Difference between revisions of "1993 AIME Problems/Problem 3"
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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. | 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 | + | 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 |
== See also == | == See also == |
Latest revision as of 23:05, 8 July 2022
Contents
Problem
The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught fish for various values of .
In the newspaper story covering the event, it was reported that
- (a) the winner caught fish;
- (b) those who caught or more fish averaged fish each;
- (c) those who caught or fewer fish averaged fish each.
What was the total number of fish caught during the festival?
Solution 1
Suppose that the number of fish is and the number of contestants is . The fishers that caught or more fish caught a total of fish. Since they averaged fish,
Similarily, those who caught or fewer fish averaged fish per person, so
Solving the two equation system, we find that and , the answer.
Solution 2
Let be the total number of fish caught by the contestants who didn't catch , or fish and let be the number of contestants who didn't catch , or fish. From , we know that . From we have . Using these two equations gets us . Plug this back into the equation to get . Thus, the total number of fish caught is - Heavytoothpaste
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 |
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