Difference between revisions of "2010 AMC 8 Problems"

Latest revision as of 20:50, 16 October 2024

2010 AMC 8 (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 1 point for each correct answer. There is no penalty for wrong answers. No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted. Rules Figures are not necessarily drawn to scale. You will have 40 minutes working time to complete the test.
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

Problem 1

At Euclid Middle School, the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest?

$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

@@ Line 1: / Line 1: @@
+{{AMC8 Problems|year=2010|}}
 ==Problem 1==
-At Euclid High School, the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are <math>11</math> students in Mrs. Germain's class, 8 in Mr. Newton, and <math>9</math> in Mrs. Young's class are taking the AMC <math>8</math> this year. How many mathematics students at Euclid High School are taking the contest?
+At Euclid Middle School, the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are <math>11</math> students in Mrs. Germain's class, <math>8</math> students in Mr. Newton's class, and <math>9</math> students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest?
 <math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math>
+[[2010 AMC 8 Problems/Problem 1 | Solution]]
 ==Problem 2==
-If <math>a @ b = \frac{a\times b}{a+b}</math>, for <math>a,b</math> positive integers, then what is <math>5 @10</math>?
+If <math>a @ b = \frac{a\times b}{a+b}</math> for <math>a,b</math> positive integers, then what is <math>5 @ 10</math>?
 <math>\textbf{(A)}\ \frac{3}{10} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{10}{3} \qquad\textbf{(E)}\ 50</math>
+[[2010 AMC 8 Problems/Problem 2 | Solution]]
 ==Problem 3==
@@ Line 16: / Line 21: @@
 pen ycycyc=rgb(0.55,0.55,0.55);
 pair A=(1,-6), B=(1,-2), D=(1,-5.8), E=(1,-5.6), F=(1,-5.4), G=(1,-5.2), H=(1,-5), J=(1,-4.8), K=(1,-4.6), L=(1,-4.4), M=(1,-4.2), N=(1,-4), P=(1,-3.8), Q=(1,-3.6), R=(1,-3.4), S=(1,-3.2), T=(1,-3), U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,-6), P_1=(14,-5), Q_1=(14,-4), R_1=(14,-3), S_1=(14,-2), C_1=(1.4,-6), D_1=(1.8,-6), G_1=(2.4,-6), H_1=(2.8,-6), I_1=(3.4,-6), J_1=(3.8,-6), K_1=(4.4,-6), L_1=(4.8,-6), M_1=(5.4,-6), N_1=(5.8,-6), T_1=(6.4,-6), U_1=(6.8,-6), V_1=(7.4,-6), W_1=(7.8,-6), Z_1=(8.4,-6), A_2=(8.8,-6), B_2=(9.4,-6), C_2=(9.8,-6), D_2=(10.4,-6), E_2=(10.8,-6), L_2=(2.4,-3.2), M_2=(2.8,-3.2), N_2=(3.4,-4), O_2=(3.8,-4), P_2=(4.4,-3.6), Q_2=(4.8,-3.6), R_2=(5.4,-3.6), S_2=(5.8,-3.6), T_2=(6.4,-3.4), U_2=(6.8,-3.4), V_2=(7.4,-3.8), W_2=(7.8,-3.8), Z_2=(8.4,-2.8), A_3=(8.8,-2.8), B_3=(9.4,-3.2), C_3=(9.8,-3.2), D_3=(10.4,-3.8), E_3=(10.8,-3.8);
