Difference between revisions of "1990 AJHSME Problems/Problem 5"
5849206328x (talk | contribs) (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) | |||
Line 5: | Line 5: | ||
<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== | + | ==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 23:05, 4 July 2013
Contents
Problem
Which of the following is closest to the product ?
Solution 1
Clearly, Since the function is strictly increasing, we can say that from which it follows that is much too small and is much too large, so is the answer.
Solution 2
Since is quite close to , or , we can look for the answer choice that is just below , which would be .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.