Two bus tickets from city A to city B and three tickets from city A to city C cost Rs 77, while three tickets from A to B and two tickets from A to C cost Rs 73. What are the individual fares from city A to B and from city A to C?

Introduction / Context:
This arithmetic reasoning question involves forming and solving simultaneous linear equations based on cost information. We have bus ticket fares from city A to B and from city A to C. Different combinations of these tickets are purchased, and their total costs are given. The goal is to find the individual fares. This type of problem is very common in bank exams, aptitude tests, and school-level algebra practice.

Given Data / Assumptions:

• Let the fare from city A to B be x rupees.
• Let the fare from city A to C be y rupees.
• Two tickets A to B plus three tickets A to C cost Rs 77, so 2x + 3y = 77.
• Three tickets A to B plus two tickets A to C cost Rs 73, so 3x + 2y = 73.
• We assume normal arithmetic with no discounts or hidden charges.

Concept / Approach:
We translate the verbal information into two linear equations in two variables. We then solve this system using either the elimination method or the substitution method. The elimination method is usually faster here: we can multiply equations to make the coefficients of one variable equal and then subtract to eliminate that variable. Once we find one fare, we substitute back to get the other.

Step-by-Step Solution:
Step 1: Write the equations clearly: 2x + 3y = 77 and 3x + 2y = 73.Step 2: Multiply the first equation by 3: 6x + 9y = 231.Step 3: Multiply the second equation by 2: 6x + 4y = 146.Step 4: Subtract the second new equation from the first new equation: (6x + 9y) - (6x + 4y) = 231 - 146.Step 5: This gives 5y = 85, so y = 17.Step 6: Substitute y = 17 into 2x + 3y = 77: 2x + 3*17 = 77, so 2x + 51 = 77.Step 7: Therefore 2x = 26 and x = 13.Step 8: The fares are Rs 13 from A to B and Rs 17 from A to C.

Verification / Alternative check:
We verify by substituting x = 13 and y = 17 back into the original statements. For two tickets to B and three to C: 2*13 + 3*17 = 26 + 51 = 77, which matches the first given total. For three tickets to B and two to C: 3*13 + 2*17 = 39 + 34 = 73, which matches the second total. Since both conditions are satisfied, our solution is fully consistent.

Why Other Options Are Wrong:
Option A (Rs. 4, Rs. 23): 2*4 + 3*23 = 8 + 69 = 77 works, but 3*4 + 2*23 = 12 + 46 = 58, not 73, so it fails the second equation.
Option C (Rs. 15, Rs. 14): 2*15 + 3*14 = 30 + 42 = 72, not 77, so it is invalid.
Option D (Rs. 17, Rs. 13): 2*17 + 3*13 = 34 + 39 = 73, which matches the second equation but not the first one.

Common Pitfalls:
Students sometimes mix up which equation corresponds to which combination of tickets, leading to wrong equations. Another common mistake is arithmetic error during elimination, especially when multiplying and subtracting equations. Writing each step clearly and double checking multiplication and subtraction prevents such errors.

Final Answer:
The fares are Rs 13 from city A to B and Rs 17 from city A to C, so the correct option is Rs. 13, Rs. 17.