AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
Personal tools

1988 AJHSME Problems/Problem 7

From AoPSWiki

Problem

2.46\times 8.163\times (5.17+4.829) is closest to

\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500

Solution

We estimate the first thing to be 2.5, the second thing to be 8, and the third thing to be 10. We now have 2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}.

See Also

1988 AJHSME (Problems • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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/1988_AJHSME_Problems/Problem_7"
Visit the AoPS Book Store.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us