Answer

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**Hint:**We will assume the speeds of the both train and bus as $x$ kmph and $y$ kmph. In the problem they have mentioned that Roohi takes $4$ hours if she travels $60$ km by train and the remaining by bus. From the above statement we will calculate the time taken by Roohi to cover $60$ km by train and remaining distance by bus by using the relation between velocity $v$, Distance $d$, time $t$ i.e. $v=\dfrac{d}{t}\Rightarrow t=\dfrac{d}{v}$. By calculating the travelling time of Roohi in train and bus, we will add up them and equate them to given time $4$ hours. Here we will get an equation in terms of $x$ and $y$. Now we will calculate the times taken by Roohi in the second case i.e. If she travels $100$ km by train and the remaining by bus, she takes $10$ minutes longer. From the above statement we can obtain another equation in terms of $x$ and $y$ as we did earlier. To find the velocities we will solve the obtained two equations.

**Complete step by step answer:**

Given that, Total distance covered by Roohi to reach home is $300$ km.

Let us take

The velocity of the train is $x$ kmph.

The velocity of the bus is $y$ kmph.

Case-1: She travels $60$ km by train and the remaining by bus, travelling time is $4$ hours. So, we have the parameters as

Distance travelled in the train is $60$ km.

Distance travelled in the bus is $300-60=240$ km.

Time travelled in the train is ${{t}_{1}}=\dfrac{60}{x}$.

Time travelled in the bus is ${{t}_{2}}=\dfrac{240}{y}$.

Total time travelled is given by summing the time travelled in train and time travelled in bus, then

$\begin{align}

& t={{t}_{1}}+{{t}_{2}} \\

& \Rightarrow t=\dfrac{60}{x}+\dfrac{240}{y}....\left( \text{i} \right) \\

\end{align}$

But we have given that the total travelling time was $4$ hours.

$\therefore t=4$

Substituting the value of $t$ from equation $\left( \text{i} \right)$, then we will have

$\begin{align}

& t=4 \\

& \Rightarrow \dfrac{60}{x}+\dfrac{240}{y}=4 \\

\end{align}$

Dividing the whole equation with $4$, then we will have

$\begin{align}

& \dfrac{60}{4x}+\dfrac{240}{4y}=\dfrac{4}{4} \\

& \Rightarrow \dfrac{15}{x}+\dfrac{60}{y}=1....\left( \text{a} \right) \\

\end{align}$

Case-2: If she travels $100$ km by train and the remaining by bus, she takes $10$ minutes longer. So, we have the parameters as

Distance travelled in the train is $100$ km.

Distance travelled in a bus is $300-100=200$ km.

Time travelled in the train is ${{t}_{1}}=\dfrac{100}{x}$.

Time travelled in the bus is ${{t}_{2}}=\dfrac{200}{y}$.

Total time travelled is given by summing the time travelled in train and time travelled in bus, then

$\begin{align}

& t={{t}_{1}}+{{t}_{2}} \\

& \Rightarrow t=\dfrac{100}{x}+\dfrac{200}{y}....\left( \text{ii} \right) \\

\end{align}$

But we have given that the total travelling time is increased by $10$ minutes to the case-2, then

$\begin{align}

& \therefore t=4+\dfrac{10}{60} \\

& \Rightarrow t=\dfrac{25}{6} \\

\end{align}$

Substituting the value of $t$ from equation $\left( \text{ii} \right)$, then we will have

$\begin{align}

& t=\dfrac{25}{6} \\

& \Rightarrow \dfrac{100}{x}+\dfrac{200}{y}=\dfrac{25}{6} \\

\end{align}$

Dividing the whole equation with $25$, then we will have

$\begin{align}

& \dfrac{100}{25x}+\dfrac{200}{25y}=\dfrac{25}{6\times 25} \\

& \Rightarrow \dfrac{4}{x}+\dfrac{8}{y}=\dfrac{1}{6}....\left( \text{b} \right) \\

\end{align}$

For solving the equation $\left( a \right)$ and $\left( b \right)$, we will substituting $\dfrac{1}{x}=p$ and $\dfrac{1}{y}=q$ , then we will get

$\begin{align}

& \left( a \right)\Rightarrow 15p+60q=1 \\

& \left( b \right)\Rightarrow 4p+8q=\dfrac{1}{6} \\

\end{align}$

Solving the above equations, we will get

$p=\dfrac{1}{60}$, $q=\dfrac{1}{80}$

$\Rightarrow x=60$, $y=80$.

**$\therefore $ The velocity of the train is $60$ kmph and velocity of bus is $80$ kmph.**

**Note:**In the problem we have solved the equations in a single step. If you don’t understand the solution, please follow the below steps to solve the equations

$15p+60q=1...\left( \text{I} \right)$

$4p+8q=\dfrac{1}{6}....\left( \text{II} \right)$

Doing $15\times \left( \text{II} \right)-4\times \left( \text{I} \right)$, then we will get

$\begin{align}

& 15\left[ 4p+8q \right]-4\left[ 15p+60q \right]=15\times \dfrac{1}{6}-4\times 1 \\

& \Rightarrow 60p+120q-60p-240q=\dfrac{5}{2}-4 \\

& \Rightarrow -120q=\dfrac{5-2\times 4}{2} \\

& \Rightarrow -120q=-\dfrac{3}{2} \\

& \Rightarrow q=\dfrac{3}{2\times 120} \\

& \Rightarrow q=\dfrac{1}{80} \\

\end{align}$

Substituting the value of $q$ in equation $\left( \text{I} \right)$, then we will have

$\begin{align}

& 4p+8\left( \dfrac{1}{80} \right)=\dfrac{1}{6} \\

& \Rightarrow 4p=\dfrac{1}{6}-\dfrac{1}{10} \\

& \Rightarrow 4p=\dfrac{10-6}{6\times 10} \\

& \Rightarrow 4p=\dfrac{4}{60} \\

& \Rightarrow p=\dfrac{1}{60} \\

\end{align}$

In this way we can solve two equations.

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