Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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Five positive consecutive integers starting with <math>a</math> have average <math>b</math>. What is the average of <math>5</math> consecutive integers that start with <math>b</math>? | Five positive consecutive integers starting with <math>a</math> have average <math>b</math>. What is the average of <math>5</math> consecutive integers that start with <math>b</math>? | ||
− | <math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D) | + | <math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7</math> |
==Solution 1== | ==Solution 1== | ||
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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> | ||
+ | |||
+ | ==Solution 3 (fast)== | ||
+ | We know from experience (or logical reasoning) that the average of <math>5</math> consecutive numbers is the <math>3^\text{rd}</math> one or the <math>1^\text{st} + 2</math>. With the logic, we find that <math>b=a+2</math>. Therefore, <math>b+2=(a+2)+2=\boxed{a+4}</math>. | ||
+ | |||
+ | ~MathFun1000 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/GonWHjrROzI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solutions== | ||
+ | ===Video Solution 1=== | ||
+ | https://youtu.be/wBdD6Ge8FuE | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ===Video Solution 2=== | ||
+ | https://youtu.be/rJytKoJzNBY | ||
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] |
Latest revision as of 03:36, 1 November 2024
- 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 (fast)
We know from experience (or logical reasoning) that the average of consecutive numbers is the one or the . With the logic, we find that . Therefore, .
~MathFun1000
Solution 4
The list of numbers is so . The new list is and the average is .
~JH. L
Video Solution (CREATIVE THINKING)
https://youtu.be/GonWHjrROzI
~Education, the Study of Everything
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.