Difference between revisions of "2024 AIME II Problems/Problem 1"
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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> | ||
-Westwoodmonster | -Westwoodmonster | ||
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==Solution 3== | ==Solution 3== | ||
Revision as of 20:40, 22 November 2024
Contents
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 
denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know 
, since there are 900 residents in total. This simplifies to
, since we know 
and 
.
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, 
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 
. Solving and , we get 
-Westwoodmonster
Solution 3
We know that there are 195 diamond rings, 367 golf clubs, and 562 garden spades, so we can calculate that there are 
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 
items. We can see that there are 
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 
.
~Callisto531
Solution 4
Let 
be the number of people who have exactly one of these things and let 
be the number of people who have exactlty four of these objects. We have 
so 
Including those who have more than one object, we have
![\[195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.\]](http://latex.artofproblemsolving.com/2/4/9/2496e6635d55828094e22a49b32d7bc9d89ad3ed.png)
This is because we count those who own exactly 
objects twice, those who own 
thrice, and those who own 
four times. Solving gives 
Solving the system 
gives 
so 
-Benedict T (countmath1)
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
