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Difference between revisions of "2022 AIME II Problems/Problem 1"

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~eamo
 
~eamo
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==Solution 2 (Kind of lame)==
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Since at the beginning, adults make up <math>\frac{5}{12}</math> of the concert, the amount of people must be a multiple of 12.
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Call the amount of people in the beginning <math>x</math>.Then <math>x</math> must be divisible by 12, in other words: <math>x</math> must be a multiple of 12.
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Since after 50 more people arrived, adults make up <math>\frac{11}{25}</math> of the concert, <math>x+50</math> is a multiple of 25. This means <math>x+50</math> must be a multiple of 5.
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Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), <math>x</math> must end in 0.
 +
 +
Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that <math>x=300</math> is the minimum number of people at the concert.
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So therefore, after 50 more people arrive, there are <math>300+50=350</math> people at the concert, and the number of adults is <math>350*\frac{11}{25}=154</math>. Therefore the answer is <math>\boxed{154}</math>.
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I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.
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~hastapasta
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==Solution 3==
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Let <math>a</math> be the number of adults before the bus arrived and <math>x</math> be the total number of people at the concert. So, <math>\frac{a}{x}=\frac{5}{12}</math>. Solving for <math>x</math> in terms of <math>a</math>, <math>x = \frac{12}{5}a</math>. After the bus arrives, let's say there are an additional <math>y</math> adults out of the 50 more people who enter the concert. From that, we get <math>\frac{a+y}{x+50}=\frac{11}{25}</math>. Replacing <math>x</math> with the value of <math>a</math>, the second equation becomes <math>\frac{a+y}{\frac{12}{5}a+50}=\frac{11}{25}</math>.
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By cross-multiplying and simplifying, we get that <math>25(y-22)=\frac{7a}{5}</math>.
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Observe that we must make sure <math>y-22</math> is positive and divisible by <math>7</math> to have an integer value of <math>a</math>. The smallest possible value of <math>y</math> that satisfies this conditions is <math>29</math>. Plugging this into the equation, <math>a = 125</math>. The question asks for the minimum number of adults that are there after the bus arrives, which is <math>a+y</math>. Thus, the answer is simply <math>125+29=\boxed{154}</math>.
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~mathical8
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==Video Solution (Mathematical Dexterity)==
 +
https://www.youtube.com/watch?v=gBIxZ6SUr_w
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==Video Solution by Power of Logic==
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https://youtu.be/ZRnMlqgAJVM
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==Video Solution by WhyMath==
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https://youtu.be/2TfPEFpopTs
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2022|n=II|before=First Problem|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:34, 19 November 2023

Problem

Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution 1

Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$, and since we are looking for the minimum amount of people, $x$ is $300$. That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.

~eamo

Solution 2 (Kind of lame)

Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.

Call the amount of people in the beginning $x$.Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12. Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.

Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.

Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.

So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\frac{11}{25}=154$. Therefore the answer is $\boxed{154}$.

I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.

~hastapasta

Solution 3

Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\frac{a}{x}=\frac{5}{12}$. Solving for $x$ in terms of $a$, $x = \frac{12}{5}a$. After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From that, we get $\frac{a+y}{x+50}=\frac{11}{25}$. Replacing $x$ with the value of $a$, the second equation becomes $\frac{a+y}{\frac{12}{5}a+50}=\frac{11}{25}$.

By cross-multiplying and simplifying, we get that $25(y-22)=\frac{7a}{5}$.

Observe that we must make sure $y-22$ is positive and divisible by $7$ to have an integer value of $a$. The smallest possible value of $y$ that satisfies this conditions is $29$. Plugging this into the equation, $a = 125$. The question asks for the minimum number of adults that are there after the bus arrives, which is $a+y$. Thus, the answer is simply $125+29=\boxed{154}$.

~mathical8

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=gBIxZ6SUr_w

Video Solution by Power of Logic

https://youtu.be/ZRnMlqgAJVM

Video Solution by WhyMath

https://youtu.be/2TfPEFpopTs

~savannahsolver

See Also

2022 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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