Difference between revisions of "1985 AJHSME Problems/Problem 3"

(New page: ==Problem== <math>\frac{10^7}{5\times 10^4}=</math> <math>\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math> ==Solution=...)
 
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==Solution==
 
==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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So the answer is (D)
  
 
==See Also==
 
==See Also==
  
 
[[1985 AJHSME Problems]]
 
[[1985 AJHSME Problems]]

Revision as of 21:49, 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 quickly make quick work of this.

$\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}$. We know that $10^3 = 10 \times 10 \times 10$. We also know that $\frac{10}{2} = 5$. So we have $\frac{10^3}{5} = \frac{10 \times 10 \times 10}{5} = 2 \times 10 \times 10 = 200$

So the answer is (D)

See Also

1985 AJHSME Problems