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

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

@@ Line 9: / Line 9: @@
 ==Solution 2 (Kind of lame)==
-Since at the beginning, adults make up <math>\frac{5}{12}</math> of the concert, the amount of adults must be a multiple of 12.
+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.
 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.
@@ Line 23: / Line 23: @@
 ~hastapasta
+==Solution 3==
+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>.
+By cross-multiplying and simplifying, we get that <math>25(y-22)=\frac{7a}{5}</math>.
+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>.
+~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

Page

Toolbox

Search