Difference between revisions of "1990 AJHSME Problems/Problem 5"

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<math>\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110</math>
 
<math>\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110</math>
  
==Solution==
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==Solution 1==
  
 
Clearly, <cmath>.4<.48017<.5</cmath>
 
Clearly, <cmath>.4<.48017<.5</cmath>
 
Since the function <math>f(x)=x^3</math> is strictly increasing, we can say that <cmath>.4^3<.48017^3<.5^3</cmath>
 
Since the function <math>f(x)=x^3</math> is strictly increasing, we can say that <cmath>.4^3<.48017^3<.5^3</cmath>
 
from which it follows that <math>\text{A}</math> is much too small and <math>\text{C}</math> is much too large, so <math>\boxed{\text{B}}</math> is the answer.
 
from which it follows that <math>\text{A}</math> is much too small and <math>\text{C}</math> is much too large, so <math>\boxed{\text{B}}</math> is the answer.
 
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==Solution 2==
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Since <math>0.48017</math> is quite close to <math>0.5</math>, or <math>\dfrac{1}{2}</math>, we can look for the answer choice that is just below <math>(\dfrac{1}{2})^3=\dfrac{1}{8}=0.125</math>, which would be <math>\boxed{\textbf{(B)}}</math>.
 
==See Also==
 
==See Also==
  
 
{{AJHSME box|year=1990|num-b=4|num-a=6}}
 
{{AJHSME box|year=1990|num-b=4|num-a=6}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:05, 4 July 2013

Problem

Which of the following is closest to the product $(.48017)(.48017)(.48017)$?

$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$

Solution 1

Clearly, \[.4<.48017<.5\] Since the function $f(x)=x^3$ is strictly increasing, we can say that \[.4^3<.48017^3<.5^3\] from which it follows that $\text{A}$ is much too small and $\text{C}$ is much too large, so $\boxed{\text{B}}$ is the answer.

Solution 2

Since $0.48017$ is quite close to $0.5$, or $\dfrac{1}{2}$, we can look for the answer choice that is just below $(\dfrac{1}{2})^3=\dfrac{1}{8}=0.125$, which would be $\boxed{\textbf{(B)}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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