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Difference between revisions of "1998 AJHSME Problems/Problem 15"

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===Problem 15===
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==Problem 15==
  
 
Estimate the population of Nisos in the year 2050.
 
Estimate the population of Nisos in the year 2050.
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The population triples every <math>25</math> years from <math>200</math>, and there are <math>50</math> years between <math>2000</math> and <math>2050</math>, so we will have <math>200\times3=600\times3=1800</math>.
 
The population triples every <math>25</math> years from <math>200</math>, and there are <math>50</math> years between <math>2000</math> and <math>2050</math>, so we will have <math>200\times3=600\times3=1800</math>.
  
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<math>1800</math> can be rounded to <math>2000=\boxed{D}</math>
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This is an underestimate, since there are actually <math>2</math> more years for the island's population to grow.
  
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Therefore, <math>1800</math> can be rounded to <math> \text{(D)}\ 2000</math>
  
 
== See also ==
 
== See also ==
−
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
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{{AJHSME box|year=1998|num-b=14|num-a=16}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 19:55, 11 May 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 15

Estimate the population of Nisos in the year 2050.

$\text{(A)}\ 600 \qquad \text{(B)}\ 800 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 2000 \qquad \text{(E)}\ 3000$

Solution

The population triples every $25$ years from $200$, and there are $50$ years between $2000$ and $2050$, so we will have $200\times3=600\times3=1800$.

This is an underestimate, since there are actually $2$ more years for the island's population to grow.

Therefore, $1800$ can be rounded to $\text{(D)}\ 2000$

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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. AMC logo.png

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