Rounded percentages

Because of the special rule for rounding that preserves symmetry, $q(p)$ and $q(100\% - p)$ will alway be equal.

Proof: If one of them was smaller, say $q(p) = q(100\% - p) - x$ , so $\frac a {q(100\% - p)-x} \approx p$ , then $100\% - p$ could also be written as $100\% - p \approx 100\% - \frac a {q(100\% - p) - x} = \frac {q(100\% - p) - x} {q(100\% - p) - x} - \frac a {q(100\% - p) - x} = \frac {q(100\% - p) - x - a} {q(100\% - p) - x}$ . But both $q(100\% - p) - x - a$ and $q(100\% - p) - x$ are integers, so this would be a better way to write $100\% - p$ as a rounded fraction and would imply $q(100\% - p) = q(100\% - p) - x = q(p)$ . This now leads to a contradiction because we assumed $q(p) < q(100\% - p)$ and got $q(p) = q(100\% - p)$ , and therefore $q(p) = q(100\% - p)$ .

Since two values of $q$ are always equal, the 10 largest values we are searching are made up of 5 pairs $(p, 100\% - p)$ . But the sum in each of these pairs is simply $100\%$ and so the sum of 5 pairs is $5 * 100\% = \boxed{500\%}$ .

By the special round rule, if $q(x\%) = b$ because $\frac{a}{b} \approx x\%, \frac{1-a}{b}\approx (100-x)\%,$ so $q((100-x)\%) \leq q(x\%)$ . By symmetry we get equality.

By assuming uniqueness of an answer (assuming that the 10th largest $q(p)$ is greater than the 11th). We pair up the top 10 to form 5 pairs adding to 100%. Hence 500%.

Brute force. The solution set is: {{3,1,29},{34,10,29},{66,19,29},{97,28,29},{49,17,35},{51,18,35},{2,1,41},{98,40,41},{1,1,67},{99,66,67}}, where the first element of a triple is the percentage, the second is the numerator and the third is the denominator. As Henry U stated, they match up in pairs.