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 $2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}$ .

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1988 AJHSME Problems/Problem 7

Problem

Solution

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