Difference between revisions of "2002 AMC 10A Problems/Problem 1"

(New page: ==Problem== The ratio <math>\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}</math> is closest to which of the following numbers? <math>\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{...)
 
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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>\text{(D)}\ 5 \qquad</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>\text{(D)}\ 5 \qquad</math>.
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==See Also==
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{{AMC10 box|year=2002|ab=A|before=First question|num-a=2}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:59, 26 December 2008

Problem

The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?

$\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 10$

Solution

We factor $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}$. As $\frac{101}{20}=5.05$, our answer is $\text{(D)}\ 5 \qquad$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AMC 10 Problems and Solutions