1988 AJHSME Problems/Problem 7

Problem

$2.46\times 8.163\times (5.17+4.829)$ is closest to

$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

Solution

We estimate the first thing to be $2.5$ , the second thing to be $8$ , and the third thing to be $10$ . We now have $2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}$ .

Note: The real answer is 200.789719 ~ xxsc

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1988 AJHSME Problems/Problem 7

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