Question

A man travels 300 km partly by train and partly by car. If the man covers 60 km by train and rests by car, it takes him 4 hours. If the man covers 100 km by train and rests by car, he takes 10 minutes longer. Find the speed of the train and the car.

Answer 1

Hint: In both the cases, we have to find the individual distance and from the formula of distance, speed and time, the speed of train and car can be calculated.
We can use the Formula: $time = \dfrac{{dis\tan ce}}{{speed}}$

Step by step answer: The total distance travelled by the man is 300 km. Let the velocity of the train be v₁ and the velocity of the car be v₂.
The following steps are:
In the first case: The distance travelled by train is 60 km. So, the distance travelled by car will be = 300 – 60 =240 km.
Let the time taken by train be t₁ and by car be t₂.
Total time taken to cover the whole distance = 4 hours.
$time = \dfrac{{dis\tan ce}}{{speed}}$
Time taken to cover the distance by train (t₁) =$\dfrac{{60}}{{{v_1}}}$
Time taken to cover the distance by car (t₂) = $\dfrac{{240}}{{{v_2}}}$
Total time = ${t_1} + {t_2}$
 $\Rightarrow$ $4 = \dfrac{{60}}{{{v_1}}} + \dfrac{{240}}{{{v_2}}}$
 $\Rightarrow$ $4 - \dfrac{{240}}{{{v_2}}} = \dfrac{{60}}{{{v_1}}}$
 $\Rightarrow$ $\dfrac{{4{v_2} - 240}}{{{v_2}}} = \dfrac{{60}}{{{v_1}}}$
 $\Rightarrow$ ${v_1} = \dfrac{{60{v_2}}}{{4{v_2} - 240}}$
In the second case: The distance travelled by train is 100 km. So, the distance travelled by car = $300 - 100 = 200$km.
Let the time taken by train be t₃ and by car be t₄.
Time = 10 minutes = $\dfrac{{10}}{{60}}$ [ we are converting the time to hours]
$\Rightarrow$ = $\dfrac{1}{6}$
Total time = $4 + \dfrac{1}{6}$ [Since the time taken is 10 minutes longer than 4 hours]
                   = $\dfrac{{25}}{6}$hours
${t_3} = \dfrac{{100}}{{{v_1}}}$ and ${t_4} = \dfrac{{200}}{{{v_2}}}$
total time = t₃ + t₄
$\Rightarrow$ $\dfrac{{25}}{6} = \dfrac{{100}}{{{v_1}}} + \dfrac{{200}}{{{v_2}}}$
$\Rightarrow$ $\dfrac{{25}}{6} - \dfrac{{200}}{{{v_2}}} = \dfrac{{100}}{{{v_1}}}$
$\Rightarrow$ $\dfrac{{25{v_2} - 1200}}{{6{v_2}}} = \dfrac{{100}}{{{v_1}}}$
$\Rightarrow$ $\dfrac{{25{v_2} - 1200}}{{6{v_2}}} = \dfrac{{100 \times \left( {4{v_2} - 240} \right)}}{{60{v_2}}}$ [Substituting the value of v₁ from the above equation]
$25{v_2} - 1200 = \dfrac{{100\left( {4{v_2} - 240} \right)}}{{10}}$ [v₂ and denominators are being reduced]
$\Rightarrow$ $25{v_2} - 1200 = 10\left( {4{v_2} - 240} \right)$ [We have reduced 100 and 10]
$\Rightarrow$ $25{v_2} - 1200 = 40{v_2} - 2400$
$
 \Rightarrow {v_1} = \dfrac{{60 \times 80}}{{4 \times 80 - 240}} \\
 \Rightarrow {v_1} = \dfrac{{4800}}{{320 - 240}} = \dfrac{{4800}}{{80}} \\
  \Rightarrow {v_1} = 60km/hr \\
$
$1200 = 15{v_2}$
$\Rightarrow$ ${v_2} = \dfrac{{1200}}{{15}} = 80km/hr$
Now, substituting the value of v₂ in v₁ in the equation of first case
${v_1} = \dfrac{{60{v_2}}}{{4{v_2} - 240}}$
$
  \Rightarrow {v_1} = \dfrac{{60 \times 80}}{{4 \times 80 - 240}} \\
  \Rightarrow {v_1} = \dfrac{{4800}}{{320 - 240}} = \dfrac{{4800}}{{80}} \\
 \therefore {v_1} = 60km/hr \\
$
Hence, the speed of the train is 60 km/hr and the speed of the car is 80 km/hr.

Note: The time given in minutes should be converted to hours because the time given in first case is in hours and distances are given in km.