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Difference between revisions of "2023 AMC 12A Problems/Problem 8"

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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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{{duplicate|[[2023 AMC 10A Problems/Problem 10|2023 AMC 10A #10]] and [[2023 AMC 12A Problems/Problem 8|2023 AMC 12A #8]]}}
  
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-Jonathan Yu
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==Problem==
 +
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an <math>11</math> on the next quiz, her mean will increase by <math>1</math>. If she scores an <math>11</math> on each of the next three quizzes, her mean will increase by <math>2</math>. What is the mean of her quiz scores currently?
 +
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math>
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 +
 
 +
==Solution 1==
 +
 
 +
Let <math>a</math> represent the amount of tests taken previously and <math>x</math> the mean of the scores taken previously.
 +
 
 +
We can write the following equations:
 +
 
 +
<cmath>\frac{ax+11}{a+1}=x+1\qquad (1)</cmath>
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<cmath>\frac{ax+33}{a+3}=x+2\qquad (2)</cmath>
 +
 
 +
Multiplying equation <math>(1)</math> by <math>(a+1)</math> and solving, we get:
 +
<cmath>ax+11=ax+a+x+1</cmath>
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<cmath>11=a+x+1</cmath>
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<cmath>a+x=10\qquad (3)</cmath>
 +
 
 +
Multiplying equation <math>(2)</math> by <math>(a+3)</math> and solving, we get:
 +
<cmath>ax+33=ax+2a+3x+6</cmath>
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<cmath>33=2a+3x+6</cmath>
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<cmath>2a+3x=27\qquad (4)</cmath>
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Solving the system of equations for <math>(3)</math> and <math>(4)</math>, we find that <math>a=3</math> and <math>x=\boxed{\textbf{(D) }7}</math>.
 +
 
 +
~ Larry Page (Yes, it's really me)
 +
 
 +
==Solution 2 (Variation on Solution 1)==
 +
 
 +
Suppose Maureen took <math>n</math> tests with an average of <math>m</math>.
 +
 
 +
If she takes another test, her new average is <math>\frac{(nm+11)}{(n+1)}=m+1</math>
 +
 
 +
Cross-multiplying: <math>nm+11=nm+n+m+1</math>, so <math>n+m=10</math>.
 +
 
 +
If she takes <math>3</math> more tests, her new average is <math>\frac{(nm+33)}{(n+3)}=m+2</math>
 +
 
 +
Cross-multiplying: <math>nm+33=nm+2n+3m+6</math>, so <math>2n+3m=27</math>.
 +
 
 +
But <math>2n+3m</math> can also be written as <math>2(n+m)+m=20+m</math>. Therefore <math>m=27-20=\boxed{\textbf{(D) }7}</math>
 +
 
 +
~Dilip ~megaboy6679 (latex)
 +
 
 +
==Solution 3 (do this if you are bored)==
 +
 
 +
Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously.
 +
 
 +
So, <math>\frac{s+11}{t+1} = \frac{s}{t}+1</math> and <math>\frac{s+33}{t+3} = \frac{s}{t}+2</math>
 +
 
 +
We can use the first equation to write <math>s</math> in terms of <math>t</math>.
 +
 
 +
We then substitute this into the second equation: <math>\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10t}{t}+2</math>
 +
 
 +
From here, we should solve for t: multiply both sides by (<math>t</math>) and then (<math>t+3</math>), combining like terms to get <math>t^2-3t=0</math>. Factorize to get <math>t=0</math> or <math>t=3</math>, and therefore <math>t=3</math> (makes sense for the problem).
 +
 
 +
We substitute this to get <math>s=21</math>.
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Therefore, the solution to the problem is <math>\frac{21}{3}=</math> <math>\boxed{\textbf{(D) }7}</math>
 +
 
 +
~ Larry Page
 +
 
 +
==Solution 4 (Testing Answer Choices)==
 +
Let's consider all the answer choices. If the average is <math>8</math>, then, we can assume that all her test choices were <math>8</math>. We can see that she must have gotten <math>8</math> twice, in order for another score of <math>11</math> to bring her average up by one. However, adding three <math>11</math>'s will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is <math>\boxed{\textbf{(D) }7}</math>
 +
~andliu766 (minor wording edit by mihikamishra)
 +
 
