1988 AJHSME Problems/Problem 7

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

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AJHSME/AMC 8 Problems and Solutions