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**Hint:**In this question, we are given the total distance and total time covered by Vishal. Some distance is covered by bus and some by airplane. We are also given speeds of buses and airplanes. We need to find the distance traveled by bus. For this, we will form two equations in terms of distance and time. We will suppose time as x hours and y hours for bus and train respectively and then form an equation. Solving them will give us time by bus which when multiplied by speed will give us distance traveled by bus. We will use the formula $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$.

**Complete step-by-step solution**

Here, the total distance is given as 1900km and total time as 5 hours covered by Vishal. The average speed of the bus is 60km/hr and the average speed of the airplane is 700km/hr. We need to find the distance traveled by bus.

Let us suppose that it took x hours to travel the distance by bus and it took y hours to travel the distance by airplane.

As we know, $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$.

So, putting values of speed of bus and time taken by bus, we get:

\[\begin{align}

& \text{Speed of bus}=\dfrac{\text{Distance travelled by bus}}{\text{Time taken by bus}} \\

& \Rightarrow 60=\dfrac{\text{Distance travelled by bus}}{\text{x}} \\

\end{align}\]

So cross multiplying we get:

Distance travelled by bus = (60x)km.

Now putting values of speed of aeroplane and time taken by aeroplane, we get:

\[\begin{align}

& \text{Speed of aeroplane}=\dfrac{\text{Distance travelled by aeroplane}}{\text{Time taken by aeroplane}} \\

& \Rightarrow 700=\dfrac{\text{Distance travelled by aeroplane}}{\text{y}} \\

\end{align}\]

Cross multiplying we get:

Distance travelled by aeroplane = (700y)km.

Now the total distance is given as 1900km.

Therefore, the total distance = distance traveled by bus + distance traveled by airplane.

\[\Rightarrow 1900=60x+700y\]

Taking 10 commons from both sides and canceling, we get:

\[\Rightarrow 6x+70y=190\]

Dividing both sides by 2, we get:

\[\Rightarrow 3x+35y=95\cdots \cdots \cdots \left( 1 \right)\]

So this is our first equation.

Since the total time taken was 5 hours. So,

Total time = time taken by bus + time taken by aeroplane.

\[\begin{align}

& \Rightarrow 5=x+y \\

& \Rightarrow x+y=5\cdots \cdots \cdots \left( 2 \right) \\

\end{align}\]

This is our second equation.

Now, we need to solve equation (1) and (2),

For this, let us use elimination methods.

Multiplying equation (2) by 3 we get,

\[\Rightarrow 3x+3y=15\ \cdots \cdots \cdots \left( 3 \right)\]

Now subtracting (3) from (1) we get,

\[\begin{align}

& \Rightarrow 3x+35y-3x-3y=95-15 \\

& \Rightarrow 32y=80 \\

& \Rightarrow y=\dfrac{80}{32} \\

& \Rightarrow y=2.5 \\

\end{align}\]

Now putting values of y in equation (2) we get,

\[\begin{align}

& \Rightarrow x+2.5=5 \\

& \Rightarrow x=5-2.5 \\

& \Rightarrow x=2.5 \\

\end{align}\]

Hence time taken by bus is equal to 2.5 hours and time taken by aeroplane is also 2.5 hours.

Let us find the distance traveled by bus.

As we know, $\text{Distance}=\text{Speed}\times \text{Time}$.

So, distance travelled by bus $\Rightarrow \text{Speed of bus}\times \text{Time taken by bus}$.

**Distance travelled by bus $\Rightarrow 60\times 2.5=150km$. Hence the required distance is 150km.**

**Note:**Students should carefully form equations using given statements. Make sure that, units of distance are in km, units of time are in hours only because we are given speed in km/hr. Students can make the mistake of altering the formula i.e. they can divide the speed by time to get the distance traveled which is wrong.

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