1988 AJHSME Problems/Problem 7

Revision as of 15:55, 21 November 2020 by Avamarora (talk | contribs) (Solution)

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 can round. We can make 2.46X8.163X(5.17+4.829) to 2X8X(5+5) wich equals 160. 160 is closest to 200 so the answer is B. ~avamarora

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png