-filldraw(C_1--E_1--F_1--D_1--cycle,ycycyc); filldraw(G_1--L_2--M_2--H_1--cycle,ycycyc); filldraw(I_1--N_2--O_2--J_1--cycle,ycycyc); filldraw(K_1--P_2--Q_2--L_1--cycle,ycycyc); filldraw(M_1--R_2--S_2--N_1--cycle,ycycyc); filldraw(T_1--T_2--U_2--U_1--cycle,ycycyc); filldraw(V_1--V_2--W_2--W_1--cycle,ycycyc); filldraw(Z_1--Z_2--A_3--A_2--cycle,ycycyc); filldraw(B_2--B_3--C_3--C_2--cycle,ycycyc); filldraw(D_2--D_3--E_3--E_2--cycle,ycycyc); D(B--A,linewidth(0.4)); D(H--(8,-5),linewidth(0.4)); D(N--(8,-4),linewidth(0.4)); D(T--(8,-3),linewidth(0.4)); D(B--(8,-2),linewidth(0.4)); D(B--S_1); D(T--R_1); D(N--Q_1); D(H--P_1); D(A--O_1); D(C_1--E_1); D(E_1--F_1); D(F_1--D_1); D(D_1--C_1); D(G_1--L_2); D(L_2--M_2); D(M_2--H_1); D(H_1--G_1); D(I_1--N_2); D(N_2--O_2); D(O_2--J_1); D(J_1--I_1); D(K_1--P_2); D(P_2--Q_2); D(Q_2--L_1); D(L_1--K_1); D(M_1--R_2); D(R_2--S_2); D(S_2--N_1); D(N_1--M_1); D(T_1--T_2); D(T_2--U_2); D(U_2--U_1); D(U_1--T_1); D(V_1--V_2); D(V_2--W_2); D(W_2--W_1); D(W_1--V_1); D(Z_1--Z_2); D(Z_2--A_3); D(A_3--A_2); D(A_2--Z_1); D(B_2--B_3); D(B_3--C_3); D(C_3--C_2); D(C_2--B_2); D(D_2--D_3); D(D_3--E_3); D(E_3--E_2); D(E_2--D_2); label("0",(0.88,-5.91),SE*lsf,fp); label("\$ 5",(0.3,-4.84),SE*lsf,fp); label("\$ 10",(0.2,-3.84),SE*lsf,fp); label("\$ 15",(0.2,-2.85),SE*lsf,fp); label("\$ 20",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95),SE*lsf,fp); label("$3$",(3.53,-5.94),SE*lsf,fp); label("$4$",(4.55,-5.94),SE*lsf,fp); label("$5$",(5.49,-5.95),SE*lsf,fp); label("$6$",(6.53,-5.95),SE*lsf,fp); label("$7$",(7.55,-5.95),SE*lsf,fp); label("$8$",(8.52,-5.95),SE*lsf,fp); label("$9$",(9.57,-5.97),SE*lsf,fp); label("$10$",(10.56,-5.94),SE*lsf,fp); label("Month",(7.14,-6.43),SE*lsf,fp);
+filldraw(C_1--E_1--F_1--D_1--cycle,ycycyc); filldraw(G_1--L_2--M_2--H_1--cycle,ycycyc); filldraw(I_1--N_2--O_2--J_1--cycle,ycycyc); filldraw(K_1--P_2--Q_2--L_1--cycle,ycycyc); filldraw(M_1--R_2--S_2--N_1--cycle,ycycyc); filldraw(T_1--T_2--U_2--U_1--cycle,ycycyc); filldraw(V_1--V_2--W_2--W_1--cycle,ycycyc); filldraw(Z_1--Z_2--A_3--A_2--cycle,ycycyc); filldraw(B_2--B_3--C_3--C_2--cycle,ycycyc); filldraw(D_2--D_3--E_3--E_2--cycle,ycycyc); D(B--A,linewidth(0.4)); D(H--(8,-5),linewidth(0.4)); D(N--(8,-4),linewidth(0.4)); D(T--(8,-3),linewidth(0.4)); D(B--(8,-2),linewidth(0.4)); D(B--S_1); D(T--R_1); D(N--Q_1); D(H--P_1); D(A--O_1); D(C_1--E_1); D(E_1--F_1); D(F_1--D_1); D(D_1--C_1); D(G_1--L_2); D(L_2--M_2); D(M_2--H_1); D(H_1--G_1); D(I_1--N_2); D(N_2--O_2); D(O_2