Difference between revisions of "2021 AMC 12B Problems/Problem 4"

(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
WLOG assume there <math>3</math> students in the morning class and <math>4</math> in the afternoon class. Then the average is <math>\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}</math>
+
WLOG, assume there are <math>3</math> students in the morning class and <math>4</math> in the afternoon class. Then the average is <math>\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}</math>
 +
 
 +
==Solution 2==
 +
Let there be <math>3x</math> students in the morning class and <math>4x</math> students in the afternoon class. The total number of students is <math>3x + 4x = 7x</math>. The average is <math>\frac{3x\cdot84 + 4x\cdot70}{7x}=76</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}76}</math>.
 +
<br><br>
 +
~ {TSun} ~

Revision as of 19:25, 11 February 2021

Problem

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class's mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$. What is the mean of the scores of all the students?

$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution

WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}$

Solution 2

Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)}76}$.

~ {TSun} ~

Invalid username
Login to AoPS