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

(Created page with '==Problem== Which of the following is closest to the product <math>(.48017)(.48017)(.48017)</math>? <math>\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qq…')
 
 
(One intermediate revision by one other user not shown)
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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>
  
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==Solution==
+
==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.
−
 
+
==Solution 2==
 +
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]]
 +
{{MAA Notice}}

Latest revision as of 00:05, 5 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 (Problems • Answer Key • Resources)
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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