AoPSWiki

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

Personal tools

Search

Toolbox

1989 AJHSME Problems/Problem 2

From AoPSWiki

Problem

$\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=$

$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$

Solution

$\begin{align*}\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\&= \frac...$

See Also

1989 AJHSME (Problems • Resources)
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1989_AJHSME_Problems/Problem_2"

Category: Introductory Algebra Problems

Visit the AoPS Book Store.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us