Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1986 AJHSME Problems/Problem 4

Revision as of 18:02, 20 January 2009 by Waffle (talk | contribs)

Problem

The product $(1.8)(40.3+.07)$ is closest to

$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

Solution

The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.

$1.8$ is about $2$, $40.3$ is about $40$, and $.07$ is about $0$. Putting this in, we get $(2)(40 + 0) = 80$

See Also

1986 AJHSME Problems

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_4&oldid=29625"