 +
==Solution 5==
 +
Let <math>n</math> be the number of existing quizzes. So after one more test, score <math>11</math> has <math>n+1</math> extra points to distribute to <math>n+1</math> quizzes. Also, after three more quizzes, there will be <math>3(n+1)</math> extra points to distribute to the <math>n+3</math> quizzes. So <math>3n+3=2(n+3)</math>. This means <math>n=3</math>. <math>n+1</math> extra points means original mean (average) is <math>7</math>
 +
 
 +
~Pratima
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=_o_QjH-jv2f3b_T0&t=2074
 +
~little-fermat
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=fQ77Xhb7x1EP_Ieh&t=2361 ~Math-X
 +
 
 +
==Video Solution by Power Solve (easy to digest!)==
 +
https://www.youtube.com/watch?v=ZUgo3-BKt30
 +
 
 +
==Video Solution (🚀 Just 3 min 🚀)==
 +
https://youtu.be/J99XkR9tK74
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=1QJ_BQOfZtg
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/VzgNmdKp8UE
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
 
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=A|num-b=9|num-a=11}}
 +
{{AMC12 box|year=2023|ab=A|num-b=7|num-a=9}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 06:27, 5 November 2024

The following problem is from both the 2023 AMC 10A #10 and 2023 AMC 12A #8, so both problems redirect to this page.

Problem

Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$


Solution 1

Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.

We can write the following equations:

\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]

Multiplying equation $(1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]

Multiplying equation $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\]

Solving the system of equations for $(3)$ and $(4)$, we find that $a=3$ and $x=\boxed{\textbf{(D) }7}$.

~ Larry Page (Yes, it's really me)

Solution 2 (Variation on Solution 1)

Suppose Maureen took $n$ tests with an average of $m$.

If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$

Cross-multiplying: $nm+11=nm+n+m+1$, so $n+m=10$.

If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$

Cross-multiplying: $nm+33=nm+2n+3m+6$, so $2n+3m=27$.

But $2n+3m$ can also be written as $2(n+m)+m=20+m$. Therefore $m=27-20=\boxed{\textbf{(D) }7}$

~Dilip ~megaboy6679 (latex)

Solution 3 (do this if you are bored)

Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.

So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$

We can use the first equation to write $s$ in terms of $t$.

We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10t}{t}+2$

From here, we should solve for t: multiply both sides by ($t$) and then ($t+3$), combining like terms to get $t^2-3t=0$. Factorize to get $t=0$ or $t=3$, and therefore $t=3$ (makes sense for the problem).

We substitute this to get $s=21$.

Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{\textbf{(D) }7}$

~ Larry Page

Solution 4 (Testing Answer Choices)

Let's consider all the answer choices. If the average is $8$, then, we can assume that all her test choices were $8$. We can see that she must have gotten $8$ twice, in order for another score of $11$ to bring her average up by one. However, adding three $11$'s will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is $\boxed{\textbf{(D) }7}$ ~andliu766 (minor wording edit by mihikamishra)

Solution 5

Let $n$ be the number of existing quizzes. So after one more test, score $11$ has $n+1$ extra points to distribute to $n+1$ quizzes. Also, after three more quizzes, there will be $3(n+1)$ extra points to distribute to the $n+3$ quizzes. So $3n+3=2(n+3)$. This means $n=3$. $n+1$ extra points means original mean (average) is $7$

~Pratima

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=_o_QjH-jv2f3b_T0&t=2074 ~little-fermat

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=fQ77Xhb7x1EP_Ieh&t=2361 ~Math-X

Video Solution by Power Solve (easy to digest!)

https://www.youtube.com/watch?v=ZUgo3-BKt30

Video Solution (🚀 Just 3 min 🚀)

https://youtu.be/J99XkR9tK74

~Education, the Study of Everything


Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=1QJ_BQOfZtg

Video Solution

https://youtu.be/VzgNmdKp8UE

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


See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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