Difference between revisions of "2007 AMC 10A Problems/Problem 6"

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<math>\text{(A)}\ 2002\ \text{and}\ 2003 \qquad \text{(B)}\ 2003\ \text{and}\ 2004 \qquad \text{(C)}\ 2004\ \text{and}\ 2005 \qquad \text{(D)}\ 2005\ \text{and}\ 2006 \qquad \text{(E)}\ 2006\ \text{and}\ 2007</math>
 
<math>\text{(A)}\ 2002\ \text{and}\ 2003 \qquad \text{(B)}\ 2003\ \text{and}\ 2004 \qquad \text{(C)}\ 2004\ \text{and}\ 2005 \qquad \text{(D)}\ 2005\ \text{and}\ 2006 \qquad \text{(E)}\ 2006\ \text{and}\ 2007</math>
  
== Solution ==
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== Solution 1 ==
 
We compute the percentage increases:
 
We compute the percentage increases:
  
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The answer is <math>\mathrm{(A)}</math>.  
 
The answer is <math>\mathrm{(A)}</math>.  
  
In fact, the answer follows directly from examining the differences between each year. The largest differences are <math>6,6,7</math>, and it is easy to see that due to the decreased starting number of students in 2002 that that will be our answer.  
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In fact, the answer follows directly from examining the differences between each year. The largest differences are <math>6</math> and <math>7</math>. Due to the decreased starting number of students between <math>2002</math> and <math>2003</math>, that interval will be our answer.
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~edited by mobius247
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== Solution 2 ==
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We make the numerator 1 and compare the denominators.
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#<math>\frac{66 - 60}{60} = \frac{1}{10}</math>
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#<math>\frac{70 - 66}{66} = \frac{1}{16.5}</math>
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#<math>\frac{76-70}{70} \approx \frac{1}{11.7}</math>
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#<math>\frac{78-76}{76} = \frac{1}{38}</math>
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#<math>\frac{85-78}{78} \approx \frac{1}{11.1}</math>
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The answer is <math>\mathrm{(A)}</math>.
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~thatmathsguy
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:55, 2 June 2023

Problem

At Euclid High School, the number of students taking the AMC 10 was $60$ in 2002, $66$ in 2003, $70$ in 2004, $76$ in 2005, $78$ and 2006, and is $85$ in 2007. Between what two consecutive years was there the largest percentage increase?

$\text{(A)}\ 2002\ \text{and}\ 2003 \qquad \text{(B)}\ 2003\ \text{and}\ 2004 \qquad \text{(C)}\ 2004\ \text{and}\ 2005 \qquad \text{(D)}\ 2005\ \text{and}\ 2006 \qquad \text{(E)}\ 2006\ \text{and}\ 2007$

Solution 1

We compute the percentage increases:

  1. $\frac{66 - 60}{60} = 10\%$
  2. $\frac{70 - 66}{66} \approx 6\%$
  3. $\frac{76-70}{70} \approx 8.6\%$
  4. $\frac{78-76}{76} \approx 2.6\%$
  5. $\frac{85-78}{78} \approx 9\%$

The answer is $\mathrm{(A)}$.

In fact, the answer follows directly from examining the differences between each year. The largest differences are $6$ and $7$. Due to the decreased starting number of students between $2002$ and $2003$, that interval will be our answer.

~edited by mobius247

Solution 2

We make the numerator 1 and compare the denominators.

  1. $\frac{66 - 60}{60} = \frac{1}{10}$
  1. $\frac{70 - 66}{66} = \frac{1}{16.5}$
  1. $\frac{76-70}{70} \approx \frac{1}{11.7}$
  1. $\frac{78-76}{76} = \frac{1}{38}$
  1. $\frac{85-78}{78} \approx \frac{1}{11.1}$


The answer is $\mathrm{(A)}$.

~thatmathsguy

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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

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