--J_1); D(J_1--I_1); D(K_1--P_2); D(P_2--Q_2); D(Q_2--L_1); D(L_1--K_1); D(M_1--R_2); D(R_2--S_2); D(S_2--N_1); D(N_1--M_1); D(T_1--T_2); D(T_2--U_2); D(U_2--U_1); D(U_1--T_1); D(V_1--V_2); D(V_2--W_2); D(W_2--W_1); D(W_1--V_1); D(Z_1--Z_2); D(Z_2--A_3); D(A_3--A_2); D(A_2--Z_1); D(B_2--B_3); D(B_3--C_3); D(C_3--C_2); D(C_2--B_2); D(D_2--D_3); D(D_3--E_3); D(E_3--E_2); D(E_2--D_2); label("0",(0.88,-5.91),SE*lsf,fp); label("$ 5$",(0.3,-4.84),SE*lsf,fp); label("$ 10$",(0.2,-3.84),SE*lsf,fp); label("$ 15$",(0.2,-2.85),SE*lsf,fp); label("$ 20$",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95),SE*lsf,fp); label("$3$",(3.53,-5.94),SE*lsf,fp); label("$4$",(4.55,-5.94),SE*lsf,fp); label("$5$",(5.49,-5.95),SE*lsf,fp); label("$6$",(6.53,-5.95),SE*lsf,fp); label("$7$",(7.55,-5.95),SE*lsf,fp); label("$8$",(8.52,-5.95),SE*lsf,fp); label("$9$",(9.57,-5.97),SE*lsf,fp); label("$10$",(10.56,-5.94),SE*lsf,fp); label("Month",(7.14,-6.43),SE*lsf,fp);
 D(A,linewidth(1pt)); D(B,linewidth(1pt)); D(D,linewidth(1pt)); D(E,linewidth(1pt)); D(F,linewidth(1pt)); D(G,linewidth(1pt)); D(H,linewidth(1pt)); D(J,linewidth(1pt)); D(K,linewidth(1pt)); D(L,linewidth(1pt)); D(M,linewidth(1pt)); D(N,linewidth(1pt)); D(P,linewidth(1pt)); D(Q,linewidth(1pt)); D(R,linewidth(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth(1pt)); D(F_1,linewidth(1pt)); D(O_1,linewidth(1pt)); D(P_1,linewidth(1pt)); D(Q_1,linewidth(1pt)); D(R_1,linewidth(1pt)); D(S_1,linewidth(1pt)); D(C_1,linewidth(1pt)); D(D_1,linewidth(1pt)); D(G_1,linewidth(1pt)); D(H_1,linewidth(1pt)); D(I_1,linewidth(1pt)); D(J_1,linewidth(1pt)); D(K_1,linewidth(1pt)); D(L_1,linewidth(1pt)); D(M_1,linewidth(1pt)); D(N_1,linewidth(1pt)); D(T_1,linewidth(1pt)); D(U_1,linewidth(1pt)); D(V_1,linewidth(1pt)); D(W_1,linewidth(1pt)); D(Z_1,linewidth(1pt)); D(A_2,linewidth(1pt)); D(B_2,linewidth(1pt)); D(C_2,linewidth(1pt)); D(D_2,linewidth(1pt)); D(E_2,linewidth(1pt)); D(L_2,linewidth(1pt)); D(M_2,linewidth(1pt)); D(N_2,linewidth(1pt)); D(O_2,linewidth(1pt)); D(P_2,linewidth(1pt)); D(Q_2,linewidth(1pt)); D(R_2,linewidth(1pt)); D(S_2,linewidth(1pt)); D(T_2,linewidth(1pt)); D(U_2,linewidth(1pt)); D(V_2,linewidth(1pt)); D(W_2,linewidth(1pt)); D(Z_2,linewidth(1pt)); D(A_3,linewidth(1pt)); D(B_3,linewidth(1pt)); D(C_3,linewidth(1pt)); D(D_3,linewidth(1pt)); D(E_3,linewidth(1pt));
 clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy>
@@ Line 25: / Line 30: @@
 \textbf{(D)}\ 89 \qquad
 \textbf{(E)}\ 100</math>
+[[2010 AMC 8 Problems/Problem 3 | Solution]]
 ==Problem 4==
-What is the sum of the mean, medium, and mode of the numbers, <math>2,3,0,3,1,4,0,3</math>?
