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 19:06, 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 | |
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All AHSME Problems and Solutions |
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