Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 AIME Problems/Problem 3"

(solution)
m (→‎Solution: -of)
Line 18: Line 18:
  
 
== Solution ==
 
== Solution ==
−
Suppose that the number of fish is <math>x</math> and the number of contestants is <math>y</math>. Of those which caught 3 or more fish (<math>y - 21</math> did that), <math>x - \left(0(9) + 1(5) + 2(7)\right) = x - 19</math> fish were caught. Since they averaged 6 fish, <math>6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126</math>. Similarily, of those which caught 12 or fewer fish averaged 5 fish per person, so <math>5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40</math>. Solving the two equation system, we find that <math>y = 175</math> and <math>x = 943</math>, the answer.
+
Suppose that the number of fish is <math>x</math> and the number of contestants is <math>y</math>. Of those which caught 3 or more fish (<math>y - 21</math> did that), <math>x - \left(0(9) + 1(5) + 2(7)\right) = x - 19</math> fish were caught. Since they averaged 6 fish, <math>6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126</math>. Similarily, those whom caught 12 or fewer fish averaged 5 fish per person, so <math>5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40</math>. Solving the two equation system, we find that <math>y = 175</math> and <math>x = 943</math>, the answer.
−
+
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1993|num-b=2|num-a=4}}
 
{{AIME box|year=1993|num-b=2|num-a=4}}

Revision as of 16:04, 26 March 2007

Problem

The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$.

$n\,$$0\,$$1\,$$2\,$$3\,$$\dots\,$$13\,$$14\,$$15\,$
$\mbox{number of contestants who caught }n\mbox{ fish}\,$$9\,$$5\,$$7\,$$23\,$$\dots\,$$5\,$$2\,$$1\,$

In the newspaper story covering the event, it was reported that

(a) the winner caught 15 fish;

(b) those who caught 3 or more fish averaged 6 fish each;

(c) those who caught 12 or fewer fish averaged 5 fish each.

What was the total number of fish caught during the festival?

Solution

Suppose that the number of fish is $x$ and the number of contestants is $y$. Of those which caught 3 or more fish ($y - 21$ did that), $x - \left(0(9) + 1(5) + 2(7)\right) = x - 19$ fish were caught. Since they averaged 6 fish, $6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126$. Similarily, those whom caught 12 or fewer fish averaged 5 fish per person, so $5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40$. Solving the two equation system, we find that $y = 175$ and $x = 943$, the answer.

See also

1993 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_3&oldid=14205"