Art of Problem Solving

1968 AHSME Problems/Problem 28: Difference between revisions

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== Solution ==
== Solution ==
<math>\fbox{}</math>
<math>\fbox{D}</math>


== See also ==
== See also ==

Revision as of 02:34, 29 September 2014

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

$\fbox{D}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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
All AHSME Problems and Solutions

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