Question

A man travels 600 km apart by train and partly by car. It takes 8 hours and 40 minutes if he travels 300 km by train and rest by car. It would take 30 min more if he travels 200 Km by train and the west by the car, find the speed of the train and by car seprately
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Answer 1

Step-by-step explanation:

YOUR ANSWER REFER TO THE ATTACHMENT::::::⬆⬆⬆

Answer 2

❍ Given : A man travels 600 km a part by train and partly by car. It takes 8 hours and 40 minutes if he travels 300 km by train and rest by car. It would take 30 minutes more if he travels 200 Km by train and the west by the car,

❍ To Find : find the speed of the train and by car seprately

$\qquad$____________________________

Let’s assume,

• ❒ The speed of the train be x km/hr

• ❒ The speed of the car = y km/hr

From the question, it’s understood that there are two parts

• ❍Part 1: When the man travels 400 km by train and the rest by car.

• ❍Part 2: When Ramesh travels 200 km by train and the rest by car.

We solve this question in two parts let solve

$\huge \qquad \qquad \frak{Part \: 1,}$

Time taken by the man to travel 400km by train

$\sf \: : \longmapsto \red{ \boxed{ \bf{ \pink{time = \dfrac{distance}{speed}}}}} \\ \\ \\ \sf : \longmapsto \: time \: = \frac{400}{x} \: hrs$

Time taken by the man to travel

$\sf : \longmapsto \: (600 \: – \: 400) \\ \\ \\ \sf: \longmapsto 200km \: by \: car \\ \\ \\ \sf: \longmapsto \dfrac{200}{y } \: hrs$

$\sf \: : \longmapsto \red{ \boxed{ \bf{ \pink{time = \dfrac{distance}{speed}}}}} \\ \\ \\ \sf : \longmapsto \: time \: = \frac{200}{y} \: hrs$

Time taken by a man to cover 600km

$\large{ \sf \: =  \dfrac{400}{x } \: hrs +  \dfrac{200}{y } \: hrs}$

Total time taken for this journey

$\sf: \longmapsto \: = 6 hours + 30 mins \\ \\ \\ \sf: \longmapsto 6 + \dfrac{1}{2} \\ \\ \large \pink{ \sf: \longmapsto \dfrac{13}{2}}$

So, by equations its

$\sf:\longrightarrow\dfrac{400}{x} + \dfrac{200}{y }= \dfrac{13}{2} \\ \\ \\ \sf:\longrightarrow200 \bigg( \frac{2}{x} + \frac{1}{y} \bigg) = \frac{13}{2} \\ \\ \\ \sf:\longrightarrow \frac{2}{x} + \frac{1}{y} = \frac{13}{2} \times 200 \\ \\ \\ \\ \large \pink{\sf:\longrightarrow \frac{2}{x} + \frac{1}{y} = \frac{13}{400}\: } - - - (1)$

$\huge \qquad \qquad \frak{Part \: 2,}$

Time taken by the man to travel 200 km by train

$\sf \: : \longmapsto \red{ \boxed{ \bf{ \pink{time = \dfrac{distance}{speed}}}}} \\ \\ \\ \sf : \longmapsto \: time \: = \frac{200}{x} \: hrs$

Time taken by the man to travel (600 – 200)

$\sf : \longmapsto \: (600 \: – \: 400) \\ \\ \\ \sf: \longmapsto 200km \: by \: car \\ \\ \\ \sf: \longmapsto \dfrac{200}{y } \: hrs$

For the part, the total time of the journey is given as 6hours 30 mins + 30 mins that is 7hrs,

$\sf: ⇒ \dfrac{200}{x} +  \dfrac{400}{y} = 7 \\ \\ \\ \sf: ⇒200 \bigg( \dfrac{1}{x}+  \dfrac{2}{y} \bigg) = 7$

$\large \pink{ \sf: ⇒ \dfrac{1}{x }+  \dfrac{2}{y} =  \dfrac{7}{200}  - - - (ii)}$

$\bf \: Taking  \: \dfrac{1}{x }= u, and  \: \dfrac{1}{y} = v,$

So, the equations (i) and (ii) becomes,

$\pink{ \sf: \longmapsto \: 2u + v = \dfrac{ 13}{400} - - - - (iii) }\\ \\ \\ \pink{\sf: \longmapsto u + 2v = \dfrac{ 7}{200} - - - - (iv)}$

Solving equation (iii) and equation (iv), by equation (iv) x 2 – equation (iii)

$\sf: ⇒3v = \dfrac{14}{200} \: – \: \dfrac{13}{400} \\ \\ \sf: ⇒3v = \dfrac{1}{400 } \times (28 – 13) \\ \\ \\ \sf: \: ⇒3v = \frac{1 \times 15}{400} \\ \\ \sf: ⇒3v = \dfrac{ 15}{400 } \\ \\ \sf: \: ⇒v = 26 \times 3 \\ \\ \sf: \: ⇒v = \dfrac{1}{80}$

$\\ \\ \large \pink{\sf : \longmapsto y = \dfrac{1}{v }= 80}$

Now, using v in (iii) we find u,

$\sf: ⇒2u + \bigg ( \dfrac{1}{80} \bigg) =  \dfrac{13}{400 }\\ \\ \sf:⇒2u = \dfrac{13}{400 }– \dfrac{1}{80} \\ \\ \sf: ⇒2u = \dfrac{8}{400 }\\ \\ \sf: ⇒u = \dfrac{1}{100} \\ \\ \sf: ⇒ x = \dfrac{1}{u }= 100$

Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.