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 | ||
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==Solution 3 (Assumption)== | ==Solution 3 (Assumption)== | ||
Revision as of 01:36, 26 February 2022
Contents
Problem
Four numbers are written in a row. The average of the first two is 
the average of the middle two is 
and the average of the last two is 
What is the average of the first and last of the numbers?
Solution 1 (Arithmetic)
Note that the sum of the first two numbers is 
the sum of the middle two numbers is 
and the sum of the last two numbers is 
It follows that the sum of the four numbers is 
so the sum of the first and last numbers is 
Therefore, the average of the first and last numbers is 
~MRENTHUSIASM
Solution 2 (Algebra)
Let 
and 
be the four numbers in that order. We are given that
and we wish to find 
We add 
and 
then subtract 
from the result: ![\[\frac{a+d}{2}=21+30-26=\boxed{\textbf{(B) } 25}.\]](http://latex.artofproblemsolving.com/f/b/f/fbfec2d8a8f3b506dc010c0fae39d1777f6b929d.png)
~MRENTHUSIASM
Solution 3 (Assumption)
We can just assume some of the numbers. For example, we could have the sequence 
(note that the problem didn't ever say that they had to be in increasing order!), and the average of the first and last numbers would be 
. We can check this with other sequences, such as 
, where the average of the first and last numbers would still be 
.
~wuwang2002
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
