Question

After travelling for 30 minutes a train meets an accident, due to which it has to stop for 45 minutes. Due to the accident its speed is also reduced to 32​ of its former value and the train reaches its destination 1 hour 30 minutes late. Had the accident occurred 60 km later, the train would have reached 30 minutes earlier. The length of journey is
 $ (a)\;{\text{90 km}} $
 $ (b)\;12{\text{0 km}} $
 $ (c)\;150{\text{ km}} $
 $ (d)\;18{\text{0 km}} $

Answer 1

Hint: In the above given equation, first of all calculate the running times and then equate them with the respective times given in the question. You will reach at some equations, solve those simultaneous equations in $ d $ and $ v $ to obtain the solution.

Complete step-by-step answer:
Let the distance be $ d $ and speed be $ v $ .
Therefore, we get the normal running time as $ vd $ ​.
In the first case, running time is $ \left( {d + 1\dfrac{1}{2}} \right){\text{hr}} $ whereas in the second case, the running time is $ \left( {\dfrac{d}{v} + 1} \right)\,{\text{hr}} $ .

Let the distance be \[d\;{\text{km}}\]and speed of the train be $ v\;{\text{km/hr}} $ .
Then, the normal time to reach the destination comes out to be $ \dfrac{d}{v}{\text{ hr}} $ .
In the first case, due to late running by $ 1\dfrac{1}{2}{\text{ hr}} $ the total time $ {t_1} = \left( {\dfrac{d}{v} + \dfrac{3}{2}} \right){\text{ hr}} $
 $ \Rightarrow {t_1} = \dfrac{{2d + 3v}}{{2v}}{\text{ hr}} $ … (1)
Similarly, in the second case, the train is late by one hour.
\[\therefore {t_2} = \dfrac{d}{v} + 1\]
\[ \Rightarrow {t_2} = \dfrac{{d + v}}{v}{\text{ hr}}\] … (2)
First case: Now the train travels 30km with a speed $ v $ , stops for 45min, that is $ \dfrac{{45}}{{60}}{\text{ hr = }}\dfrac{3}{4}{\text{ hr}} $ for accident.
∴Total time taken upto the point of accident is $ \left( {\dfrac{{30}}{v} + \dfrac{3}{4}} \right){\text{hr}} $ … (a)
The rest of the distance is \[\left( {d - 30} \right){\text{ km}}\]and the train runs with a speed of $ \dfrac{2}{3}v $ .
∴Time taken to reach the destination is $ (d - 30) \times \dfrac{3}{{2v}}\;{\text{hr}} $
 $ = \dfrac{{3d - 90}}{{2v}}\;{\text{hr}} $ … (b)
Therefore, the total time taken by the train can be obtained by adding equation (a) and (b), $ \Rightarrow {t_1} = \left( {\dfrac{{30}}{v} + \dfrac{3}{4} + \dfrac{{3d - 90}}{v}} \right){\text{ hr}} $
Taking $ {t_1} $ ​ from equation (1) we have,
 $ \dfrac{{30}}{v} + \dfrac{3}{4} + \dfrac{{3d - 90}}{{2v}} = \dfrac{{2d + 3v}}{v} $
 $ 2d - 3v = 60 $ … (3)
Second case: The accident happens 60 km later.
That is, it runs (60+30) km = 90km with a speed of $ v{\text{ km/hr}} $ .
∴Time taken is $ \dfrac{{90}}{{v}}{\text{ hr}} $ .
Then it stops for $ \dfrac{3}{4}{\text{ hr}} $ .
∴Total time taken upto this point is $ \left( {\dfrac{{90}}{v} + \dfrac{3}{4}} \right)\;{\text{hr = }}\dfrac{{360 + 3v}}{{4v}}\;{\text{hr}} $ … (a)
After the accident the distance left $ = (d - 90)\;{\text{km}} $ and the speed= $ = \dfrac{{2v}}{3}{\text{ km/hr}} $ .
∴The total time taken to cover $ (d - 90)\;{\text{km}} $ is $ (d - 90) \times \dfrac{3}{{2v}}{\text{ hr}} $
 $ = \dfrac{{3d - 270}}{{2v}}{\text{ hr}} $ … (b)
 Therefore, the total time taken to reach the destination can be obtained by adding equation (a) and (b), $ \Rightarrow {t_2} = \left[ {\dfrac{{360 + 3v}}{{4v}} + \dfrac{{3d - 270}}{{2v}}} \right]{\text{hr}} $ .
Replacing $ {t_2} $ from equation 2, we get,
 $ \left[ {\dfrac{{360 + 3v}}{{4v}} + \dfrac{{3d - 270}}{{2v}}} \right] = \dfrac{{d + v}}{v} $
 $ \Rightarrow 2d - v = 180 $ … (4)
Multiplying equation (4) with 3 and subtracting from equation (3) we get $ 4d = 480 $
 $ \Rightarrow d = 120\;{\text{km}} $
So, the correct solution is the option $ (b) $ .

Note: Simplify the simultaneous equations carefully and do not mingle the conditions provided for both the cases. While formulating the equations, make sure you consider the late running time when needed. Also, the unit system so chosen must also be considered carefully.