Question

Ramesh travels 760km to his home party by train and partly by car. He takes 8 hours if he travels 160km by train and the rest by car. He takes 12 minutes more if he travels 240km by train and the rest by car. Find the speed of the train and car respectively.

Answer 1

Hint:
 Assume ‘u’ and ‘v’ as the speed of train and car respectively. Apply the formula: - Time = Distance / Speed for car and train separately and add these times and equate with the given total time in the question. Form two linear equations in ‘u’ and ‘v’ and solve them to get their values. To convert the time given in minutes into hours, use the conversion: - 60 minutes = 1 hour.

Complete step by step answer:
Here, we have been provided with the total time taken by Ramesh to cover a distance of 760km partly by train and partly by car. We have to determine the speeds of the train and car.
Now, let us assume ‘u’ and ‘v’ as the speed of train and car respectively. Let us check the two situations one – by – one.
(1) In the first condition it is given that Ramesh travels 160km by train and the rest by car. It takes 8 hours for him to cover the distance of 760km. So, we have,
\[\Rightarrow \] Distance travelled by the train = 160km
\[\Rightarrow \] Distance travelled by the car = 760 – 160 = 600km
Therefore, applying the formula: - Time = Distance / Speed, we have,
\[\Rightarrow \] Time taken by the train = \[\dfrac{160}{u}\]
\[\Rightarrow \] Time taken by the car = \[\dfrac{600}{v}\]
\[\because \] Total time taken = 8 hours
\[\Rightarrow \] Time taken by (train + car) = 8 hours
\[\Rightarrow \dfrac{160}{u}+\dfrac{600}{v}=8\]
\[\Rightarrow \dfrac{20}{u}+\dfrac{75}{v}=1\] - (1)
(2) In the second condition it is given that Ramesh travels 240km by train and the rest by car. It takes 8 hours 12 minutes for him to cover the same distance of 760km.
So, we have,
\[\Rightarrow \] Distance travelled by the train = 240km
\[\Rightarrow \] Distance travelled by the car = 760 – 240 = 520km
Again, applying the formula: - Time = Distance / Speed, we have,
\[\Rightarrow \] Time taken by the train = \[\dfrac{240}{u}\]
\[\Rightarrow \] Time taken by the car = \[\dfrac{520}{v}\]
\[\because \] Total time taken = 8 hours 12 minutes
Converting minutes into hours, we have,
\[\because \] 60 minutes = 1 hour
\[\Rightarrow \] 1 minute = \[\dfrac{1}{60}\] hour
\[\Rightarrow \] 12 minutes = \[\dfrac{12}{60}\] hour
\[\Rightarrow \] 12 minutes = \[\dfrac{1}{5}\] hour
So, total time taken = \[\left( 8+\dfrac{1}{5} \right)\] hours = \[\dfrac{41}{5}\] hours
\[\Rightarrow \] Time taken by (train + car) = \[\dfrac{41}{5}\] hours
\[\Rightarrow \dfrac{240}{u}+\dfrac{520}{v}=\dfrac{41}{5}\] - (2)
Substituting \[\dfrac{1}{u}=x\] and \[\dfrac{1}{v}=y\] in equations (1) and (2), we get,
(1) \[\Rightarrow 20x+75y=1\] - (3)
(2) \[\Rightarrow 240x+520y=\dfrac{41}{5}\] - (4)
Multiplying equation (3) with 12 and subtracting equation (4) from it, we get,
\[\begin{align}
  & \Rightarrow \left( 240x+900y \right)-\left( 240x+520y \right)=12-\dfrac{41}{5} \\
 & \Rightarrow 380y=\dfrac{60-41}{5} \\
 & \Rightarrow 380y=\dfrac{19}{5} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow y=\dfrac{19}{5\times 380} \\
 & \Rightarrow y=\dfrac{1}{100} \\
\end{align}\]
Substituting the value of y in equation (3), we get,
\[\begin{align}
  & \Rightarrow 20x+75\times \dfrac{1}{100}=1 \\
 & \Rightarrow 20x+\dfrac{3}{4}=1 \\
 & \Rightarrow 20x=1-\dfrac{3}{4} \\
 & \Rightarrow 20x=\dfrac{1}{4} \\
\end{align}\]
Now, substituting back the assumed values of x and y, we get,
\[\because x=\dfrac{1}{80}\]
\[\Rightarrow \dfrac{1}{u}=\dfrac{1}{80}\]
\[\Rightarrow \] u = 80 km/hr
Similarly, we have,
\[\begin{align}
  & \because y=\dfrac{1}{100} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{1}{100} \\
\end{align}\]
\[\Rightarrow \] v = 100 km/hr
Hence, the speed of the train and the car is 80kmph and 100kmph respectively.

Note:
One may note that equation (1) and (2) are not linear in ‘u’ and ‘v’ but in \[\left( \dfrac{1}{u} \right)\] and \[\left( \dfrac{1}{v} \right)\]. You may directly solve them algebraically without assuming them as x and y. We have used elimination methods to solve the equations, you may use the substitution method also to get the same answer. You must remember the formula: - Distance = Speed \[\times \] Time to solve the question. Remember that you must convert the time into a particular unit, either in minutes or in hour for the calculations.