From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is 77. But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is 124. What are the fares from Delhi to station A and to station B?

Answer

To solve the problem, we need to set up a system of equations based on the information given. Let's denote:

- \( x \) as the fare from Delhi to station A.
- \( y \) as the fare from Delhi to station B.

From the problem, we have the following two conditions:

1. Buying 2 tickets for station A and 3 tickets for station B costs 77 rupees:
\[ 2x + 3y = 77 \quad \text{(Equation 1)} \]

2. Buying 3 tickets for station A and 5 tickets for station B costs 124 rupees:
\[ 3x + 5y = 124 \quad \text{(Equation 2)} \]

Now, we will solve these equations step by step.

### Step 1: Multiply the equations to align coefficients
To eliminate one of the variables, we can multiply the equations so that the coefficients of \( x \) are the same. We will multiply Equation 1 by 3 and Equation 2 by 2:

- Multiply Equation 1 by 3:
\[ 3(2x + 3y) = 3(77) \implies 6x + 9y = 231 \quad \text{(Equation 3)} \]

- Multiply Equation 2 by 2:
\[ 2(3x + 5y) = 2(124) \implies 6x + 10y = 248 \quad \text{(Equation 4)} \]

### Step 2: Subtract the equations
Now, we will subtract Equation 3 from Equation 4:
\[ (6x + 10y) - (6x + 9y) = 248 - 231 \]
This simplifies to:
\[ y = 17 \]

### Step 3: Substitute \( y \) back to find \( x \)
Now that we have \( y = 17 \), we can substitute this value back into Equation 1 to find \( x \):
\[ 2x + 3(17) = 77 \]
This simplifies to:
\[ 2x + 51 = 77 \]
Subtracting 51 from both sides gives:
\[ 2x = 26 \]
Dividing by 2:
\[ x = 13 \]

### Final Answer
Thus, the fares are:
- Fare from Delhi to station A (\( x \)) = 13 rupees
- Fare from Delhi to station B (\( y \)) = 17 rupees