1986 AJHSME Problems/Problem 20

Problem

The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to

$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$

Solution

$\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}$ which is obviously closest to $3\rightarrow\boxed{\text{D}}$ .

(The original expression is approximately equal to $3.44921198$ .)

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1986 AJHSME Problems/Problem 20

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