Question

a man travells 370 km partly by train and partly by car. if he covers 250 km by train and the rest by car it takes him 4 hours. but if he travels 130 km by train and the rest by car he takes 14 minutes longer. find the speed of the train and that of the car

Answer 1

Let the speed of the car be C kmph
Let the speed of the train be T kmph

4 hours = 250 km / T kmph + (370 - 250) km / C kmph

         4 = 250/T + 120/C  -- equation 1

4 hrs 18 minutes = 130 km / T kmph + (370 - 130)km / C kmph
       258/60 = 4.3 = 130 / T + 240 / C      --- equation 2

Multiply equation 1 by 2 and subtract equation 2 from it.

       8 - 4.3 = 500/T - 130/T + 240/C - 240/C

       3.7 = 370 / T

       T = 370/3.7 = 100 kmph

Substitute the value of T in equation 1 to get,

       4 = 250/T + 120/C  -- equation 1
       4 = 250/100 + 120/C
       4 - 2.5 = 120/C

       C = 120/1.5 = 80 kmph

The train runs at 100 kmph and the car runs at 80 kmph

Answer 2

$\huge\underline\mathtt \green{SOLUTION:-}$

$\underline{\star\:\large{\textit{1st Case:}}}$

$:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{250}{Train} + \dfrac{(370 - 250)}{Car} = 4\\\\\\:\implies\tt\dfrac{250}{Train} + \dfrac{120}{Car} = 4 \qquad {\sf\dfrac{\quad}{}\:eq.(1)}$

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$\underline{\star\:\large{\textit{2nd Case:}}}$

$:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{(370 - 130)}{Car} = 4hr \:18min.\\\\\\:\implies\tt\dfrac{130}{Train} + \dfrac{240}{Car} = 4{}^{18}\! /{}_{60}\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{240}{Car} = \dfrac{43}{10} \qquad {\sf\dfrac{\quad}{}\:eq.(2)}$

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$\sf Let\:\:\frac{1}{Train} = x\:\:and\:\:\frac{1}{Car}=y$

☯$\underline{\textsf{Multiplying eq.(3) with 2 :}}$

$\longrightarrow\sf 250x +120y = 4\qquad\dfrac{\quad}{}\:eq.(3) \times 2\\\\\longrightarrow\sf 130x +240y = \dfrac{43}{10}\qquad\dfrac{\quad}{} \:eq.(4)$

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☯$\underline{\textsf{Subtracting eq.(4) from eq.(3) :}}$

$\dashrightarrow\tt\:\:500x + 240y = 8\\\\\dashrightarrow\tt\:\:130x + 240y = \dfrac{43}{10} \\ \dfrac{\qquad \qquad \qquad \qquad \qquad \quad}{}\\\dashrightarrow\tt\:\:(500x - 130x) = \bigg(8 - \dfrac{43}{10} \bigg)\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{(80 - 43)}{10}\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{37}{10}\\\\\\\dashrightarrow\tt\:\:x = \dfrac{37}{3700}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt x =\dfrac{1}{100} = \dfrac{1}{Train}}}}$

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☯$\underline{\textsf{Putting value of x in eq.(3) :}}$

$\dashrightarrow\tt\:\:250x+120y=4\\\\\\\dashrightarrow\tt\:\:250 \times \dfrac{1}{100} + 120y = 4\\\\\\\dashrightarrow\tt\:\:\dfrac{5}{2} + 120y = 4\\\\\\\dashrightarrow\tt\:\:120y = 4 - \dfrac{5}{2}\\\\\\\dashrightarrow\tt\:\:120y =\dfrac{(8 - 5)}{2}\\\\\\\dashrightarrow\tt\:\:120y = \dfrac{3}{2}\\\\\\\dashrightarrow\tt\:\:y = \dfrac{3}{240}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt y = \dfrac{1}{80} = \dfrac{1}{Car}}}}$

⠀

$\begin{cases}\textsf{Speed of Train = \textbf{100 Km/hr}}\\\textsf{Speed of Car = \textbf{80 Km/hr}}\end{cases}$