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"

m (→‎Solution 3 (Assumption): This is a test-efficient solution. I elaborated more on the assumption process.)
 
(21 intermediate revisions by 10 users not shown)
Line 28: Line 28:
 
==Solution 3 (Assumption)==
 
==Solution 3 (Assumption)==
  
−
We can just assume some of the numbers. For example, let the first two numbers both be <math>21</math>. It follows that the third number is <math>31</math>, and the fourth number is <math>29</math>.
+
We can just assume some of the numbers. For example, let the first two numbers both be <math>21.</math> It follows that the third number is <math>31,</math> and the fourth number is <math>29.</math> Therefore, the average of the first and last numbers is <math>\dfrac{21+29}2=\dfrac{50}2=\boxed{\textbf{(B) } 25}.</math>
  
−
The average of the first and last numbers would be <math>\dfrac{21+29}2=\dfrac{50}2=\boxed{\textbf{(B) } 25}</math>. We can check this with other sequences, such as <math>20,22,30,30</math>, where the average of the first and last numbers would still be <math>25</math>.
+
We can check this with other sequences, such as <math>20,22,30,30,</math> where the average of the first and last numbers is still <math>25.</math>
  
 
~wuwang2002
 
~wuwang2002
 +
 +
==Solution 4==
 +
 +
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>
 +
 +
If we subtract <math>(2)</math> from <math>(3)</math>, we get <cmath>\frac{c+d}{2}-\frac{b+c}{2}=\frac{d-b}{2}=4.</cmath>
 +
So if we multiply by <math>2</math> then we get <math>d=b+8.</math> Looking at our question: <cmath>\frac{a+d}{2}=\frac{a+b+8}{2}=\frac{a+b}{2}+\frac{8}{2}=21+4=\boxed{\textbf{(B) } 25}.</cmath>
 +
 +
~funnyarmadillo58_aops
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/oUEa7AjMF2A?si=2goytlOr7qxq69I9&t=2801
 +
 +
~Math-X
 +
 +
==Video Solution (🚀1 min solution 🚀)==
 +
https://youtu.be/oVG6zqPVPfM
 +
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
Line 38: Line 63:
  
 
~Interstigation
 
~Interstigation
 +
 +
==Video Solution by Ismail.maths==
 +
 +
https://www.youtube.com/watch?v=38JjGdGI5a0
 +
 +
~Ismail.maths93
 +
 +
==Video Solution==
 +
https://youtu.be/hs6y4PWnoWg
 +
 +
~STEMbreezy
 +
 +
==Video Solution==
 +
https://youtu.be/tpzsowaoQRc
 +
 +
~savannahsolver
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/Hh6DrF178lc
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=15|num-a=17}}
 
{{AMC8 box|year=2022|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:31, 21 December 2024

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

Solution 3 (Assumption)

We can just assume some of the numbers. For example, let the first two numbers both be $21.$ It follows that the third number is $31,$ and the fourth number is $29.$ Therefore, the average of the first and last numbers is $\dfrac{21+29}2=\dfrac{50}2=\boxed{\textbf{(B) } 25}.$

We can check this with other sequences, such as $20,22,30,30,$ where the average of the first and last numbers is still $25.$

~wuwang2002

Solution 4

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 $\frac{a+d}{2}.$

If we subtract $(2)$ from $(3)$, we get \[\frac{c+d}{2}-\frac{b+c}{2}=\frac{d-b}{2}=4.\] So if we multiply by $2$ then we get $d=b+8.$ Looking at our question: \[\frac{a+d}{2}=\frac{a+b+8}{2}=\frac{a+b}{2}+\frac{8}{2}=21+4=\boxed{\textbf{(B) } 25}.\]

~funnyarmadillo58_aops

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

https://youtu.be/oUEa7AjMF2A?si=2goytlOr7qxq69I9&t=2801

~Math-X

Video Solution (🚀1 min solution 🚀)

https://youtu.be/oVG6zqPVPfM

~Education, the Study of Everything

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1394

~Interstigation

Video Solution by Ismail.maths

https://www.youtube.com/watch?v=38JjGdGI5a0

~Ismail.maths93

Video Solution

https://youtu.be/hs6y4PWnoWg

~STEMbreezy

Video Solution

https://youtu.be/tpzsowaoQRc

~savannahsolver

Video Solution by Dr. David

https://youtu.be/Hh6DrF178lc

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=237499"