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

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So the answer is <math>\boxed{\text{D}}</math>
 
So the answer is <math>\boxed{\text{D}}</math>
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==Video Solution==
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https://youtu.be/KW5HexBjEHU
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 07:38, 7 January 2023

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 $\boxed{\text{D}}$

Video Solution

https://youtu.be/KW5HexBjEHU

~savannahsolver

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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