+What is the sum of the mean, median, and mode of the numbers <math>2,3,0,3,1,4,0,3</math>?
 <math> \textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9 </math>
+[[2010 AMC 8 Problems/Problem 4 | Solution]]
 ==Problem 5==
-Alice needs to replace a light bulb located <math>10</math> centimeters below the ceiling of her kitchen. The ceiling is <math>2.4</math> meters above the floor. Alice is <math>1.5</math> meters tall and can reach <math>46</math> centimeters above her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
+Alice needs to replace a light bulb located <math>10</math> centimeters below the ceiling in her kitchen. The ceiling is <math>2.4</math> meters above the floor. Alice is <math>1.5</math> meters tall and can reach <math>46</math> centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
 <math> \textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40 </math>
+[[2010 AMC 8 Problems/Problem 5 | Solution]]
 ==Problem 6==
-Which of the following has the greatest number of line of symmetry?
+Which of the following figures has the greatest number of lines of symmetry?
+<math> \textbf{(A)}\ \text{equilateral triangle}</math>
+<math>\textbf{(B)}\ \text{non-square rhombus} </math>
+<math>\textbf{(C)}\ \text{non-square rectangle}</math>
+<math>\textbf{(D)}\ \text{isosceles trapezoid}</math>
+<math>\textbf{(E)}\ \text{square} </math>
-<math> \textbf{(A)}\ \text{ Equilateral Triangle}</math>
+[[2010 AMC 8 Problems/Problem 6 | Solution]]
-<math>\textbf{(B)}\ \text{Non-square rhombus} </math>
-<math>\textbf{(C)}\ \text{Non-square rectangle}</math>
-<math>\textbf{(D)}\ \text{Isosceles Triangle}</math>
-<math>\textbf{(E)}\ \text{Square} </math>
 ==Problem 7==
-Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than one dollar?
+Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
 <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99 </math>
+[[2010 AMC 8 Problems/Problem 7 | Solution]]
 ==Problem 8==
-As Emily is riding her bike on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. 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> miles per hour. Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson?
+As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. 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> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson?
 <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math>
+[[2010 AMC 8 Problems/Problem 8 |  Solution]]
 ==Problem 9==
-Ryan got <math>80\%</math> of the problems on a <math>25</math>-problem test, <math>90\%</math> on a <math>40</math>-problem test, and <math>70\%</math> on a <math>10</math>-problem test. What percent of all problems did Ryan answer correctly?
+Ryan got <math>80\%</math> of the problems correct on a <math>25</math>-problem test, <math>90\%</math> on a <math>40</math>-problem test, and <math>70\%</math> on a <math>10</math>-problem test. What percent of all the problems did Ryan answer correctly?
 <math> \textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86 </math>
+[[2010 AMC 8 Problems/Problem 9 | Solution]]
 ==Problem 10==
-<math>6</math> pepperoni circles will exactly fit across the diameter of a <math>12</math>-inch pizza when placed. If a total of <math>24</math> circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered with pepperoni?
+Six pepperoni circles will exactly fit across the diameter of a <math>12</math>-inch pizza when placed. If a total of <math>24</math> circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
 <math> \textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78 </math>
+[[2010 AMC 8 Problems/Problem 10 | Solution]]
 ==Problem 11==
-The top of one tree is <math>16</math> feet higher than the top of another tree. The height of the <math>2</math> trees are at a ratio of <math>3:4</math>. In feet, how tall is the taller tree?
