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 2)
(→‎Solution 3)
 
Line 12: Line 12:
 
== Solution 1==
 
== Solution 1==
 
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.
 
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.
−
== Solution 2==
+
== Solution 2 ==
 
Let <math>f</math> be the total number of fish caught by the contestants who didn't catch <math>0, 1, 2, 3, 13, 14</math>, or <math>15</math> fish and let <math>a</math> be the number of contestants who didn't catch <math>0, 1, 2, 3, 13, 14</math>, or <math>15</math> fish. From <math>\text{(b)}</math>, we know that <math>\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9</math>. From <math>\text{(c)}</math> we have <math>\frac{f+69+14+5}{a+44}=5\implies f=5a+132</math>. Using these two equations gets us <math>a=123</math>. Plug this back into the equation to get <math>f=747</math>. Thus, the total number of fish caught is <math>5+14+69+f+65+28+15=\boxed{943}</math> - Heavytoothpaste
 
Let <math>f</math> be the total number of fish caught by the contestants who didn't catch <math>0, 1, 2, 3, 13, 14</math>, or <math>15</math> fish and let <math>a</math> be the number of contestants who didn't catch <math>0, 1, 2, 3, 13, 14</math>, or <math>15</math> fish. From <math>\text{(b)}</math>, we know that <math>\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9</math>. From <math>\text{(c)}</math> we have <math>\frac{f+69+14+5}{a+44}=5\implies f=5a+132</math>. Using these two equations gets us <math>a=123</math>. Plug this back into the equation to get <math>f=747</math>. Thus, the total number of fish caught is <math>5+14+69+f+65+28+15=\boxed{943}</math> - Heavytoothpaste
 +
 +
== Solution 3 ==
 +
Let <math>x</math> be the average number of fish caught among <math>n</math> of those who caught <math>4</math> to <math>12</math> fishes. According to statement B:
 +
<cmath>\frac{xn + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15}{n + 31} = 6</cmath>
 +
<cmath>\implies xn - 6n = 9</cmath>
 +
According to statement C:
 +
<cmath>\frac{xn + 0 \cdot 9 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23}{n + 44} = 5</cmath>
 +
<cmath>\implies xn - 5n = 132</cmath>
 +
We then solve for <math>n</math> and then <math>xn</math>. Note <math>xn</math> is the number of fish caught by the <math>n</math> people we defined above.
 +
<cmath>n = 123</cmath>
 +
<cmath>xn = 132 + 5(123) = 747</cmath>
 +
Thus, the total number of fish caught is
 +
<cmath>747 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15 \cdot 1 = \boxed{943}</cmath> ~Totient4Breakfast
  
 
== See also ==
 
== See also ==

Latest revision as of 10:23, 19 March 2025

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 1

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.

Solution 2

Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish. From $\text{(b)}$, we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$. From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f=5a+132$. Using these two equations gets us $a=123$. Plug this back into the equation to get $f=747$. Thus, the total number of fish caught is $5+14+69+f+65+28+15=\boxed{943}$ - Heavytoothpaste

Solution 3

Let $x$ be the average number of fish caught among $n$ of those who caught $4$ to $12$ fishes. According to statement B: \[\frac{xn + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15}{n + 31} = 6\] \[\implies xn - 6n = 9\] According to statement C: \[\frac{xn + 0 \cdot 9 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23}{n + 44} = 5\] \[\implies xn - 5n = 132\] We then solve for $n$ and then $xn$. Note $xn$ is the number of fish caught by the $n$ people we defined above. \[n = 123\] \[xn = 132 + 5(123) = 747\] Thus, the total number of fish caught is \[747 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15 \cdot 1 = \boxed{943}\] ~Totient4Breakfast

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=245427"