Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
| − | We are given that <cmath> | + | We are given that <cmath>b=\frac{a+a+1+a+2+a+3+a+4}{5}</cmath> |
| − | b | + | <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> | 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> | ||
Revision as of 11:27, 9 February 2014
- 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 ?
$\textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}}\ a+6\qquad\textbf{(E)}\ a+7$ (Error compiling LaTeX. Unknown error_msg)
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 ![\[b=\frac{a+a+1+a+2+a+3+a+4}{5}\]](http://latex.artofproblemsolving.com/e/2/a/e2a6864d90f09646e0670de52b58eca0e6c0b1d9.png)
![\[\implies b =a+2\]](http://latex.artofproblemsolving.com/d/0/3/d0319fc0f9cb638d8026e4c6e9d642c5ea7d3727.png)
We are asked to find the average of the 5 consecutive integers starting from in terms of . By substitution, this is ![\[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]](http://latex.artofproblemsolving.com/5/1/6/5169a0e7b4ba46f5065ee9b32f4356bd2a7b90ef.png)
Thus, the answer is
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. 
