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Difference between revisions of "1985 AJHSME Problems/Problem 3"
m (→‎Solution)
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==Solution==
 
==Solution==
  
−
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>
+
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>
  
−
So the answer is (D)
+
We know that <math>10^3 = 10 \times 10 \times 10</math>, so
 +
 
 +
<cmath>\begin{align*}
 +
\frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\
 +
&= 2\times 10\times 10 \\
 +
&= 200 \\
 +
\end{align*}</cmath>
 +
 
 +
So the answer is <math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1985 AJHSME Problems]]
 
[[1985 AJHSME Problems]]

Revision as of 21:54, 12 January 2009

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

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