Difference between revisions of "2014 AMC 10A Problems/Problem 10"
(→Problem) |
|||
| Line 25: | Line 25: | ||
{{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]] | ||
Revision as of 10:00, 20 May 2015
- 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 ![\[b=\frac{a+a+1+a+2+a+3+a+4}{5}\]](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. 