+The top of one tree is <math>16</math> feet higher than the top of another tree. The heights of the two trees are in the ratio <math>3:4</math>. In feet, how tall is the taller tree?
 <math> \textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112 </math>
+[[2010 AMC 8 Problems/Problem 11 | Solution]]
+==Problem 12==
+Of the <math>500</math> balls in a large bag, <math>80\%</math> are red and the rest are blue. How many of the red balls must be removed from the bag so that <math>75\%</math> of the remaining balls are red?
+<math> \textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150 </math>
+[[2010 AMC 8 Problems/Problem 12 | Solution]]
 ==Problem 13==
-The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shorter side is <math>30\%</math> of the perimeter. What is the length of the longest side?
+The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is <math>30\%</math> of the perimeter. What is the length of the longest side?
 <math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
+[[2010 AMC 8 Problems/Problem 13 | Solution]]
 ==Problem 14==
@@ Line 79: / Line 112: @@
 <math> \textbf{(A)}\ 67 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210 </math>
+[[2010 AMC 8 Problems/Problem 14 | Solution]]
 ==Problem 15==
-A jar contains <math>5</math> different colors of gumdrops. <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> red, <math>10\%</math> yellow, and the other <math>30</math> gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
+A jar contains five different colors of gumdrops: <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> red, <math>10\%</math> yellow, and the other <math>30</math> gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
 <math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 </math>
+[[2010 AMC 8 Problems/Problem 15 | Solution]]
 ==Problem 16==
@@ Line 89: / Line 126: @@
 <math> \textbf{(A)}\  \frac{\sqrt{\pi}}{2}  \qquad\textbf{(B)}\ \sqrt{\pi} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ 2\pi \qquad\textbf{(E)}\ \pi^{2}</math>
+[[2010 AMC 8 Problems/Problem 16 | Solution]]
 ==Problem 17==
@@ Line 94: / Line 133: @@
 <asy>
-import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);
+import graph; size(300);
-draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt)); draw((1,1)--(5,1),linewidth(1.2pt)+linetype("2pt 2pt")); draw((1,1)--(1,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((2,2)--(2,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((3,2)--(3,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((4,2)--(4,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((5,1)--(5,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((0,0)--(5,1.5),linewidth(1.2pt));
+real lsf = 0.5;
-dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);
+pen dp = linewidth(0.7) + fontsize(10);
+defaultpen(dp);
+pen ds = black;
+pen xdxdff = rgb(0.49,0.49,1);
+draw((0,0)--(6,0),linewidth(1.2pt));
+draw((0,0)--(0,1),linewidth(1.2pt));
+draw((0,1)--(1,1),linewidth(1.2pt));
+draw((1,1)--(1,2),linewidth(1.2pt));
+draw((1,2)--(5,2),linewidth(1.2pt));
+draw((5,2)--(5,1),linewidth(1.2pt));
+draw((5,1)--(6,1),linewidth(1.2pt));
+draw((6,1)--(6,0),linewidth(1.2pt));
+draw((1,1)--(5,1),linewidth(1.2pt));
+draw((1,1)--(1,0),linewidth(1.2pt));
+draw((2,2)--(2,0),linewidth(1.2pt));
+draw((3,2)--(3,0),linewidth(1.2pt));
+draw((4,2)--(4,0),linewidth(1.2pt));
+draw((5,1)--(5,0),linewidth(1.2pt));
+draw((0,0)--(5,1.5),linewidth(1.2pt));
+dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf);
+dot((0,1),ds);
+dot((1,1),ds);
+dot((1,2),ds);
+dot((5,2),ds);
+label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds);
+label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds);
+dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds);
+dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds);
+dot((4,2),ds); dot((5,1.5),ds);
+label("$Q$", (5.14,1.51),NE*lsf);
+clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);
 </asy>
@@ Line 104: / Line 173: @@
 \textbf{(D)}\ \frac 23 \qquad
 \textbf{(E)}\ \frac 34</math>
+[[2010 AMC 8 Problems/Problem 17 | Solution]]
 ==Problem 18==
-A decorative window is made up of a rectangle with semicircles at either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>. And <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircle.
