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• 2009 AMC 8 Problems
  {{AMC8 Problems|year=2009}} ==Problem 1==
  18 KB (2,551 words) - 18:46, 27 February 2024
• 2010 AMC 8 Problems/Problem 4
  ==Problem== What is the sum of the mean, median, and mode of the numbers <math>2,3,0,3,1,4,0,3</math>?
  752 bytes (102 words) - 20:00, 5 February 2023
• 2010 AMC 8 Problems/Problem 21
  ==Problem== ...emaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realiz
  2 KB (350 words) - 22:35, 21 April 2024
• 2010 AMC 8 Problems/Problem 24
  ==Problem== ==Solution 1==
  2 KB (324 words) - 16:08, 31 December 2023
• 2010 AMC 8 Problems/Problem 23
  ==Problem== ...lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf);
  2 KB (298 words) - 00:29, 18 December 2023
• 2010 AMC 8 Problems
  {{AMC8 Problems|year=2010|}} ==Problem 1==
  18 KB (2,768 words) - 21:05, 9 January 2024
• 2011 AMC 8 Problems
  {{AMC8 Problems|year=2011|}} ==Problem 1==
  16 KB (2,371 words) - 17:34, 9 January 2024
• 2010 AMC 8 Problems/Problem 22
  ==Problem== ==Solution 1==
  2 KB (338 words) - 09:06, 10 March 2023
• 2010 AMC 8 Problems/Problem 2
  ==Problem== <math>\textbf{(A)}\ \frac{3}{10} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{10}{3} \qquad\textbf{(E)}\
  669 bytes (94 words) - 19:59, 5 February 2023
• 2010 AMC 8 Problems/Problem 13
  ==Problem== ==Solution 1(algebra solution)==
  1 KB (212 words) - 20:54, 22 January 2023
• 2010 AMC 8 Problems/Problem 25
  ==Problem== ...3</math> at a time. For example, Jo could climb <math>3</math>, then <math>1</math>, then <math>2</math>. In how many ways can Jo climb the stairs?
  4 KB (595 words) - 13:38, 4 April 2024
• 2010 AMC 8 Problems/Problem 19
  == Problem == ...+ fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);
  2 KB (279 words) - 09:04, 10 March 2023
• 2010 AMC 8 Problems/Problem 11
  == Problem == == Solution 1(algebra solution)==
  1 KB (205 words) - 20:52, 22 January 2023
• 2010 AMC 8 Problems/Problem 17
  ==Problem== pen xdxdff = rgb(0.49,0.49,1);
  3 KB (459 words) - 21:04, 23 December 2023
• 2010 AMC 8 Problems/Problem 18
  ==Problem== ...((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0));
  2 KB (301 words) - 09:04, 10 March 2023
• 2010 AMC 8 Problems/Problem 3
  ==Problem== ...h(0.7); pointpen=black; pen fp = fontsize(10); pointfontpen=fp; real xmin=-1.33,xmax=11.05,ymin=-9.01,ymax=-0.44;
  5 KB (847 words) - 19:59, 5 February 2023
• 2010 AMC 8 Problems/Problem 6
  ==Problem== An isosceles trapezoid has <math>1</math> line of symmetry.
  900 bytes (126 words) - 18:16, 15 October 2023
• 2010 AMC 8 Problems/Problem 7
  ==Problem== ...> dimes, and <math>4</math> pennies. Thus, the correct answer is <math>3+2+1+4=\boxed{\textbf{(B)}\ 10}</math>.
  1 KB (183 words) - 14:33, 28 December 2023
• 2010 AMC 8 Problems/Problem 12
  ==Problem== ==Solution 1(logical solution)==
  2 KB (269 words) - 14:46, 19 August 2023
• 2010 AMC 8 Problems/Problem 8
  ==Problem== ...After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math>
  1 KB (224 words) - 15:26, 22 December 2023