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Difference between revisions of "2022 AMC 8 Problems/Problem 16"
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~wuwang2002
 
~wuwang2002
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==Solution 4==
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Let <math>a,b,c,</math> and <math>d</math> be the four numbers in that order. We are given that
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<cmath>\begin{align*}
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\frac{a+b}{2} &= 21, &(1) \\
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\frac{b+c}{2} &= 26, &(2) \\
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\frac{c+d}{2} &= 30, &(3)
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\end{align*}</cmath>
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and we wish to find <math>\frac{a+d}{2}.</math>
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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>
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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>
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~funnyarmadillo58_aops
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/oUEa7AjMF2A?si=2goytlOr7qxq69I9&t=2801
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~Math-X
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 +
==Video Solution (🚀1 min solution 🚀)==
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https://youtu.be/oVG6zqPVPfM
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~STEMbreezy
 
~STEMbreezy
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==Video Solution==
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https://youtu.be/tpzsowaoQRc
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~savannahsolver
  
 
==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 13:04, 5 October 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

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

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