Difference between revisions of "1989 AHSME Problems/Problem 2"

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<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=</math>
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== Problem ==
  
(A) <math>\frac{1}{5}</math>\hskip.5in (B)<math>\frac{1}{4}</math>\hskip.5in    (C)<math>\frac{2}{7}</math>\hskip.5in    (D)<math>\frac{5}{12}</math> \hskip.5in  (E)<math>\frac{7}{12}</math>
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<math> \sqrt{\frac{1}{9}+\frac{1}{16}}= </math>
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<math> \textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12} </math>
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== Solution ==
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<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}</math> and hence the answer is <math>\fbox{D}</math>.
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== See also ==
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{{AHSME box|year=1989|num-b=1|num-a=3}} 
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[[Category: Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:39, 22 October 2014

Problem

$\sqrt{\frac{1}{9}+\frac{1}{16}}=$

$\textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12}$

Solution

$\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}$ and hence the answer is $\fbox{D}$.


See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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