1986 AJHSME Problems/Problem 4

Problem

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

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

$(1.8)(40.37)\approx (1.8)(40)=72.$

$74$ is the closest to $72$ , so the answer is $\boxed{\text{C}}$ .

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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