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Difference between revisions of "2024 AIME II Problems/Problem 1"

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==Problem==
 
==Problem==
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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.
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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.
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==Video Solution & More by MegaMath==
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https://www.youtube.com/watch?v=3ZuOIsVZhdA
  
 
==Solution 1==
 
==Solution 1==
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<math>w+z=229</math>, since we know <math>x=437</math> and <math>y=234</math>.  
 
<math>w+z=229</math>, since we know <math>x=437</math> and <math>y=234</math>.  
  
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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.  
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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 not 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>  
 
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>  
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==Solution 2==
 
==Solution 2==
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Let <math>a,b,c</math> 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 <math>x,y,z</math> 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 <math>n</math> denote the number of people that own all <math>4</math> items.
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hi guys welcoem back
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plz subsribce clooll
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<math>a+x+y+n=195</math> (the number of people that got diamond rings), <math>b+x+z+n=367</math> (the number of people that got golf clubs), <math>c+y+z+n=562</math> (the number of people that got garden spades). We also know <math>a+b+c=437</math> (the number of people that own two objects), and <math>x+y+z=234</math> (the number of people that own three objects). Adding the first three equations gives <cmath>a+b+c+2(x+y+z)+3n=1124.</cmath>
 
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Substituting the second two equations gives <math>437+2\cdot 234+3n=1124</math>, so <math>n=\boxed{073}.</math>
 
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~nezha33
 
  
 
==Solution 3==
 
==Solution 3==
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~Callisto531
 
~Callisto531
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==Solution 4==
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Let <math>a</math> be the number of people who have exactly one of these things and let <math>b</math> be the number of people who have exactlty four of these objects. We have <math>a + 437 + 234 + d = 900,</math> so <math>a + d = 229.</math>
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Including those who have more than one object, we have
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<cmath>195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.</cmath>
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This is because we count those who own exactly <math>2</math> objects twice, those who own <math>3</math> thrice, and those who own <math>4</math> four times. Solving gives <math>a + 4d = 448.</math>
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Solving the system <math>a + 4d = 448, a + d = 229</math> gives <math>3d = 219,</math> so <math>d = \boxed{\textbf{(073)}}.</math>
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-Benedict T (countmath1)
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==Video Solution==
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https://youtu.be/n60sIVSYWlc
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AIME box|year=2024|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2024|n=II|before=First Problem|num-a=2}}
  
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[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:45, 24 October 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.

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=3ZuOIsVZhdA

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 not 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

hi guys welcoem back plz subsribce clooll

Solution 3

We know that there are 195 diamond rings, 367 golf clubs, and 562 garden spades, so we can calculate that there are $195+367+562=1124$ items, with the exclusion of candy hearts which is irrelevant to the question. There are 437 people who owns 2 items, which means 1 item since candy hearts are irrelevant, and there are 234 people who own 2 items plus a bag of candy hearts, which means that the 234 people collectively own $234*2=468$ items. We can see that there are $1124-437-468=219$ items left, and since the question is asking us for the people who own 4 items, which means 3 items due to the irrelevance of candy hearts, we simply divide 219 by 3 and get $219/3=\boxed{073}$.

~Callisto531

Solution 4

Let $a$ be the number of people who have exactly one of these things and let $b$ be the number of people who have exactlty four of these objects. We have $a + 437 + 234 + d = 900,$ so $a + d = 229.$


Including those who have more than one object, we have \[195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.\] This is because we count those who own exactly $2$ objects twice, those who own $3$ thrice, and those who own $4$ four times. Solving gives $a + 4d = 448.$


Solving the system $a + 4d = 448, a + d = 229$ gives $3d = 219,$ so $d = \boxed{\textbf{(073)}}.$


-Benedict T (countmath1)

Video Solution

https://youtu.be/n60sIVSYWlc

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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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