Question

Ved travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and rests by car. Find the speed of the train and the car.

Answer 1

Hint:
We can take the speed of the train and car as two variables. Then we can find the time taken in each mode by dividing the distance with the respective speed. Then we can take the sum of the time taken in each case and equate it to the given time. Then we will have 2 equations with 2 variables. Then we can solve them to get the required speeds.

Complete step by step solution:
Let x be the speed of the train and y be the speed of the car.
It is given that the distance to his home is 600km.
Let us consider the 1st case.
It is given that he travels 120km by train.
We know that, ${\text{time taken = }}\dfrac{{dis\tan ce}}{{speed}}$
Then the time taken to travel 120 km in the train is given by,
${t_{train}} = \dfrac{{120}}{x}$
Now we can find the distance travelled in a car. It is given by the total distance minus the distance travelled in the train.
$ \Rightarrow {d_{car}} = 600 - 120$
$ \Rightarrow {d_{car}} = 480$
Then the time taken to travel 480km in the car is given by,
${t_{car}} = \dfrac{{480}}{y}$
It is given that in the 1st case, the total time taken is 8 hours. So we can write,
${t_{train}} + {t_{car}} = 8$
On substituting the values, we get,
$ \Rightarrow \dfrac{{120}}{x} + \dfrac{{480}}{y} = 8$
On dividing both sides with 8, we get,
$ \Rightarrow \dfrac{{15}}{x} + \dfrac{{60}}{y} = 1$… (1)
Now we can consider the 2nd case.
It is given that he travels 200km by train.
We know that, ${\text{time taken = }}\dfrac{{dis\tan ce}}{{speed}}$
Then the time taken to travel 200 km in the train is given by,
${t_{train}} = \dfrac{{200}}{x}$
Now we can find the distance travelled in a car. It is given by the total distance minus the distance travelled in the train.
$ \Rightarrow {d_{car}} = 600 - 200$
$ \Rightarrow {d_{car}} = 400$
Then the time taken to travel 480km in the car is given by,
${t_{car}} = \dfrac{{400}}{y}$
It is given that in the 2nd case, the total time taken is 20 minutes longer than in case 1. So, the total time taken will be 8 hours and 20 minutes. We can write 20 minutes as a fraction.
$ \Rightarrow 8hr\,20\min = 8 + \dfrac{{20}}{{60}}hr$
So we have,
\[ \Rightarrow 8hr\,20\min = 8 + \dfrac{1}{3}hr\]
On converting into improper fraction we get,
\[ \Rightarrow 8hr\,20\min = \dfrac{{25}}{3}hr\]
Then the total time taken in case 2 is given by,
$ \Rightarrow {t_{train}} + {t_{car}} = \dfrac{{25}}{3}$
On substituting the values, we get,
$ \Rightarrow \dfrac{{200}}{x} + \dfrac{{400}}{y} = \dfrac{{25}}{3}$
On dividing throughout with 25, we get,
$ \Rightarrow \dfrac{8}{x} + \dfrac{{16}}{y} = \dfrac{1}{3}$
On multiplying both sides with 3, we get,
$ \Rightarrow \dfrac{{24}}{x} + \dfrac{{48}}{y} = 1$
Now we write the equation in terms of $\dfrac{1}{x}$.
$ \Rightarrow \dfrac{{24}}{x} = 1 - \dfrac{{48}}{y}$
On cross multiplication we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{24}}\left( {1 - \dfrac{{48}}{y}} \right)$
On simplification we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{24}} - \dfrac{2}{y}$.. (2)
On substituting equation (2) in (1), we get,
$ \Rightarrow 15\left( {\dfrac{1}{{24}} - \dfrac{2}{y}} \right) + \dfrac{{60}}{y} = 1$
On opening the bracket we get,
$ \Rightarrow \dfrac{{15}}{{24}} - \dfrac{{30}}{y} + \dfrac{{60}}{y} = 1$
On simplification, we get,
$ \Rightarrow \dfrac{5}{8} + \dfrac{{60 - 30}}{y} = 1$
On rearranging we get,
$ \Rightarrow \dfrac{{30}}{y} = 1 - \dfrac{5}{8}$
On simplification we get,
$ \Rightarrow \dfrac{{30}}{y} = \dfrac{{8 - 5}}{8}$
So we have,
\[ \Rightarrow \dfrac{{30}}{y} = \dfrac{3}{8}\]
On cross multiplying we get,
\[ \Rightarrow y = \dfrac{{30 \times 8}}{3}\]
On simplification we get,
\[ \Rightarrow y = 80\]
Now we can substitute in the equation (2), we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{24}} - \dfrac{2}{{80}}$
On taking LCM we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{{80 - 48}}{{80 \times 24}}$
On simplification we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{{32}}{{80 \times 24}}$
On taking the reciprocal on both sides, we get,
$ \Rightarrow x = \dfrac{{80 \times 24}}{{32}}$
Hence we have,
$ \Rightarrow x = 60$

Therefore, the speed of the train is Therefore, the speed of the train is $60 \text{km/h}$ and the speed of the car is $80 \text{km/h}$.

Note:
We took the variables as the speed as we needed to find the speed of the train and car. We can only equate the total time taken to the sum of the time taken in the car and train in each case. While solving we must not take the LCM of the fractions as it will make the equation complex and difficult to solve. We must only solve for the reciprocal of the variable and then take its reciprocal to get the required answer. We can also solve the equations by giving substitutions for the variable, which is given by,
We have equations,
$\dfrac{{24}}{x} + \dfrac{{48}}{y} = 1$
$\dfrac{{15}}{x} + \dfrac{{60}}{y} = 1$
Let $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$ . Then the equations will become,
$ \Rightarrow 24u + 48v = 1$
$ \Rightarrow 15u + 60v = 1$
On cross multiplication, we get,
$ \Rightarrow \dfrac{u}{{60\left( { - 1} \right) - 48\left( { - 1} \right)}} = \dfrac{v}{{15\left( { - 1} \right) - 24\left( { - 1} \right)}} = \dfrac{1}{{15 \times 48 - 24 \times 60}}$
On simplification we get,
$ \Rightarrow \dfrac{u}{{ - 60 + 48}} = \dfrac{{ - v}}{{ - 15 + 24}} = \dfrac{1}{{720 - 1440}}$
On solving further we get,
$ \Rightarrow \dfrac{u}{{ - 12}} = \dfrac{v}{{ - 9}} = \dfrac{1}{{ - 720}}$
On taking the 1st and last terms, we get,
$ \Rightarrow \dfrac{u}{{ - 12}} = \dfrac{1}{{ - 720}}$
On cross multiplication we get,
$ \Rightarrow u = \dfrac{{ - 12}}{{ - 720}}$
On simplification we get,
$ \Rightarrow u = \dfrac{1}{{60}}$
On taking the 2nd and last terms, we get,
$ \Rightarrow \dfrac{v}{{ - 9}} = \dfrac{1}{{ - 720}}$
On cross multiplication we get,
$ \Rightarrow v = \dfrac{{ - 9}}{{ - 720}}$
On simplification we get,
$ \Rightarrow v = \dfrac{1}{{80}}$
On re substituting, we get,
$ \Rightarrow x = 60$
\[ \Rightarrow y = 80\]
Therefore, the speed of the train is $60 \text{km/h}$ and the speed of the car is $80 \text{km/h}$.