Henry Ernest Dudeney/Modern Puzzles/83 - A Critical Vote/Solution

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Modern Puzzles by Henry Ernest Dudeney: $83$

A Critical Vote
A meeting of a charitable society was held to decide whether the members should expand their operations.
It was arranged that during the count those in favour of the motion should remain standing,
and those who voted against should sit down.
"Ladies and gentlemen," said the chairman in due course, "I have the pleasure to announce that the motion is carried by a majority exactly equal to exactly a quarter of the opposition."
"Excuse me, sir," called somebody from the back, "but some of us over here could not sit down, because there are not enough chairs."
"Then those who wanted to sit down but couldn't are to hold up their hands ... I find there are a dozen of you, so the motion is lost by a majority of one."
Now, how many people voted at that meeting?


Solution

There were $207$ people at the meeting, of which $103$ voted in favour and $104$ against.


Proof

Let $n$ be the total number of people at the meeting.

Let $a$ be the number of people in favour of the motion.

Let $b$ be the number of people against the motion.

Let $m$ be the number of people who wanted to vote against the motion but could not, as they could not sit down.

We have:

\(\text {(1)}: \quad\) \(\ds n\) \(=\) \(\ds a + b\) from the definition of the problem
\(\text {(2)}: \quad\) \(\ds \paren {a + m} - \paren {b - m}\) \(=\) \(\ds \dfrac 1 4 \paren {b - m}\) "... the motion is carried by a majority exactly equal to exactly a quarter of the opposition."
\(\text {(3)}: \quad\) \(\ds m\) \(=\) \(\ds 12\) "Then those who wanted to sit down but couldn't are to hold up their hands ... I find there are a dozen of you ..."
\(\text {(4)}: \quad\) \(\ds b\) \(=\) \(\ds a + 1\) "... the motion is lost by a majority of one."
\(\text {(5)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 4 \paren {a - b + 2 m}\) \(=\) \(\ds b - m\) simplifying $(2)$
\(\ds \leadsto \ \ \) \(\ds 4 \paren {a - \paren {a + 1} + 24}\) \(=\) \(\ds \paren {a + 1} - 12\) substituting from $(3)$ and $(4)$ into $(5)$
\(\ds \leadsto \ \ \) \(\ds 96 - 4\) \(=\) \(\ds a - 11\) simplifying
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 103\) simplifying
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 104\) from $(4)$
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 103 + 104 = 207\) from $(1)$

$\blacksquare$


Sources

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