For the Pupils of the Canton High School to Solve.
In June, 1880, the population of Canton was 12,258; ten years later it was 26,189; now, December, 1892, two and a half years later, the city is taking a special census. Supposing the rate of increase for these two and a half years to be the same as the average rate of increase for the ten years ending June, 1890, what is the population of Canton now? Publish the solution in the NEWS DEMOCRAT Friday evening, giving the answer to the nearest integer.
3. Publish the solution in the NEWS DEMOCRAT Friday evening, giving the answer to the nearest integer. So Papa Charles isn't going to make like King Solomon and offer to cut people up.
4. C. F. STOKEY, One of the Enumerators. I don't know what "One of the Enumerators" means. Conceivably it's a joke, or a reference to a club of some sort.
5. I hesitate to stick my neck out, but I think the answer is 33,630. If anybody else tries the problem, I'm sure they'll come up with a more correct answer.
1880 population: 12258
1890 population: 26189
Difference between 1890 and 1880: 13931
10 year Rate of increase: Difference divided by 1880 population: 1.136482297
1880 population plus 10 year rates times 1880 population makes 1890 population: 26189
2 1/2 year rate of increase: 10 year rate divided by 4: 0.2841205743
So 1892 population should be 1890 population plus 2 1/2 year rate times 1890 population: 33629.83372
Rounding the nearest integer makes 33630.
A quick google didn't tell me what Canton's actual population was in December 1892, or why they were doing a special census.