Difference between revisions of "2002 AMC 10A Problems/Problem 1"
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==Solution== | ==Solution== | ||
We factor <math>\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}</math> as <math>\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}</math>. As <math>\frac{101}{20}=5.05</math>, our answer is <math>\boxed{\textbf{(D)}\ 5 }</math>. | We factor <math>\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}</math> as <math>\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}</math>. As <math>\frac{101}{20}=5.05</math>, our answer is <math>\boxed{\textbf{(D)}\ 5 }</math>. | ||
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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/VEW3sYvitGI | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
Latest revision as of 23:38, 18 July 2024
Problem
The ratio 
is closest to which of the following numbers?
Solution
We factor as 
. As 
, our answer is 
.
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See Also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
