Question

roohi travels 300 kilometre travel to her home by train and partly by bus she takes 4 hours if he travels 60 kilometre by train and the remaining by bus. if she Travels 100 KM by train and the remaining by bus she takes 10 minutes longer.find the speed of the train and the buses separately ​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

Let, the speed of the Train is = X Km/h

and the speed of the bus = Y km/h

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•Let's Head to the problem now

$\huge \star$ according to the condition (1)

Roohi travels 300 kilometre travel to her home by train and she took 4 hours in total journey.

whereas she traveled partly by train and partly by bus

Now, if he travels 60 kilometre by train

$\therefore Required\: Time\: is\:</p><p>T_{train}=\frac{60}{X}\:hours$

Roohi traveled the remaining path by bus .

$\therefore$She traveled by bus =(300-60)Km=240 Km

$\therefore Required\: Time\: is\:</p><p>T_{bus}=\frac{240}{Y}\:hours$

Now total time wasted=4 hours, so

$\implies\frac{60}{X}+\frac{240}{Y} =4</p><p>\\ \implies\frac{60Y+240X}{XY}=4\\</p><p> \implies 60Y+240X=4XY\\</p><p> \implies \boxed{\bf \red{15Y+60X=XY} }...............(1)\:\: \\(\because dividing\: by \:4 \:on\: both\:sides)$

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$\huge \star$ according to the condition (2)

Roohi travels 300 kilometre travel to her home by train and she took 4 hours and 10 minutes in total journey.

whereas she traveled partly by train and partly by bus

Now, if he travels 100 kilometre by train

$\therefore Required\: Time\: is\:</p><p>T_{train}=\frac{100}{X}\:hours$

Roohi traveled the remaining path by bus .

$\therefore$She traveled by bus =(300-100)Km=200 Km

$\therefore Required\: Time\: is\:</p><p>T_{bus}=\frac{200}{Y}\:hours$

Now total time wasted

=4 hours and 10 minutes

$\bf\rightarrow 4\frac{10}{60}\:hours\\ \rightarrow \frac{25}{6}\:hours$

$\implies\frac{100}{X}+\frac{200}{Y} =\frac{25}{6}</p><p>\\ \implies\frac{100Y+200X}{XY}=\frac{25}{6}\\</p><p> \implies 600Y+1200X=25XY\\</p><p> \implies \boxed{\bf \red{24Y+48X=XY} }...............(2)\:\: \\(\because dividing\: by \:25\:on\: both\:sides)$

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Now comparing equation (1)&(2),

$\bf\implies 15Y+60X=24Y+48X\\</p><p>\bf\implies60X-48X=24Y-15Y\\ \bf\implies 12X=9Y\\ \bf\implies 4X=3Y \\ \bf\implies \red{Y=\frac{4}{3}X}$

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$\therefore$ Putting the value of $Y=(4/3)X$ in equation (2), we get...

$\implies\cancel{24}\times\frac{4}{\cancel3}X+48X=X\frac{4}{3}X\\ \implies32X+48X=\frac{4X{}^{2}}{3}\\ \implies 240X=4X{}^{2}\\ \implies X{}^{2}-60X=0\\ </p><p>\implies X(X-60)=0$

$\bf either\\ \rightarrow X=0\:\: \\(X=0 Not \:taken \:into \:consideration)\\ (\because velocity\: here \:can\: not\: be \:Zero)\\ \bf or\\</p><p> \implies X-60=0\\ \implies \boxed{\red{X=60 km.h{}^{-1}}}$

$\bf\therefore \implies Y=\frac{4}{3}X\\</p><p>\implies Y=\frac{4}{\cancel3}\times\cancel{60}\:\\ \implies\boxed{\red{ Y=80 \:km.h{}^{-1}}}$

$\therefore$the speed of the Train is=60km/h

$\therefore$the speed of the bus is=80km/h

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Answer 2

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Let the speed of train and bus be u km/h and v km/h respectively.

According to the question,

$\frac{60}{u} + \frac{240}{v} = 4$....(i)

$\frac{100}{u} + \frac{200}{v} = 4 + \frac{10}{60} = \frac{25}{6}$....(ii)

Let $\frac{1}{u} = p\: and \frac{1}{v} = q$

The given equations reduce to:

60p + 240q = 4 ....(iii)

100p + 200q = $\frac{25}{6}$

600p + 1200q = 25....(iv)

Multiplying equation (iii) by 10, we obtain:

600p + 2400q = 40....(v)

Subtracting equation (iv) from equation (v), we obtain:

1200q = 15

q = $\frac{15}{1200} = \frac{1}{80}$

Substituting the value of q in equation (iii), we obtain:

60p + 3 = 4

60p = 1

p = $\frac{1}{60}$

:. p = $\frac{1}{u} = \frac{1}{60}$, q = $\frac{1}{v} = \frac{1}{80}$

u = 60 km/h , v = 80 km/h

Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.

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