Difference between revisions of "2023 AMC 10A Problems/Problem 10"

Revision as of 19:47, 9 November 2023

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 equation $(ax+11)/a+1 = x+1$ and $(ax+33)/a+3 = x+2$ .

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, $x=7$ and $a=3$ . (D)

~walmartbrian ~Shontai ~andyluo

@@ Line 4: / Line 4: @@
 ==Solution 1==
-Let a represent the amount of tests taken previously and x the mean of the scores taken previously. We can write the equation (ax+11)/a+1 = x+1 and (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, x=7 and a=3. (D)
+Let <math>a</math> represent the amount of tests taken previously and <math>x</math> the mean of the scores taken previously.
-~walmartbrian ~Shontai
+We can write the equation <math>(ax+11)/a+1 = x+1</math> and <math>(ax+33)/a+3 = x+2</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>. (D)
+~walmartbrian ~Shontai ~andyluo

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Difference between revisions of "2023 AMC 10A Problems/Problem 10"

Revision as of 19:47, 9 November 2023

Solution 1