Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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<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> | <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== | + | ==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>. | 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>. | 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> \begin{aligned}\frac{a+a+1+a+2+a+3+a+4}5 & =b\rightarrow \\ | ||
| + | b & =a+2\end{aligned}</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> | ||
==See Also== | ==See Also== | ||
Revision as of 00:36, 7 February 2014
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
\begin{aligned}\frac{a+a+1+a+2+a+3+a+4}5 & =b\rightarrow \\
b & =a+2\end{aligned} (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
