
A railway half-ticket costs half the full fare but the reservation charges are the same on a half-ticket as on a full ticket. One reserved full first class ticket from station A to station B costs Rs. 2125. Also, one reserved a full first class ticket and a half first class ticket from A to B costs Rs. 3200. Find the full fare from station A to station B and also the reservation charges for a ticket.
Answer
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Hint: Assume that the full fare of the ticket is Rs. $x$ and reservation charges on a ticket is Rs. $y$. Form two sets of linear equations in two variables with the help of provided conditions and solve them to get the answer.
Complete step-by-step answer:
Let us assume that the full fare of the ticket is Rs. $x$ and reservation charges on a ticket is Rs. $y$. It is given that, half-ticket costs half the full fare and the reservation charges are the same on a half-ticket as on a full ticket. Therefore,
Cost of a full ticket $=x+y$,
And, Cost of a half ticket $=\dfrac{x}{2}+y$.
Now, one reserved full first class ticket from station A to station B costs Rs. 2125. Therefore, mathematically,
$x+y=2125........................(i)$
Also, one reserved full first class ticket and a half first class ticket from A to B costs Rs. 3200. Therefore, mathematically,
$\begin{align}
& x+y+\dfrac{x}{2}+y=3200 \\
& \dfrac{3x}{2}+2y=3200...................(ii) \\
\end{align}$
Now, multiplying equation (i) by 2 and subtracting it from equation (ii) we get,
$\begin{align}
& \dfrac{3x}{2}+2y-(2x+2y)=3200-2\times 2125 \\
& \dfrac{3x}{2}-2x=3200-4250 \\
& \dfrac{-x}{2}=-1050 \\
\end{align}$
Cancelling the minus sign and cross multiplying, we get,
$\begin{align}
& x=2\times 1050 \\
& x=2100 \\
\end{align}$
Substituting the value of $x$ in equation (i) we get,
$\begin{align}
& y=2125-2100 \\
& y=25 \\
\end{align}$
Hence, full fare and reservation charge for a ticket from station A to B are Rs. 2100 and Rs. 25 respectively.
Note: One may get confused in understanding such a long question. The main problem here is to form linear equations with the help of provided information. So, we have to go thoroughly through the question.
Complete step-by-step answer:
Let us assume that the full fare of the ticket is Rs. $x$ and reservation charges on a ticket is Rs. $y$. It is given that, half-ticket costs half the full fare and the reservation charges are the same on a half-ticket as on a full ticket. Therefore,
Cost of a full ticket $=x+y$,
And, Cost of a half ticket $=\dfrac{x}{2}+y$.
Now, one reserved full first class ticket from station A to station B costs Rs. 2125. Therefore, mathematically,
$x+y=2125........................(i)$
Also, one reserved full first class ticket and a half first class ticket from A to B costs Rs. 3200. Therefore, mathematically,
$\begin{align}
& x+y+\dfrac{x}{2}+y=3200 \\
& \dfrac{3x}{2}+2y=3200...................(ii) \\
\end{align}$
Now, multiplying equation (i) by 2 and subtracting it from equation (ii) we get,
$\begin{align}
& \dfrac{3x}{2}+2y-(2x+2y)=3200-2\times 2125 \\
& \dfrac{3x}{2}-2x=3200-4250 \\
& \dfrac{-x}{2}=-1050 \\
\end{align}$
Cancelling the minus sign and cross multiplying, we get,
$\begin{align}
& x=2\times 1050 \\
& x=2100 \\
\end{align}$
Substituting the value of $x$ in equation (i) we get,
$\begin{align}
& y=2125-2100 \\
& y=25 \\
\end{align}$
Hence, full fare and reservation charge for a ticket from station A to B are Rs. 2100 and Rs. 25 respectively.
Note: One may get confused in understanding such a long question. The main problem here is to form linear equations with the help of provided information. So, we have to go thoroughly through the question.
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