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Difference between revisions of "1985 AJHSME Problems/Problem 3"
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==Solution==
 
==Solution==
  
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We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we quickly make quick work of this.<br><br><math>\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}</math>. We know that <math>10^3 = 10 \times 10 \times 10</math>. We also know that <math>\frac{10}{2} = 5</math>. So we have <math>\frac{10^3}{5} = \frac{10 \times 10 \times 10}{5} = 2 \times 10 \times 10 = 200</math>
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We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: <cmath>\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}</cmath>
  
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So the answer is (D)
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We know that <math>10^3 = 10 \times 10 \times 10</math>, so
 +
 
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<cmath>\begin{align*}
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\frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\
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&= 2\times 10\times 10 \\
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&= 200 \\
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\end{align*}</cmath>
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So the answer is <math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
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[[1985 AJHSME Problems]]
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{{AJHSME box|year=1985|num-b=2|num-a=4}}
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[[Category:Introductory Algebra Problems]]
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 +
 
 +
{{MAA Notice}}

Revision as of 17:01, 3 July 2013

Problem

$\frac{10^7}{5\times 10^4}=$


$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution

We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]

We know that $10^3 = 10 \times 10 \times 10$, so

\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}

So the answer is $\boxed{\text{D}}$

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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