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# An election percent problem

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In Ms. Pham's third grade class, an election with two candidates was held in which the losing candidate received 31% of the vote, expressed to the nearest whole percent. (If the percent is half-way between two integers, it is rounded up. For example, 47.5% is rounded up to 48%) Knowing that each student cast a vote for one or the other candidate, what is the minimum number of votes that could have been cast in the election?

Jul 18, 2019

#1
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The losing candidate could actually have received as little as 30.5%  of the vote  [ this would be rounded to 31% ]

If there were 100 votes cast, the losing candidate would have received  30.5 votes  [ not possible]

But, doubling this number of votes  = 200   and 30.5%  of 200  - 61 votes

So....the least number of votes that could have been cast  = 200   Jul 18, 2019
edited by CPhill  Jul 18, 2019
#3
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I think you misunderstood the question; it is asking the minimum total number of votes cast in the election

Guest Jul 18, 2019
#2
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CPhill: Where is the mistake in my approach?
Try dividing the number of students that voted for the loser who got 31% of the vote:
2/0.31 =~ 6 students. But:
2/6 =33% - so this doesn't work.
3/0.31 =~10 students. But:
3/10 =30% - so this doesn't work either.........and so on.
You will find that:
4 / 0.31 =~13 students. But:
4 / 13 =30.77% = ~31% - so this works. Therefore, the minimum number of students who voted = 13

Jul 18, 2019
#4
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I was taught, and I believe it is proper, that when the decimal fraction is exactly .5 then if the number preceding it is odd it rounds up, and if the number preceding it is even it rounds down.  Over the long term, there would be about the same number of odds and evens.

.

Jul 19, 2019
#5
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I was taught, and I believe it is proper, that when the decimal fraction is exactly .5 then if the number preceding it is odd it rounds up, and if the number preceding it is even it rounds down.

Good for you. You were taught correctly.  The rounding method you describe is still used in statistical analysis.

The floor function for odd numbers and the ceiling function for even numbers that are followed by an exact value of (5) are used to process collected data to prevent rounding bias for statistical analysis, and other data processing reasons. One notable exception is the data used for interest payouts for financial instruments, such as dividends on stocks, bank accounts, etc.  An ending “5” will always use the ceiling function regardless if the preceding number is even or odd. The reasons for this exception are interesting.

This however is a type of Diophantine question –meaning that it requires integers for solutions. Though derived from a statistical perspective, the nature of the question presented is minimally statistical, and the data used to solve it is not statistical –the variables are not random. The writer of the question included a parenthetical statement to clarify the exception for this question.

GA

GingerAle  Jul 20, 2019