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> ...) |
5849206328x (talk | contribs) m |
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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]] |
Revision as of 14:40, 1 June 2009
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 |