1968 AHSME Problems/Problem 28

Revision as of 05:29, 27 January 2019 by Timneh (talk | contribs) (Solution)

Problem

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$, then a possible value for the ratio $a/b$, to the nearest integer, is:

$\text{(A) } 5\quad \text{(B) } 8\quad \text{(C) } 11\quad \text{(D) } 14\quad \text{(E) none of these}$


Solution

$\fbox{D}$


$\frac{a+b}{2}=2\cdot\sqrt{ab}$

$\frac{a}{b} +1=4\cdot\sqrt{\frac{a}{b}}$

setting $x=\sqrt{\frac{a}{b}}$ we get a quadratic equation is$x^2+1=4x$ with solutions $x=\frac{4\pm \sqrt{16-4}}{2}$

$x^2=\frac{a}{b}=(4+3)+4\sqrt{3}~14$.

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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