac{{(60)}}{{\text{x}}} + \dfrac{{\left( {{\text{6y - 360}}} \right)}}{{{\text{5x}}}} = \dfrac{{{\text{y + 1}}{\text{.2x}}}}{{\text{x}}} \\
   \Rightarrow \dfrac{{5(60) + \left( {{\text{6y - 360}}} \right)}}{{{\text{5x}}}} = \dfrac{{{\text{y + 1}}{\text{.2x}}}}{{\text{x}}} \\
   \Rightarrow \dfrac{{{\text{6y - 60}}}}{{\text{x}}} = \dfrac{{{\text{5(y + 1}}{\text{.2x)}}}}{{\text{x}}} \\
   \Rightarrow {\text{y - 6x = 60 - - - - - - - }}\left( 1 \right) \\
$
Second Case –
Accident at 60 km further, i.e. 120 km away and reached 1 hour late.
Time taken till the accident is given by,${\text{t = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}} = \dfrac{{120}}{{\text{x}}}$
Accident happens 60 km further away from the initial, i.e. 120 km from anantpur station, so the remaining distance = y – 120.
Also given the speed of train after accident is$\dfrac{{{5^{{\text{th}}}}}}{6}$of the actual i.e.$\left( {\dfrac{5}{6} \times {\text{x}}} \right)$
Therefore the time taken to cover the remaining distance = $\dfrac{{{\text{remaining distance}}}}{{{\text{speed after accident}}}}$
Time taken to cover the remaining = ${\text{6 }} \times {\text{ }}\dfrac{{\left( {{\text{y - 120}}} \right)}}{{{\text{5x}}}}$
Here the delay = 1 hour
Now According to the question, we have
The sum of times taken before and after accident = time taken to cover the distance normally + delay
$
   \Rightarrow \dfrac{{120}}{{\text{x}}} + 6 \times \dfrac{{\left( {{\text{y - 120}}} \right)}}{{{\text{5x}}}} = \dfrac{{\text{y}}}{{\text{x}}} + 1 \\
   \Rightarrow \dfrac{{120}}{{\text{x}}} + \dfrac{{\left( {{\text{6y - 720}}} \right)}}{{{\text{5x}}}} = \dfrac{{\text{y}}}{{\text{x}}} + 1 \\
   \Rightarrow \dfrac{{600 + \left( {{\text{6y - 720}}} \right)}}{{{\text{5x}}}} = \dfrac{{{\text{y + x}}}}{{\text{x}}} \\
   \Rightarrow \dfrac{{600 + \left( {{\text{6y - 720}}} \right)}}{{\text{x}}} = \dfrac{{{\text{5(y + x)}}}}{{\text{x}}} \\
   \Rightarrow \dfrac{{\left( {{\text{6y - 120}}} \right)}}{{\text{x}}} = \dfrac{{{\text{5(y + x)}}}}{{\text{x}}} \\
   \Rightarrow {\text{y - 5x = 120 - - - - - - - }}\left( 2 \right) \\
$
Now we solve the equations (1) and (2), to solve for the values of ‘x’ and ‘y’
y – 6x = 60 and y – 5x = 120.
We get, y = 120 + 5x
Substituting this is the first equation we get,
120 + 5x – 6x = 60
⟹x = 60
Putting this in the above, we get
⟹y – 5(60) = 120
⟹y = 420.
We took x as the speed of the train and y as the distance between the two stations.
Therefore, the normal speed of train is 60 Km/hr and the distance between anantpur and barmula is 420 km.

So, the correct answer is “Option A”.

Note: In order to solve this type of questions the key is to know the formula of speed of a body and time taken for the body to travel a distance at a uniform speed.
Considering the speed and distance as two variables, establishing a relation between the variables in each case with the given data is the way to deal with such problems. We solve the two relations for the unknowns.
We have to be very careful while taking in the data from the question as it can be very tricky, all the quantities are almost similar and interdependent in the problems.