Difference between revisions of "2012 AMC 8 Problems/Problem 22"
Bharatputra (talk | contribs) |
Bharatputra (talk | contribs) |
||
| Line 2: | Line 2: | ||
<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | <math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | First, we find that the minimum value of the median of <math> R </math> will be <math> 3 </math>. | ||
| + | |||
| + | We then experiment with sequences of numbers to determine other possible medians. | ||
| + | |||
| + | Median: <math> 4 </math> | ||
| + | |||
| + | Sequence: <math> -1, 0, 2, 3, 4, 6, 9, 10, 14 </math> | ||
| + | |||
| + | Median: <math> 5 </math> | ||
| + | |||
| + | Sequence: <math> 0, 2, 3, 4, 5, 6, 9, 10, 14 </math> | ||
| + | |||
| + | Median: <math> 6 </math> | ||
| + | |||
| + | Sequence: <math> 0, 2, 3, 4, 6, 9, 10, 14, 15 </math> | ||
| + | |||
| + | Median: <math> 7 </math> | ||
| + | |||
| + | Sequence: <math> 2, 3, 4, 6, 7, 8, 9, 10, 14 </math> | ||
| + | |||
| + | Median: <math> 8 </math> | ||
| + | |||
| + | Sequence: <math> 2, 3, 4, 6, 8, 9, 10, 14, 15 </math> | ||
| + | |||
| + | Median: <math> 9 </math> | ||
| + | |||
| + | Sequence: <math> 2, 3, 4, 6, 9, 14, 15, 16, 17 </math> | ||
| + | |||
| + | Any number greater than <math> 9 </math> also cannot be a median of set <math> R </math>. | ||
| + | |||
| + | There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>. | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=21|num-a=23}} | {{AMC8 box|year=2012|num-b=21|num-a=23}} | ||
Revision as of 13:15, 24 November 2012
Let 
be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?
Solution
First, we find that the minimum value of the median of will be 
.
We then experiment with sequences of numbers to determine other possible medians.
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Any number greater than also cannot be a median of set .
There are then 
possible medians of set .
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
