Question

Aman travels 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 18 minutes longer. Find the speed of the train
and that of the car.

Answer 1

Answer:

Train speed is 100 km/hr.

Car speed is 80 km/hr.

Step-by-step explanation:

Let the speed of the train be T km/hr and the speed of the car be C km/hr.

Then:

 ( time to do 250 km by train ) + ( time to do 120 km by car ) = 4 hr

           ⇒ 250 / T  +  120 / C  =  4      ...(1)

 ( time to do 130 km by train ) + ( time to do 240 km by car ) = 4 hr 18 mins

           ⇒ 130 / T  +  240 / C  =  4.3      ...(2)

[ Notice that above we have used 18 mins = 18/60 hr  = 3/10 hr  =  0.3 hr ]

Multiplying equation (1) by 2 (preparing to eliminate the C) gives

               500 / T  +  240 / C  =  8      ...(3)

Subtracting (2) from (3) gives

               370 / T = 8 - 4.3 = 3.7

          ⇒  T  =  370 / 3.7  =  100.

Substituting this into (1)  (any of the equations would do) gives

        250 / 100  +  120 / C  =  4

   ⇒  120 / C  =  4 - 2.5  =  1.5

   ⇒  C  =  120 / 1.5  =  80

Hope this helps!

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

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

⠀

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