Election Votes Distribution

CAT 2022 Slot 2 · Quantitative Ability · Easy · Arithmetic

This is an easy Quantitative Ability question from the CAT 2022 Slot 2 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

In an election, there were four candidates and 80% of the registered voters casted 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 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

  1. A.

    40192

  2. B.

    60288

  3. C.

    50240

  4. D.

    62800

Answer

D

Explanation

Let total casted votes be VV. Candidate 1 receives 30%30\% of V=0.30VV = 0.30 V. Remaining votes = 70%70\% of V=0.70VV = 0.70 V. These are divided among Candidates 2, 3, 4 in ratio 1:2:31 : 2 : 3. Shares: 16×70%=11.67%\frac{1}{6} \times 70\% = 11.67\%, 26×70%=23.33%\frac{2}{6} \times 70\% = 23.33\%, 36×70%=35%\frac{3}{6} \times 70\% = 35\%. So the votes percentages of the 4 candidates are 30%,11.67%,23.33%,35%30\%, 11.67\%, 23.33\%, 35\%. The winner receives 35%35\% of VV. The second highest candidate receives 30%30\% of VV. Difference = 35%30%=5%35\% - 30\% = 5\% of V=2512V = 2512. 0.05V=2512    V=502400.05 V = 2512 \implies V = 50240. Since 80% of registered voters (RR) casted votes: 0.80R=50240    R=502400.8=628000.80 R = 50240 \implies R = \frac{50240}{0.8} = 62800.

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