Difference between revisions of "2022 AMC 8 Problems/Problem 16"
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==Solution== | ==Solution== | ||
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| + | Note that the sum of the first two numbers is <math>21\cdot2=42,</math> the sum of the middle two numbers is <math>26\cdot2=52,</math> and the sum of the last two numbers is <math>30\cdot2=60.</math> | ||
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| + | It follows that the sum of the four numbers is <math>42+60=102,</math> so the sum of the first and last numbers is <math>102-52=50.</math> Therefore, the average of the first and last numbers is <math>50\div2=\boxed{\textbf{(B) } 25}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 12:35, 28 January 2022
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
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
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. 
