Difference between revisions of "2022 AMC 8 Problems/Problem 16"

Revision as of 02:13, 26 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

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

Video Solution

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

~Interstigation

@@ Line 28: / Line 28: @@
 ==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>. Therefore, the average of the first and last numbers is <math>\dfrac{21+29}2=\dfrac{50}2=\boxed{\textbf{(B) } 25}</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>
-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>.
+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

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