When IS 16% Achievable?

Relevant wiki: Rational Numbers - Problem Solving

One possible ratio that would yield exactly $16\%$ would be $\frac{16}{100}$ . After simplification, we get $\frac{4}{25}$ , which means that also $25$ students could be in the group and $4$ of them would get A. Is there a fraction that is roughly the same but has smaller denominator? If yes, it has to have smaller numerator, too. That is, the numerator would have be $1,2$ or $3$ .

Let us check whether it is possible: For $3$ , we have $\frac{3}{n}$ between $15.5\%$ and $16.5\%$ . This gives $0.155<\frac{3}{n}<0.165$ , and from this we have $0.155n<3<0.165n$ . Thus, any $n$ such that $\frac{3}{0.165}$ < $18.2$ <n< $19.4$ < $\frac{3}{0.155}$ ) would suffice. Therefore, we have $\frac{3}{19}=15.7\%$ a possible solution.

Is there a solution for $2$ or $1$ ? If yes, $0.155<\frac{2}{n}<0.165$ or $0.155<\frac{1}{n}<0.165$ , respectivelly. From this, $12<\frac{2}{0.165}<n<\frac{2}{0.155}<13$ and $6<\frac{1}{0.165}<n<\frac{1}{0.155}<7$ , respectivelly. There is no integer between $12$ and $13$ and between $5$ and $6$ . This means there is no such solution for $2$ and $1$ . Therefore, $19$ students is the smallest group.

P.S.: I was inspired by Hejný, Milan. Teória vyučovania matematiky. 2. vyd. Bratislava: Slovenské pedagogické nakladateľstvo, 1990. 554 s. ISBN 80-08-01344-3.

You try 1 to 7.

1/1 == 100%

1/2 == 50%

1/3==33%

1/4==25%

...

1/6==17%

Then go to 2.

2/7 ==30%

2/9==22%

2/11==18%

Then so on.

Then, 3/19.