+A decorative window is made up of a rectangle with semicircles on either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>, and <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles?
 <asy>
@@ Line 114: / Line 185: @@
 <math> \textbf{(A)}\ 2:3 \qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi \qquad\textbf{(D)}\ 9: \pi \qquad\textbf{(E)}\ 30 : \pi</math>
+[[2010 AMC 8 Problems/Problem 18 | Solution]]
 ==Problem 19==
@@ Line 126: / Line 199: @@
 <math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math>
+[[2010 AMC 8 Problems/Problem 19 | Solution]]
 ==Problem 20==
@@ Line 131: / Line 206: @@
 <math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20 </math>
+[[2010 AMC 8 Problems/Problem 20 | Solution]]
 ==Problem 21==
-Hui is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, she read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> more. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> more. She then realizes she has <math>62</math> pages left, which she finishes the next day. How many pages are in this book?
+Hui is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining 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 realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book?
 <math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math>
+[[2010 AMC 8 Problems/Problem 21 | Solution]]
 ==Problem 22==
@@ Line 141: / Line 220: @@
 <math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math>
+[[2010 AMC 8 Problems/Problem 22 | Solution]]
 ==Problem 23==
-Semicircles <math>POQ</math> and <math>ROS</math> pass through the center of circle <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>?
+Semicircles <math>POQ</math> and <math>ROS</math> pass through the center of circle <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>?
 <asy>
-import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf);
+import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf);
 dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds);
-clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
+clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy>
-</asy>
+<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math>
-<math> \textbf{(A)}\ \frac{\sqrt 2}4 \qquad\textbf{(B)}\ \frac 12 \qquad\textbf{(C)}\ \frac{2}{\pi} \qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math>
+[[2010 AMC 8 Problems/Problem 23 | Solution]]
 ==Problem 24==
 What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?
-<math> \textbf{(A)}\ 2^2^4<10^8<5^1^2 </math>
+<math> \textbf{(A)}\ 2{}^2{}^4<10^8<5{}^1{}^2 </math>
-<math> \textbf{(B)}\ 2^2^4<5^1^2<10^8 </math>
-<math> \textbf{(C)}\ 5^1^2<2^2^4<10^8 </math>
+<math> \textbf{(B)}\ 2{}^2{}^4<5{}^1{}^2<10^8 </math>
-<math> \textbf{(D)}\ 10^8<5^1^2<2^2^4</math>
-<math> \textbf{(E)}\ 10^8<2^2^4<5^1^2 </math>
+<math> \textbf{(C)}\ 5{}^1{}^2<2{}^2{}^4<10^8 </math>
+<math> \textbf{(D)}\ 10^8<5{}^1{}^2<2{}^2{}^4</math>
+<math> \textbf{(E)}\ 10^8<2{}^2{}^4<5{}^1{}^2 </math>
+[[2010 AMC 8 Problems/Problem 24 | Solution]]
 ==Problem 25==
-Everyday at school, Jo climbs a flight of <math>6</math> stairs. Joe can take the stairs <math>1,2</math>, or <math>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?
+Everyday at school, Jo climbs a flight of <math>6</math> stairs. Jo can take the stairs <math>1</math>, <math>2</math>, or <math>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?
 <math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math>
+[[2010 AMC 8 Problems/Problem 25 | Solution]]
+==See Also==
+{{AMC8 box|year=2010|before=[[2009 AMC 8 Problems|2009 AMC 8]]|after=[[2011 AMC 8 Problems|2011 AMC 8]]}}
+* [[AMC 8]]
+* [[AMC 8 Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by 2009 AMC 8	Followed by 2011 AMC 8
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All AJHSME/AMC 8 Problems and Solutions

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Difference between revisions of "2010 AMC 8 Problems"

Latest revision as of 20:50, 16 October 2024

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

See Also