Difference between revisions of "1989 AJHSME Problems/Problem 4"
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<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math> | <math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | <math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math>so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath> | + | <math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math> so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath> |
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| + | ==Solution 2== | ||
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| + | Using scientific notation estimated to the first significant digit, we see the numerator rounds to <math>\frac{4*10^2}{2*10^{-1}}</math>. | ||
| + | Subtracting the exponents gives <math>2-(-1)=3</math>. The answer choice that has <math>3</math> zeros to the right of the significant digit is <math>\boxed{\text{E}}</math> | ||
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| + | ~megaboy6679 | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1989|num-b=3|num-a=5}} | {{AJHSME box|year=1989|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:36, 28 October 2024
Contents
Problem
Estimate to determine which of the following numbers is closest to 
.
Solution 1
is around 
and 
is around 
so the fraction is approximately ![\[\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}\]](http://latex.artofproblemsolving.com/3/c/3/3c3e40f02aa475637780a3d0cc61e858e0027587.png)
Solution 2
Using scientific notation estimated to the first significant digit, we see the numerator rounds to 
.
Subtracting the exponents gives 
. The answer choice that has 
zeros to the right of the significant digit is 
~megaboy6679
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
