IPMAT Indore 2024 (MCQ) - The number of votes obtained by A in S5 is | PYQs + Solutions | AfterBoards
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IPMAT Indore 2024 (MCQ) PYQs

IPMAT Indore 2024

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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.\newline - S2 and S3 were won by C while A won only S1. \newline - Number of votes obtained by B in S1, S2, S3, S4, and S5 are distinct natural numbers in increasing order.

The number of votes obtained by A in S5 is

Correct Option: 3
ABCTotal20S1 (Winner: A)19102020S2 (Winner: C)2020S3 (Winner: C)2020S4 (Winner: B)2020S5 (Winner: B)20Total493516100\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.
ABCTotal20S1 (Winner: A)19102020S2 (Winner: C)2020S3 (Winner: C)2020S4 (Winner: B)91102020S5 (Winner: B)812020Total493516100\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). \newline B is increasing, hence 11 in S4, and then 12 in S5.
ABCTotal20S1 (Winner: A)19102020S2 (Winner: C)7582020S3 (Winner: C)6682020S4 (Winner: B)91102020S5 (Winner: B)812020Total493516100\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, we can see that A obtained 8 votes in S5.

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