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

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== Problem ==
 
== Problem ==
 
The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>.  
 
The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>.  
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<center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\
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<table border="1">
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\hline \text{number of contestants who caught} \ n \ \text{fish} & 9 & 5 & 7 & 23 & \dots & 5 & 2 & 1 \\
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<tr><td><math>n\,</math></td><td><math>0\,</math></td><td><math>1\,</math></td><td><math>2\,</math></td><td><math>3\,</math></td><td><math>\dots\,</math></td><td><math>13\,</math></td><td><math>14\,</math></td><td><math>15\,</math></td></tr>
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\hline \end{array}</math></center>
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<tr><td><math>\mbox{number of contestants who caught }n\mbox{ fish}\,</math></td><td><math>9\,</math></td><td><math>5\,</math></td><td><math>7\,</math></td><td><math>23\,</math></td><td><math>\dots\,</math></td><td><math>5\,</math></td><td><math>2\,</math></td><td><math>1\,</math></td></tr>
 
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</table>
 
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In the newspaper story covering the event, it was reported that  
 
In the newspaper story covering the event, it was reported that  
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:(a) the winner caught <math>15</math> fish;
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:(b) those who caught <math>3</math> or more fish averaged <math>6</math> fish each;
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:(c) those who caught <math>12</math> or fewer fish averaged <math>5</math> fish each.
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What was the total number of fish caught during the festival?
  
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(a) the winner caught 15 fish;
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== Solution 1==
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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.
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(b) those who caught 3 or more fish averaged 6 fish each;
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== Solution 2==
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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
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(c) those who caught 12 or fewer fish averaged 5 fish each.
 
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What was the total number of fish caught during the festival?
 
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== Solution ==
 
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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}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:05, 8 July 2022

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

See also

1993 AIME (Problems • Answer Key • Resources)
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