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1986 AJHSME Problems/Problem 4

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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 $(2)(40 + 0) = 80$

$80$ is about $84$

$\boxed{\text{D}}$

See Also

1986 AJHSME (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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/1986_AJHSME_Problems/Problem_4"

Category: Introductory Algebra Problems

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