Answer
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Hint: Use two variables for full fare and reservation charge. Form two equations by using the given conditions. Solve it to get the basic full fare of one reserved seat.
Complete step-by-step answer:
Let full fare cost be Rs.‘x’ and reservation charge cost be Rs.‘y’.
It is given that one reserved first class ticket from Mumbai to Ahmedabad costs Rs.216.
Hence, it will include full fare and reservation both as we are talking about full tickets for a passenger.
Hence, equation can be written as,
$x+y=216\ldots \ldots (1)$.
Now, it is also given that one full and one half ticket of reserved first class costs Rs.327.
So, as it is given in the first line of the problem that half ticket is summation of half of full fare and reservation.
Hence, equation from this condition can be written as
$\left( x+y \right)+\left( \dfrac{x}{2}+y \right)=327$
where y is reservation charge and x is full fare as supposed in the starting of the solution.
Simplifying the above relation, we get
$\begin{align}
& \dfrac{3x}{2}+2y=327 \\
& 3x+4y=654\ldots \ldots (2) \\
\end{align}$
Now, we can solve equation (1) and (2) to get values of ‘x’ and ‘y’.
Let us solve both equations by elimination approach.
On multiplying by ‘3’ in equation (1), we get
$3x+3y=648\ldots \ldots (3)$
Now, subtract equation (2) and (3), we get
\[\begin{align}
& \text{ }3x+4y=654 \\
& \underline{-\left( 3x+3y=648 \right)} \\
& \text{ }0+y=6 \\
\end{align}\]
or y = 6Rs
Now, multiplying by ‘4’ in equation (1), we get
$4x+4y=864\ldots \ldots (4)$
Now, subtracting equation (2) and (4) , we get
\[\begin{align}
& \text{ }4x+4y=864 \\
& \underline{-\left( 3x+4y=654 \right)} \\
& \text{ }x+0=210 \\
\end{align}\]
Or, x = 210Rs
Hence, basic first class full fare is Rs.210.
Note: One can go wrong while writing the equation for the second condition. He/she may divide the reservation charge by ‘2’ or may not divide the full fare, as it is mentioned that half ticket includes half of full fare and reservation charge. Need to take with it. One can solve both the equations formed by substitution and cross-multiplication as well. Answer will remain the same.
Complete step-by-step answer:
Let full fare cost be Rs.‘x’ and reservation charge cost be Rs.‘y’.
It is given that one reserved first class ticket from Mumbai to Ahmedabad costs Rs.216.
Hence, it will include full fare and reservation both as we are talking about full tickets for a passenger.
Hence, equation can be written as,
$x+y=216\ldots \ldots (1)$.
Now, it is also given that one full and one half ticket of reserved first class costs Rs.327.
So, as it is given in the first line of the problem that half ticket is summation of half of full fare and reservation.
Hence, equation from this condition can be written as
$\left( x+y \right)+\left( \dfrac{x}{2}+y \right)=327$
where y is reservation charge and x is full fare as supposed in the starting of the solution.
Simplifying the above relation, we get
$\begin{align}
& \dfrac{3x}{2}+2y=327 \\
& 3x+4y=654\ldots \ldots (2) \\
\end{align}$
Now, we can solve equation (1) and (2) to get values of ‘x’ and ‘y’.
Let us solve both equations by elimination approach.
On multiplying by ‘3’ in equation (1), we get
$3x+3y=648\ldots \ldots (3)$
Now, subtract equation (2) and (3), we get
\[\begin{align}
& \text{ }3x+4y=654 \\
& \underline{-\left( 3x+3y=648 \right)} \\
& \text{ }0+y=6 \\
\end{align}\]
or y = 6Rs
Now, multiplying by ‘4’ in equation (1), we get
$4x+4y=864\ldots \ldots (4)$
Now, subtracting equation (2) and (4) , we get
\[\begin{align}
& \text{ }4x+4y=864 \\
& \underline{-\left( 3x+4y=654 \right)} \\
& \text{ }x+0=210 \\
\end{align}\]
Or, x = 210Rs
Hence, basic first class full fare is Rs.210.
Note: One can go wrong while writing the equation for the second condition. He/she may divide the reservation charge by ‘2’ or may not divide the full fare, as it is mentioned that half ticket includes half of full fare and reservation charge. Need to take with it. One can solve both the equations formed by substitution and cross-multiplication as well. Answer will remain the same.
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