IPMAT Indore 2023 · Quantitative Ability · Question 9

Step 1 — name the total. Let N be the number of voters on the list. Keep every quantity as a percentage of N — that is what makes this clean.

Step 2 — track the votes.

Turned up: 85\% of N=0.85N

Invalid: 50

Valid votes: 0.85N-50

Winner: 0.44N (given directly as a percentage of the list, not of those who voted)

Loser: \left( 0.85N-50 \right)-0.44N=0.41N-50

Step 3 — use the margin.

\text{Winner}-\text{Loser}=0.44N-\left( 0.41N-50 \right)=0.03N+50

Setting this equal to 200:

0.03N+50=200 \Rightarrow 0.03N=150 \Rightarrow N=5000

Step 4 — check. Turnout 4250, invalid 50, valid 4200. Winner =0.44\left( 5000 \right)=2200, loser =2000. Margin 200 ✔ and 2200+2000=4200 ✔

Answer: 5000

Shortcut. Everything is a percentage of the list — so make the loser one too.

\text{Loser}=\underset{\text{turnout}}{\underset{⏟}{85\%}}-\underset{\text{winner}}{\underset{⏟}{44\%}}\text{of}N - 50=41\%N-50

The margin is then 3\% of N plus the 50 invalid votes (because those 50 were subtracted from the loser’s side only):

0.03N+50=200 \Rightarrow N=5000

The one thing to get right is that 44% is of the whole list, not of the valid votes. Misreading it as a share of valid votes is the error the question is built around.