Difference between revisions of "2002 AMC 10A Problems/Problem 1"
m (box) |
(→Solution: fix) |
||
Line 5: | Line 5: | ||
==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{\text{(D)}\ 5 | + | 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{\text{(D)}\ 5 }</math>. |
==See Also== | ==See Also== |
Revision as of 18:40, 26 December 2008
Problem
The ratio is closest to which of the following numbers?
Solution
We factor as
. As
, our answer is
.
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 |