Question

In a panchayat election, there were four candidates, and 80% of the total voters cast their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion of 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was

• A. 62800
• B. 40192
• C. 50240
• D. 60820

Answer 1

The Correct Option is A

Let the total number of registered voters be \( V \). The number of votes cast is \( 0.8V \). The winner received 30% of the cast votes, so:
\[ \text{Winner's Votes} = 0.30 \times 0.8V = 0.24V \]
The remaining votes are split in the ratio 1 : 2 : 3. Let the total remaining votes be \( R \). Then:
\[ R = 0.8V - 0.24V = 0.56V \]
The votes for the other candidates are in the ratio 1 : 2 : 3, so the total number of parts is \( 1 + 2 + 3 = 6 \). Each part is:
\[ \text{Each part} = \frac{0.56V}{6} \]
The votes for the second highest candidate is:
\[ \text{Second Highest Votes} = 2 \times \frac{0.56V}{6} = \frac{0.56V}{3} \]
The difference in votes between the winner and the second highest candidate is:
\[ 0.24V - \frac{0.56V}{3} = 2512 \]
Solving for \( V \):
\[ \frac{72V}{300} - \frac{0.56V}{3} = 2512 \quad \Rightarrow \quad \frac{0.24V \times 3}{3} - \frac{0.56V}{3} = 2512 \]
Rearranged correctly:
\[ 0.24V - \frac{0.56V}{3} = 2512 \]
Multiply both sides by 3:
\[ 3 \times 0.24V - 0.56V = 3 \times 2512 \]
\[ 0.72V - 0.56V = 7536 \]
\[ 0.16V = 7536 \quad \Rightarrow \quad V = \frac{7536}{0.16} = 47100 \]

In election problems involving percentages and ratios, break the total into parts based on the given proportions and solve accordingly.