Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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+ | {{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #9]] and [[2014 AMC 10A Problems|2014 AMC 10A #10]]}} | ||
==Problem== | ==Problem== | ||
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== | ||
+ | Let <math>a=1</math>. Our list is <math>\{1,2,3,4,5\}</math> with an average of <math>15\div 5=3</math>. Our next set starting with <math>3</math> is <math>\{3,4,5,6,7\}</math>. Our average is <math>25\div 5=5</math>. | ||
+ | |||
+ | Therefore, we notice that <math>5=1+4</math> which means that the answer is <math>\boxed{\textbf{(B)}\ a+4}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We are given that <cmath>b=\frac{a+a+1+a+2+a+3+a+4}{5}</cmath> | ||
+ | <cmath>\implies b =a+2</cmath> | ||
+ | |||
+ | We are asked to find the average of the 5 consecutive integers starting from <math>b</math> in terms of <math>a</math>. By substitution, this is <cmath>\frac{a+2+a+3+a+4+a+5+a+6}5=a+4</cmath> | ||
+ | |||
+ | Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math> | ||
+ | |||
+ | ==Video Solutions== | ||
+ | ===Video Solution One=== | ||
+ | https://youtu.be/wBdD6Ge8FuE | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ===Video Solution Two=== | ||
+ | https://youtu.be/rJytKoJzNBY | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2014|ab=A|num-b=9|num-a=11}} | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] |
Revision as of 00:17, 26 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.