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Difference between revisions of "1989 AJHSME Problems/Problem 13"

(New page: ==Problem== <math>\frac{9}{7\times 53} =</math> <math>\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \tex...)
 
 
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{{AJHSME box|year=1989|num-b=12|num-a=14}}
 
{{AJHSME box|year=1989|num-b=12|num-a=14}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:03, 5 July 2013

Problem

$\frac{9}{7\times 53} =$

$\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \text{(D)}\ \frac{.9}{7\times .53} \qquad \text{(E)}\ \frac{.09}{.07\times .53}$

Solution

\begin{align*} \frac{9}{7\times 53} &= \frac{.9\times 10}{.7\times 10\times 53} \\ &= \frac{.9}{.7\times 53} \rightarrow \boxed{\text{A}} \end{align*}

See Also

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
All AJHSME/AMC 8 Problems and Solutions

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