Difference between revisions of "1985 AJHSME Problems/Problem 3"
5849206328x (talk | contribs) (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== | ||
− | {{ | + | 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> |
+ | |||
+ | 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== | ||
− | [[ | + | {{AJHSME box|year=1985|num-b=2|num-a=4}} |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Revision as of 17:01, 3 July 2013
Problem
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:
We know that , so
So the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.