1989 AJHSME Problems/Problem 4

Revision as of 22:36, 28 October 2024 by Megaboy6679 (talk | contribs) (Solution 2)

Problem

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$.

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution 1

$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}\]

Solution 2

Using scientific notation estimated to the first significant digit, we see the numerator rounds to $\frac{4*10^2}{2*10^{-1}}$. Subtracting the exponents gives $2-(-1)=3$. The answer choice that has $3$ zeros to the right of the significant digit is $\boxed{\text{E}}$

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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