1986 AJHSME Problems/Problem 4

Problem

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

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

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 $84$

$\boxed{\text{D}}$

edit: the answer is C. The reason for this is that 1.8 and 2 are close, but when multiplied by 40, it's a big difference (80 and 72). So you could just do 1.8(40.37) which is basically 1.8*40=72. 74 is the closest to 72.

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1986 AJHSME Problems/Problem 4

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