If a + 1/b = 1 and b + 1/c = 1 for non zero real numbers a, b, and c, then what is the value of the expression c + 1/a?

Introduction / Context:
This question tests your skill in solving simple algebraic equations involving reciprocals and then combining the results to evaluate another expression. Rather than solving for each variable explicitly, you can manipulate the given equalities to find c + 1/a directly. This type of problem appears frequently in aptitude tests where clever algebraic manipulation is rewarded over brute-force calculation.

Given Data / Assumptions:

• a + 1/b = 1.
• b + 1/c = 1.
• a, b, and c are non zero real numbers so that all reciprocals are defined.
• We must find the value of c + 1/a.

Concept / Approach:
We first express a in terms of b from the first equation and c in terms of b from the second equation. Then we compute 1/a and add it to c. A key simplification step is to recognise that terms like 1/(1 − b) and b/(b − 1) are related by a sign, which ultimately causes a cancellation that leads to a neat constant value. This approach avoids unnecessarily solving a full system of three equations.

Step-by-Step Solution:
From a + 1/b = 1, rearrange to get a = 1 − 1/b.Write a over a common denominator: a = (b − 1)/b.Then 1/a = b/(b − 1).From b + 1/c = 1, rearrange to 1/c = 1 − b.Thus c = 1 / (1 − b).Now consider c + 1/a = 1 / (1 − b) + b / (b − 1).Note that 1 − b = −(b − 1), so 1 / (1 − b) = −1 / (b − 1).Therefore c + 1/a = −1 / (b − 1) + b / (b − 1) = (b − 1) / (b − 1) = 1.

Verification / Alternative check:
Choose a specific value of b that keeps denominators non zero, for example b = 2. From a + 1/b = 1, we have a + 1/2 = 1, so a = 1/2. From b + 1/c = 1, we get 2 + 1/c = 1, so 1/c = −1 and c = −1. Now compute c + 1/a: c + 1/a = −1 + 1/(1/2) = −1 + 2 = 1. This numerical check agrees with the algebraic result.

Why Other Options Are Wrong:

• Values 0, −1, 2, and 3 would require different algebraic relationships between a, b, and c that are not implied by the given equations.
• The structure of the expressions guarantees a cancellation that leads to a constant 1, not 0 or any other integer.

Common Pitfalls:

• Handling signs incorrectly when writing 1/(1 − b) in terms of (b − 1).
• Trying to solve for each variable separately instead of focusing directly on c + 1/a.
• Algebraic errors when combining fractions with denominators (1 − b) and (b − 1).

Final Answer:
1