Difference between revisions of "2024 AIME II Problems/Problem 1"
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− | + | ==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 <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 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> | ||
+ | -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 <math>195+367+562=1124</math> 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 <math>234*2=468</math> items. We can see that there are <math>1124-437-468=219</math> 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 <math>219/3=\boxed{073}</math>. | ||
+ | |||
+ | ~Callisto531 | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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> | ||
+ | |||
+ | |||
+ | Including those who have more than one object, we have | ||
+ | <cmath>195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.</cmath> | ||
+ | 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> | ||
+ | |||
+ | |||
+ | Solving the system <math>a + 4d = 448, a + d = 229</math> gives <math>3d = 219,</math> so <math>d = \boxed{\textbf{(073)}}.</math> | ||
+ | |||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/n60sIVSYWlc | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|n=II|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:45, 24 October 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 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 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
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.