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Difference between revisions of "1988 AJHSME Problems/Problem 6"

(New page: ==Problem== <math>\frac{(.2)^3}{(.02)^2} =</math> <math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20</math> ==Solution== Convert...)
 
(→‎Solution 2)
 
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==Solution==
 
==Solution==
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Converting the decimals to fractions gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>.
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Converting the [[decimal|decimals]] to [[fraction|fractions]] gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>.
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==Solution 2==
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We expand <math>\frac{(0.2)^3}{(0.02)^2}</math>, and get <math>\frac{(0.2) \times (0.2) \times (0.2)}{(0.02) \times (0.02)}</math>. The two <math>0.02</math>'s "cancel" out with the two <math>0.2</math>'s, leaving the fraction as: <math>(10) \times (10) \times (0.2)</math>. Using basic calculations, we compute this expression to get <math>20\Rightarrow \mathrm{(E)}</math>.
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~sakshamsethi
  
 
==See Also==
 
==See Also==
  
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[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=5|num-a=7}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:04, 28 July 2020

Problem

$\frac{(.2)^3}{(.02)^2} =$

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$

Solution

Converting the decimals to fractions gives us $\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}$.

Solution 2

We expand $\frac{(0.2)^3}{(0.02)^2}$, and get $\frac{(0.2) \times (0.2) \times (0.2)}{(0.02) \times (0.02)}$. The two $0.02$'s "cancel" out with the two $0.2$'s, leaving the fraction as: $(10) \times (10) \times (0.2)$. Using basic calculations, we compute this expression to get $20\Rightarrow \mathrm{(E)}$.

~sakshamsethi

See Also

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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. AMC logo.png

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