Difference between revisions of "1964 AHSME Problems/Problem 37"

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==Solution==
 
==Solution==
 
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<math>\boxed{ \textbf{(D)} }</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 18:47, 26 January 2021

Problem

Given two positive numbers $a$, $b$ such that $a<b$. Let $A.M.$ be their arithmetic mean and let $G.M.$ be their positive geometric mean. Then $A.M.$ minus $G.M.$ is always less than:

$\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}$

$\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$

Solution

$\boxed{ \textbf{(D)} }$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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
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