IPMAT Indore 2024 (MCQ) - Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency | PYQs + Solutions

IPMAT Indore 2024

DILR

Arrangements

Hard

In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituency there was a clear winner. The following additional information is available:

- Total number of votes obtained by A, B, and C across all constituencies are 49, 35 and 16 respectively.

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- S2 and S3 were won by C while A won only S1.

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- Number of votes obtained by B in S1, S2, S3, S4, and S5 are distinct natural numbers in increasing order.

Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency

Correct Option: 4

\begin{array}{|c|c|c|c|c|c|} \hline &&A&B&C&Total \\ \hline 20 & S _1 \text{ (Winner: A)} & 19& 1 & 0 &20\\ \hline 20 & S_2 \text{ (Winner: C)} & & & &20 \\ \hline 20 & S_3 \text{ (Winner: C)}& & & & 20 \\ \hline 20 & S_4 \text{ (Winner: B)}& & & &20 \\ \hline 20 & S_5 \text{ (Winner: B)}& & & &20\\ \hline & Total & 49 & 35 & 16 &100\\ \hline \end{array}

Since B has distinct natural (i.e. not zero) numbers of votes, we can start by giving it 1 vote in S1. Since we know A wins S1, we can give it 19 votes.

\begin{array}{|c|c|c|c|c|c|} \hline &&A&B&C&Total \\ \hline 20 & S _1 \text{ (Winner: A)} & 19& 1 & 0 &20\\ \hline 20 & S_2 \text{ (Winner: C)} & & & &20 \\ \hline 20 & S_3 \text{ (Winner: C)}& & & & 20 \\ \hline 20 & S_4 \text{ (Winner: B)}& 9&11 &0 &20 \\ \hline 20 & S_5 \text{ (Winner: B)}& 8& 12& 0 &20\\ \hline & Total & 49 & 35 & 16 &100\\ \hline \end{array}

Since we know that S4 and S5 are won by B, we can give C zero votes and give majority to B and remaining to A (since C has a total of only 16 votes).

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B is increasing, hence 11 in S4, and then 12 in S5.

\begin{array}{|c|c|c|c|c|c|} \hline &&A&B&C&Total \\ \hline 20 & S _1 \text{ (Winner: A)} &19 & 1 &0 &20\\ \hline 20 & S_2 \text{ (Winner: C)} &7 &5 &8 &20 \\ \hline 20 & S_3 \text{ (Winner: C)}&6 &6 &8 &20 \\ \hline 20 & S_4 \text{ (Winner: B)}& 9 &11 &0 &20 \\ \hline 20 & S_5 \text{ (Winner: B)}& 8 & 12 & 0 &20\\ \hline & Total & 49 & 35 & 16 &100\\ \hline \end{array}

Then, we can distribute C equally among S2 and S3 (8 each). We have to distribute them such that C is the clear winner of both constituences, while also making sure B has votes in increasing order. 5 and 6 work for B. Any other values would tamper with the total given values for A. Hence, the set above is our set.

From the set, A got the lowest votes in S3, being 6 votes,

IPMAT Indore 2024 (MCQ) PYQs

Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency

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