Question

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 first class ticket from the station \[A\] to \[B\] costs \[\text{Rs}.2530\]. Also, one reserved first class ticket and one reserved first class half ticket from \[A\] to \[B\] costs \[\text{Rs}.3810\]. Find the full first class fare from station \[A\] to \[B\] and also the reservation charges for a ticket.

Answer 1

Hint:
Here we will assume the reservation charge and cost of a full ticket of first class to be any variable. Then we will form the two equations using the given condition. After solving the equations using the elimination method, we will get the required answers.

Complete step by step solution:
Let the reservation charges be \[{\rm{Rs}}.x\] and let the cost of a full ticket of first class be \[{\rm{Rs}}.y\].
Therefore, the cost of a half ticket of first class becomes \[{\rm{Rs}}.\dfrac{y}{2}\].
It is given that one reserved first class ticket from the station \[A\] to \[B\] costs Rs. 2530.
So mathematically, we can write it as
\[x + y = 2530\] …………… \[\left( 1 \right)\]
It is also given that the cost of one reserved first class ticket and one reserved first class half ticket from \[A\] to \[B\] is \[\text{Rs}.3810\].
We know that the cost of railway half ticket is equal to half of the cost of full ticket, but the reservation charges are the same on a half ticket as well as on a full ticket so we get
So mathematically, we can write it as
\[\left( {x + y} \right) + \left( {x + \dfrac{y}{2}} \right) = 3810\]
On adding the like terms, we get
\[ \Rightarrow 2x + \dfrac{{3y}}{2} = 3810\] …………….. \[\left( 2 \right)\]
Now, we will multiply the equation \[\left( 1 \right)\] by \[\left( 2 \right)\].
\[2x + 2y = 5060\]
Subtracting equation \[\left( 2 \right)\] from \[\left( 1 \right)\], we get
 \[\begin{array}{l}2x + 2y - \left( {2x + \dfrac{{3y}}{2}} \right) = 5060 - 3810\\ \Rightarrow 2x + 2y - 2x - \dfrac{{3y}}{2} = 1250\end{array}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{y}{2} = 1250\]
Now, we will multiply 2 on both sides of the equation. Therefore, we get
\[ \Rightarrow 2 \times \dfrac{y}{2} = 2 \times 1250\]
On further simplification, we get
\[ \Rightarrow y = 2500\]
 Now, we will substitute the value of \[y\] in equation \[\left( 1 \right)\].
\[ \Rightarrow x + 2500 = 2530\]
On subtracting 2500 form sides, we get
\[ \Rightarrow x + 2500 - 2500 = 2530 - 2500\]
On further simplification, we get
\[ \Rightarrow x = 30\]

Hence, the reservation charges is equal to \[{\rm{Rs}}.30\] and cost of full ticket of first class is equal to \[{\rm{Rs}}.2500\]

Note:
Here we need to add the reservation charge to both full and half ticket cost individually otherwise we will get the incorrect final value. We have formed two linear equations using the assumed variable. A linear equation is an equation which has the highest degree of variable as 2 and has only one solution. In order to find the value of the variable, the number of equations should be the same as the number of variables. For example, in this question, we have two equations and two variables so we can find the solution easily.