Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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+ | ==Solution 4== | ||
+ | The list of numbers is <math>\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}</math> so <math>b=a+2</math>. The new list is <math>\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}</math> and the average is <math>a+4 \Longrightarrow \boxed{\textbf{(B) } a+4}</math>. | ||
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+ | ~JH. L | ||
==Video Solutions== | ==Video Solutions== |
Revision as of 04:52, 20 June 2022
- 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
Solution 3
We know from experience that the average of consecutive numbers is the
one or the
. With the logic, we find that
.
.
~MathFun1000
Solution 4
The list of numbers is so
. The new list is
and the average is
.
~JH. L
Video Solutions
Video Solution 1
~savannahsolver
Video Solution 2
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.