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

Revision as of 16:00, 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:

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

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 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:
-<math>\frac{ax+11}{a+1} = x+1</math> and <math>\frac{ax+33}{a+3} = x+2</math>.
+<cmath>\frac{ax+11}{a+1}=x+1\qquad (1)</cmath>
+<cmath>\frac{ax+33}{a+3}=x+2\qquad (2)</cmath>
-Expanding, <math>ax+11 = ax+a+x+1</math> and <math>ax+33 = ax+2a+3x+6</math>.
+Multiplying <math>(1)</math> by <math>(a+1)</math> and solving, we get:
+<cmath>ax+11=ax+a+x+1</cmath>
+<cmath>11=a+x+1</cmath>
+<cmath>a+x=10\qquad (3)</cmath>
-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>.
+Multiplying <math>(2)</math> by <math>(a+3)</math> and solving, we get:
+<cmath>ax+33=ax+2a+3x+6</cmath>
+<cmath>33=2a+3x+6</cmath>
+<cmath>2a+3x=27\qquad (4)</cmath>
-~walmartbrian ~Shontai ~andyluo
+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>.
+~walmartbrian ~Shontai ~andyluo ~megaboy6679
 ==Solution 2 (Variation on Solution 1)==

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