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Difference between revisions of "2022 AMC 8 Problems/Problem 16"
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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>
 
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
 +
 +
==Video Solution==
 +
https://youtu.be/Ij9pAy6tQSg?t=1394
 +
 +
~Interstigation
  
 
==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}}

Revision as of 23:47, 18 February 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

Video Solution

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

~Interstigation

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

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