Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 8 Problems/Problem 16"
Line 5: Line 5:
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28</math>
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28</math>
  
−
==Solution==
+
==Solution 1 (Arithmetic)==
  
 
Note that the sum of the first two numbers is <math>21\cdot2=42,</math> the sum of the middle two numbers is <math>26\cdot2=52,</math> and the sum of the last two numbers is <math>30\cdot2=60.</math>
 
Note that the sum of the first two numbers is <math>21\cdot2=42,</math> the sum of the middle two numbers is <math>26\cdot2=52,</math> and the sum of the last two numbers is <math>30\cdot2=60.</math>
Line 11: Line 11:
 
It follows that the sum of the four numbers is <math>42+60=102,</math> so the sum of the first and last numbers is <math>102-52=50.</math> Therefore, the average of the first and last numbers is <math>50\div2=\boxed{\textbf{(B) } 25}.</math>
 
It follows that the sum of the four numbers is <math>42+60=102,</math> so the sum of the first and last numbers is <math>102-52=50.</math> Therefore, the average of the first and last numbers is <math>50\div2=\boxed{\textbf{(B) } 25}.</math>
  
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Algebra)==
 +
 +
Let <math>a,b,c,</math> and <math>d</math> be the four numbers in that order. We are given that
 +
<cmath>\begin{align*}
 +
\frac{a+b}{2} &= 21, &(1) \\
 +
\frac{b+c}{2} &= 26, &(2) \\
 +
\frac{c+d}{2} &= 30, &(3)
 +
\end{align*}</cmath>
 +
and we wish to find <math>\frac{a+d}{2}.</math>
 +
 +
We add <math>(1)</math> and <math>(3),</math> then subtract <math>(2)</math> from the result: <cmath>\frac{a+d}{2}=21+30-26=\boxed{\textbf{(B) } 25}.</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  

Revision as of 22:36, 28 January 2022

Problem

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?

$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$

Solution 1 (Arithmetic)

Note that the sum of the first two numbers is $21\cdot2=42,$ the sum of the middle two numbers is $26\cdot2=52,$ and the sum of the last two numbers is $30\cdot2=60.$

It follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of the first and last numbers is $50\div2=\boxed{\textbf{(B) } 25}.$

~MRENTHUSIASM

Solution 2 (Algebra)

Let $a,b,c,$ and $d$ be the four numbers in that order. We are given that \begin{align*} \frac{a+b}{2} &= 21, &(1) \\ \frac{b+c}{2} &= 26, &(2) \\ \frac{c+d}{2} &= 30, &(3) \end{align*} and we wish to find $\frac{a+d}{2}.$

We add $(1)$ and $(3),$ then subtract $(2)$ from the result: \[\frac{a+d}{2}=21+30-26=\boxed{\textbf{(B) } 25}.\] ~MRENTHUSIASM

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 AJHSME/AMC 8 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_16&oldid=170658"