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

First, solve the sum of $ 40.3 $ and $.07$

$ \Rightarrow $ $ 40.3 + .07 = 40.37 $

Multiply it with $ 1.8 $

$ \Rightarrow $ $ (1.8)(40.37) $

Evalutate the result

$ \Rightarrow $ $ 72.666 $

Since we found the answer to be $ 72.666 $, the closest option is $\boxed{\text{C) 74}}$ .

Therefore, the answer is $\boxed{\text{C}}$ .

~ Solution by abcde26

Better Solution

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

Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ 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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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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1986 AJHSME Problems/Problem 4

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Problem

Solution

Better Solution

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