Difference between revisions of "1986 AJHSME Problems/Problem 20"

(New page: ==Problem== The value of the expression <math>\frac{(304)^5}{(29.7)(399)^4}</math> is closest to <math>\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 ...)
 
 
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==Solution==
 
==Solution==
  
{{Solution}}
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<cmath> \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}</cmath>
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Which is closest to <math>3\rightarrow\boxed{\text{D}}</math>.
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(The original expression is approximately equal to <math>3.44921198</math>.)
  
 
==See Also==
 
==See Also==
  
[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=19|num-a=21}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:28, 3 July 2013

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 closest to $3\rightarrow\boxed{\text{D}}$.

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

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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