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

Revision as of 08:28, 12 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:

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

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

Multiplying $(2)$ 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 $x=\boxed{\textbf{(D) }7}$ .

~walmartbrian ~Shontai ~andyluo ~megaboy6679

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

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

Solution 4(Trial and Error)

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

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

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

Video Solution

https://youtu.be/VzgNmdKp8UE

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

@@ Line 66: / Line 66: @@
 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
 ==Video Solution by Math-X (First understand the problem!!!)==

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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