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)
Line 11: Line 11:
  
 
== Solution ==
 
== Solution ==
−
Suppose that the number of fish is <math>x</math> and the number of contestants is <math>y</math>. The <math>y-(9+5+7)=y-21</math> fishers that caught <math>3</math> or more fish caught a total of <math>x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19</math> fish. Since they averaged <math>6</math> fish, <center><math>6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.</math></center> Similarily, those whom caught <math>12</math> or fewer fish averaged <math>5</math> fish per person, so <center><math>5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40.</math></center> Solving the two equation system, we find that <math>y = 175</math> and <math>x = \boxed{943}</math>, the answer.
+
Suppose that the number of fish is <math>x</math> and the number of contestants is <math>y</math>. The <math>y-(9+5+7)=y-21</math> fishers that caught <math>3</math> or more fish caught a total of <math>x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19</math> fish. Since they averaged <math>6</math> fish, <center><math>6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.</math></center> Similarily, those who caught <math>12</math> or fewer fish averaged <math>5</math> fish per person, so <center><math>5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40.</math></center> Solving the two equation system, we find that <math>y = 175</math> and <math>x = \boxed{943}</math>, the answer.
  
 
== See also ==
 
== See also ==

Revision as of 21:42, 30 May 2021

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 $n\,$.

$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ \hline \text{number of contestants who caught} \ n \ \text{fish} & 9 & 5 & 7 & 23 & \dots & 5 & 2 & 1 \\ \hline \end{array}$

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 $y$. The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,

$6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.$

Similarily, those who 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 $x = \boxed{943}$, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_3&oldid=154532"