Difference between revisions of "1998 AJHSME Problems/Problem 17"
m (→Solution 2) |
Hashtagmath (talk | contribs) (→Problem 17) |
||
Line 7: | Line 7: | ||
</center> | </center> | ||
− | + | ==Problem 17== | |
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support? | In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support? |
Revision as of 16:52, 17 April 2021
Don't Crowd the Isles
Problems 15, 16, and 17 all refer to the following:
In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.
Problem 17
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
Solution 1
We can divide the total area by how much will be occupied per person:
can stay on the island at its maximum capacity.
We can divide this by the current population the island in the year to see by what factor the population increases:
-fold increase in population.
Thus, the population increases by a factor . This is very close to , and so there are about triplings of the island's population.
It takes years to triple the island's population four times in succession.
Solution 2
We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. can live on the island at its maximum capacity.
After the year , it will not be possible for the next increase to occur, because when is tripled, it is way more than the maximum capacity.
Thus, the answer is , or
See also
1998 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.