Difference between revisions of "1965 AHSME Problems/Problem 26"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | |
| + | We shall begin by eliminating some options through counterexamples. If <math>a=b=c=d=e</math>, then <math>m=k=l=a</math>, so <math>p=a</math>, and <math>m=p</math>. Answers (C) and (D) do not allow for <math>m=p</math>, so they can be eliminated. If we set <math>a=60</math> and <math>b=c=d=e=0</math>, then <math>m=12</math>, <math>k=30</math>, <math>l=0</math>, and <math>p=15</math>. Here, <math>m<p</math>, so we can throw out options (A) and (B) as well. Now, we are left with only option <math>\fbox{\textbf{(E) }none of these}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 20:06, 18 July 2024
Problem
For the numbers 
define 
to be the arithmetic mean of all five numbers;
to be the arithmetic mean of 
and 
; 
to be the arithmetic mean of 
, and 
;
and 
to be the arithmetic mean of and . Then, no matter how 
, and are chosen, we shall always have:
Solution
We shall begin by eliminating some options through counterexamples. If 
, then 
, so 
, and 
. Answers (C) and (D) do not allow for , so they can be eliminated. If we set 
and 
, then 
, 
, 
, and 
. Here, 
, so we can throw out options (A) and (B) as well. Now, we are left with only option 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
