Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls

Question

I faced a very confusing question during my preparation for Mathematics Olympiad, here is the question:

The percentage of boys in a gathering, rounded to an integer, is $51$% and the percentage of girls in this gathering, rounded to an integer, is $49$%.

What is the minimum possible number of participants in this gathering?

Could anyone help me?

Answer 1

Let there be $b$ boys and $g$ girls, so $n=b+g$ in total. As $\frac12$ is exactly $50\,\%$ and $\frac{17}{33}=0.\overline{51}>0.515$ would be rounded to $52\,\%$, we conclude that the rational proportion $\frac bn$ is strictly between $\frac12$ and $\frac{17}{33}$. Hence the respective differences $$\frac bn-\frac12 = \frac{2b-n}{2n} \quad\text{and}\quad \frac{17}{33}-\frac bn = \frac{17n - 33b}{33n} $$ have a positive (and yet integer) numerator. Thus $$2b-n\ge1\quad\text{and}\quad 17n-33b\ge 1.$$ It follows that $$ 33(2b-n)+2(17n-33b)\ge 33+2$$ or simplified $$ n\ge 35.$$

(We also get $b=17(2b-n)+(17n-33b)\ge 18$; and indeed one verifies that with $b=18$ and $g=35-18=17$, we find suitable percentages).

Answer 2

The answer is $\boxed{35}$.

If the percentage of boys are rounded off to $51\%$, then we see that the percentage of boys is actually in the range $50.5 \leq \text{BoysPercentage} < 51.5$.

Hence, we focus on fractions of the form $\dfrac{n+1}{2n}, \dfrac{n+2}{2n}, \dfrac{n+2}{2n+1}, \dfrac{n+3}{2n+1}$.

A brief calculation using some logical deductions give that answer is $35$, where you have $18$ boys and $17$ girls.

If you want my logical deductions then please let me know.

Answer 3

Since an answer has been given, here is something which shows how I thought the problem through. What you need to learn is how to look, how to simplify, how to tease out what is important. This is not a perfect solution, but is how I really did it, in case that is helpful to you.

So I began by observing that with $100$ children I can get exact percentages, so the answer I want will be at most $100$.

Next I thought that with $50$ children I can get $48$% and $52$% but never the figures I want. That led me to think that the answer would likely be an odd number of children.

And then that seemed obvious anyway because clearly you want the number of boys to be one more than the number of girls for the tightest result, and if there were an even number of children the smallest non-zero difference would always be $2$.

So the difference between boys and girls must be one child. And to round correctly we need the girls to be at least $48.5$% and the boys to be at most $51.5$%. Now whatever your rounding convention, if it is consistent, one of these boundary figures will be rounded away from $50$%, so we have that one child must be less that the difference between the two - ie $3$%.

Well that means that the number of children must be at least $\dfrac {100}3\gt 33$. And the least odd number greater than $33$ is $35$. You know that this must work because the problem is symmetric about $50$%, but best to check just to make sure that you've not made a mistake.

None of the steps is difficult, but it is hard to see everything at the beginning. There is sometimes no alternative but getting stuck in and trying things out. Notice how even my first rather trivial and in the end mainly irrelevant observation at least brings the problem down to size.

Answer 4

Let $x$ be the number of boys , and $y$ be the number of girls.

From the text

$$\frac{50.5}{100} < \frac{x}{x+y} < \frac{51.5}{100} \iff $$

$$ 50.5y < 49.5x < 51.5y + x \iff $$

$$48.5x < 51.5y \ \ \land \ \ 50.5y< 49.5x \iff $$ $$\fbox{$y + \frac{2}{99}y < x < y+\frac{6}{97}y$}$$

So, it's easy to see that we need at least $\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} y =\ceil{\frac{97}{6}} = 17$ so to obtain an interval for $x$ containing a natural number. Therefore, we get $$17 + \frac{34}{99} < x < 18+\frac{5}{97} \implies x = 18$$