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

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https://www.youtube.com/watch?v=38JjGdGI5a0
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by Ismail.maths
 
==See Also==  
 
==See Also==  
 
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Revision as of 22:07, 3 December 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

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

by Ismail.maths

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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