In a village three people contested for the post of village Pradhan. Due to their own interest, all the voters voted and no one vote was invalid. The losing candidate got 30% votes. What could be the minimum absolute margin of votes by which the winning candidate led by the nearest rival, if each candidate got an integral per cent of votes?

### Concept & Logic This problem combines percentages with number properties (specifically, integers and divisibility). To find the minimum absolute margin of votes, we must minimize the difference in percentage between the top two candidates while ensuring that the percentage values and the total number of votes yield whole numbers (since partial votes are impossible). ### Step-by-Step Solution * **Given**: Three candidates (let's call them A, B, and C). Total votes = $100\%$. The losing candidate (C) got $30\%$ of the votes. * **Deduction**: 1. The remaining votes are shared between the winner (A) and the runner-up (B): $$A + B = 100\% - 30\% = 70\%$$ 2. To find the *minimum* margin, candidates A and B must have the closest possible percentage of votes. Since A is the winner, A's percentage must be strictly greater than B's percentage ($A > B$). 3. The closest integers that sum to $70$ with $A > B$ are: $$A = 36\%$$ $$B = 34\%$$ 4. The minimum percentage margin between the winner and the nearest rival is: $$\text{Margin \%} = 36\% - 34\% = 2\%$$ 5. We need the absolute margin of *votes*, not just the percentage. Let the total number of voters be $N$. $$\text{Margin in votes} = 2\% \text{ of } N = \frac{2}{100} \times N = \frac{N}{50}$$ 6. Because every candidate received an integer number of votes, $N$ must be a multiple of $50$ (this ensures $36\%$, $34\%$, and $30\%$ of $N$ all result in integers). 7. To get the absolute *minimum* margin, we take the smallest valid total population, which is $N = 50$. 8. Substitute $N = 50$ back into our margin equation: $$\text{Minimum Margin} = \frac{50}{50} = 1 \text{ vote}$$ ### Exam Strategy & Shortcut Once you deduce the $70\%$ is split into $36\%$ and $34\%$, you know the margin is $2\%$. Convert $2\%$ to a fraction immediately: $\frac{1}{50}$. For a fraction to represent a whole number of people (votes), the denominator gives you the minimum possible total population size. Thus, minimum total votes = $50$. If the population is $50$, and the margin is $\frac{1}{50}$ of the population, the margin is simply $1$. ### Common Pitfall The most dangerous trap is assuming the total number of votes is a standard $100$. If $N=100$, the margin would be $2$ votes, leading you to wrongly select option (b). The prompt asks for the *minimum absolute margin*, meaning you must find the smallest logically possible voting population that satisfies all integer constraints. ### Final Answer **Therefore, the correct answer is 1.**