Difference between revisions of "1988 AJHSME Problems/Problem 3"
5849206328x (talk | contribs) (New page: ==Problem== <math>\frac{1}{10}+\frac{2}{20}+\frac{3}{30} = </math> <math>\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6</math> ...) |
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==Solution== | ==Solution== | ||
| − | Each of the fractions | + | Each of the [[Fraction|fractions]] simplify to <math>\frac{1}{10}</math>, so this [[sum]] is |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ | \frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ | ||
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==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1988|num-b=2|num-a=4}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:54, 4 July 2013
Problem
Solution
Each of the fractions simplify to 
, so this sum is
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
