Difference between revisions of "1988 AJHSME Problems/Problem 3"

Latest revision as of 23:54, 4 July 2013

Problem

$\frac{1}{10}+\frac{2}{20}+\frac{3}{30} =$

$\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6$

Solution

Each of the fractions simplify to $\frac{1}{10}$ , so this sum is $\begin{align*} \frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ &= .3 \rightarrow \boxed{\text{D}} \end{align*}$

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AJHSME box|year=1988|num-b=2|num-a=4}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1988 AJHSME Problems/Problem 3"

Latest revision as of 23:54, 4 July 2013

Problem

Solution

See Also