Difference between revisions of "1985 AJHSME Problems/Problem 3"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | So the answer is <math>\boxed{\ | + | So the answer is <math>\boxed{\textbf{(D)}\ 200}</math> |
| + | ~[[shunyipanda]] (Minor edit) | ||
== Solution 2 (Brute Force) == | == Solution 2 (Brute Force) == | ||
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Thus the answer is <math>200</math>, or <math>\boxed{\textbf{(D)}\ 200}</math> | Thus the answer is <math>200</math>, or <math>\boxed{\textbf{(D)}\ 200}</math> | ||
| + | |||
| + | ==Solution 3 (Reasoning)== | ||
| + | As we can see, the numerator is way larger than the denominator so answer choices <math>\boxed{\textbf{(A)}\ .002}</math>, <math>\boxed{\textbf{(B)}\ .2}</math>, and <math>\boxed{\textbf{(C)}\ 20}</math> are eliminated. <math>10^7/10^4</math> is 1000, but in this question the expression, <math>10^7/5 \times 10^4</math>, is smaller than 1000 since the denominator is larger than 10^4. Therefore the answer is <math>\boxed{\textbf{(D)}\ 200}</math>. | ||
| + | ~[[shunyipanda]] | ||
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== | ||
Revision as of 20:19, 22 October 2025
Contents
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
~shunyipanda (Minor edit)
Solution 2 (Brute Force)
is
is
Thus the answer is
, or
Solution 3 (Reasoning)
As we can see, the numerator is way larger than the denominator so answer choices
,
, and
are eliminated.
is 1000, but in this question the expression,
, is smaller than 1000 since the denominator is larger than 10^4. Therefore the answer is
.
~shunyipanda
Video Solution by BoundlessBrain!
Video Solution
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.