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1986 AJHSME Problems/Problem 4

Revision as of 19:03, 20 January 2009 by Waffle (talk | contribs) (→‎Solution)

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 $84$

See Also

1986 AJHSME Problems

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