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

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Expanding, <math>ax+11 = ax+a+x+1</math> and <math>ax+33 = ax+2a+3x+6</math>.  
 
Expanding, <math>ax+11 = ax+a+x+1</math> and <math>ax+33 = ax+2a+3x+6</math>.  
  
This gives us <math>a+x = 10</math> and <math>2a+3x = 27</math>. Solving for each variable, <math>x=7</math> and <math>a=3</math>. The answer is <math>\boxed{\textbf{(D) }7}</math>
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This gives us <math>a+x = 10</math> and <math>2a+3x = 27</math>. Solving for each variable, <math>a=3</math> and <math>x=\boxed{\textbf{(D) }7}</math>.
  
 
~walmartbrian ~Shontai ~andyluo
 
~walmartbrian ~Shontai ~andyluo
  
==Solution 1 with slight variation==
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==Solution 2 (Variation on Solution 1)==
  
n tests with an average of m
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Suppose Maureen took n tests with an average of m
  
When she takes another test her new average, m+1, is (nm + 11)/(n+1)
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When she takes another test her new average, <math>m+1</math>, is <math>\frac{(nm + 11)}{(n+1)}</math>
  
Cross-multiplying, nm + 11 = nm + n + m + 1; n+m = 10 -- I'll use this in a moment
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Cross-multiplying, nm + 11 = nm + n + m + 1; n+m = 10 -- I'll use this in a moment.
  
 
When she takes 3 more tests, the situation is (nm + 33)/(n+3) = m+2
 
When she takes 3 more tests, the situation is (nm + 33)/(n+3) = m+2
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~Dilip
 
~Dilip
  
==Solution 2==
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==Solution 3==
  
 
Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously.
 
Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously.

Revision as of 14:50, 10 November 2023

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:

$\frac{ax+11}{a+1} = x+1$ and $\frac{ax+33}{a+3} = x+2$.

Expanding, $ax+11 = ax+a+x+1$ and $ax+33 = ax+2a+3x+6$.

This gives us $a+x = 10$ and $2a+3x = 27$. Solving for each variable, $a=3$ and $x=\boxed{\textbf{(D) }7}$.

~walmartbrian ~Shontai ~andyluo

Solution 2 (Variation on Solution 1)

Suppose Maureen took n tests with an average of m

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

Cross-multiplying, nm + 11 = nm + n + m + 1; n+m = 10 -- I'll use this in a moment.

When she takes 3 more tests, the situation is (nm + 33)/(n+3) = m+2

Cross-multiplying, nm + 33 = nm + 2n + 3m + 6; 2n + 3m = 27

But 2n + 3m, which = 27, is also 2(n+m) + m = 20 + m, so m = $\boxed{\textbf{(D) 7}}$

~Dilip

Solution 3

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: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10}{t}+2$

From here, we solve for t, getting $t=3$.

We substitute this to get $s=21$.

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

~milquetoast


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

https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X



See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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