Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math> | Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math> | ||
− | ==Video Solution== | + | ==Video Solutions== |
+ | ===Video Solution One=== | ||
https://youtu.be/wBdD6Ge8FuE | https://youtu.be/wBdD6Ge8FuE | ||
~savannahsolver | ~savannahsolver | ||
+ | |||
+ | ===Video Solution Two=== | ||
+ | https://youtu.be/rJytKoJzNBY | ||
==See Also== | ==See Also== |
Latest revision as of 23:17, 25 November 2020
- The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10, so both problems redirect to this page.
Contents
Problem
Five positive consecutive integers starting with have average . What is the average of consecutive integers that start with ?
Solution 1
Let . Our list is with an average of . Our next set starting with is . Our average is .
Therefore, we notice that which means that the answer is .
Solution 2
We are given that
We are asked to find the average of the 5 consecutive integers starting from in terms of . By substitution, this is
Thus, the answer is
Video Solutions
Video Solution One
~savannahsolver
Video Solution Two
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AMC 10 Problems and Solutions |
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.