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Difference between revisions of "2024 AIME II Problems/Problem 1"
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==Solution 1==
 
==Solution 1==
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Let <math>w,x,y,z</math> denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know <math>w+x+y+z=900</math>, since there are 900 residents in total. This simplifies to <math>w+z=229</math>, since we know <math>x=437</math> and <math>y=234</math>. Now, we set an equation of the total number of items. We know there are 195 rings, 367 clubs, 562 spades, and 900 candy hearts. Adding these up, there are 2024 (wow! the year!) items in total. Thus, <math>w+2x+3y+4z=2024</math> since we are adding the number of items each group of people contributes, and this must be equal to the total number of items. Plugging in x and y once more, we get <math>w+4z=448</math>. Solving <math>w+z=229</math> and <math>w+4z=448</math>, we get <math>z=\boxed{073}</math>  
+
Let <math>w,x,y,z</math> denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know <math>w+x+y+z=900</math>, since there are 900 residents in total. This simplifies to  
 +
 
 +
<math>w+z=229</math>, since we know <math>x=437</math> and <math>y=234</math>.  
 +
 
 +
Now, we set an equation of the total number of items. We know there are 195 rings, 367 clubs, 562 spades, and 900 candy hearts. Adding these up, there are 2024 (wow! the year!) items in total. Thus, <math>w+2x+3y+4z=2024</math> since we are adding the number of items each group of people contributes, and this must be equal to the total number of items.  
 +
 
 +
Plugging in x and y once more, we get <math>w+4z=448</math>. Solving <math>w+z=229</math> and <math>w+4z=448</math>, we get <math>z=\boxed{073}</math>  
 
-Westwoodmonster
 
-Westwoodmonster
  

Revision as of 21:54, 8 February 2024

Problem

Among the 900 residents of Aimeville, there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts. There are 437 residents who own exactly two of these things, and 234 residents who own exactly three of these things. Find the number of residents of Aimeville who own all four of these things :D.

Solution 1

Let $w,x,y,z$ denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know $w+x+y+z=900$, since there are 900 residents in total. This simplifies to

$w+z=229$, since we know $x=437$ and $y=234$.

Now, we set an equation of the total number of items. We know there are 195 rings, 367 clubs, 562 spades, and 900 candy hearts. Adding these up, there are 2024 (wow! the year!) items in total. Thus, $w+2x+3y+4z=2024$ since we are adding the number of items each group of people contributes, and this must be equal to the total number of items.

Plugging in x and y once more, we get $w+4z=448$. Solving $w+z=229$ and $w+4z=448$, we get $z=\boxed{073}$ -Westwoodmonster

Solution 2

Let $a,b,c$ denote the number of residents that own only a diamond ring and a bag of candy hearts, the number of residents that own only a golf club and a bag of candy hearts, and the number of residents that own only a garden spade and a bag of candy hearts, respectively. Let $x,y,z$ denote the number of residents that own only a diamond ring, a golf club, and a bag of candy hearts; the number of residents that own only a diamond ring, a garden spade, and a bag of candy hearts; and the number of residents that own only a golf club, a garden spade, and a bag of candy hearts. Let $n$ denote the number of people that own all $4$ items.

$a+x+y+n=195$ (the number of people that got diamond rings), $b+x+z+n=367$ (the number of people that got golf clubs), $c+y+z+n=562$ (the number of people that got garden spades). We also know $a+b+c=437$ (the number of people that own two objects), and $x+y+z=234$ (the number of people that own three objects). Adding the first three equations gives \[a+b+c+2(x+y+z)+3n=1124.\]

Substituting the second two equations gives $437+2\cdot 234+3n=1124$, so $n=\boxed{073}.$ ~nezha33

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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