Difference between revisions of "2020 AMC 12B Problems"
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==Problem 1== | ==Problem 1== | ||
− | + | What is the value in simplest form of the following expression?<cmath>\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}</cmath> | |
+ | |||
+ | <math>\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}</math> | ||
[[2020 AMC 12B Problems/Problem 1|Solution]] | [[2020 AMC 12B Problems/Problem 1|Solution]] | ||
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==Problem 2== | ==Problem 2== | ||
− | + | What is the value of the following expression? | |
+ | <cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}</cmath><math>\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80} </math> | ||
[[2020 AMC 12B Problems/Problem 2|Solution]] | [[2020 AMC 12B Problems/Problem 2|Solution]] | ||
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==Problem 3== | ==Problem 3== | ||
− | + | The ratio of <math>w</math> to <math>x</math> is <math>4 : 3</math>, the ratio of <math>y</math> to <math>z</math> is <math>3 : 2</math>, and the ratio of <math>z</math> to <math>x</math> is <math>1 : 6</math>. What is the ratio of <math>w</math> to <math>y</math>? | |
+ | |||
+ | <math>\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3 </math> | ||
[[2020 AMC 12B Problems/Problem 3|Solution]] | [[2020 AMC 12B Problems/Problem 3|Solution]] | ||
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==Problem 4== | ==Problem 4== | ||
− | + | The acute angles of a right triangle are <math>a^{\circ}</math> and <math>b^{\circ}</math>, where <math>a>b</math> and both <math>a</math> and <math>b</math> are prime numbers. What is the least possible value of <math>b</math>? | |
+ | |||
+ | <math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad\textbf{(E) }11</math> | ||
[[2020 AMC 12B Problems/Problem 4|Solution]] | [[2020 AMC 12B Problems/Problem 4|Solution]] | ||
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==Problem 5== | ==Problem 5== | ||
− | + | Teams <math>A</math> and <math>B</math> are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team <math>A</math> has won <math>\tfrac{2}{3}</math> of its games and team <math>B</math> has won <math>\tfrac{5}{8}</math> of its games. Also, team <math>B</math> has won <math>7</math> more games and lost <math>7</math> more games than team <math>A.</math> How many games has team <math>A</math> played? | |
+ | |||
+ | <math>\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63</math> | ||
[[2020 AMC 12B Problems/Problem 5|Solution]] | [[2020 AMC 12B Problems/Problem 5|Solution]] | ||
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==Problem 6== | ==Problem 6== | ||
− | + | For all integers <math>n \geq 9,</math> the value of | |
+ | <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? | ||
+ | |||
+ | <math>\textbf{(A) } \text{a multiple of }4 \qquad \textbf{(B) } \text{a multiple of }10 \qquad \textbf{(C) } \text{a prime number} \\ \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}</math> | ||
[[2020 AMC 12B Problems/Problem 6|Solution]] | [[2020 AMC 12B Problems/Problem 6|Solution]] | ||
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==Problem 7== | ==Problem 7== | ||
− | + | Two nonhorizontal, non vertical lines in the <math>xy</math>-coordinate plane intersect to form a <math>45^{\circ}</math> angle. One line has slope equal to <math>6</math> times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines? | |
+ | |||
+ | <math>\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6</math> | ||
[[2020 AMC 12B Problems/Problem 7|Solution]] | [[2020 AMC 12B Problems/Problem 7|Solution]] | ||
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==Problem 8== | ==Problem 8== | ||
− | + | How many ordered pairs of integers <math>(x, y)</math> satisfy the equation<cmath>x^{2020}+y^2=2y?</cmath> | |
+ | <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}</math> | ||
[[2020 AMC 12B Problems/Problem 8|Solution]] | [[2020 AMC 12B Problems/Problem 8|Solution]] | ||
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==Problem 9== | ==Problem 9== | ||
− | + | A three-quarter sector of a circle of radius <math>4</math> inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches? | |
+ | <asy> | ||
+ | |||
+ | draw(Arc((0,0), 4, 0, 270)); | ||
+ | draw((0,-4)--(0,0)--(4,0)); | ||
+ | |||
+ | label("$4$", (2,0), S); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math> | ||
[[2020 AMC 12B Problems/Problem 9|Solution]] | [[2020 AMC 12B Problems/Problem 9|Solution]] | ||
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==Problem 10== | ==Problem 10== | ||
− | + | In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math> | |
+ | |||
+ | <math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math> | ||
[[2020 AMC 12B Problems/Problem 10|Solution]] | [[2020 AMC 12B Problems/Problem 10|Solution]] | ||
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==Problem 11== | ==Problem 11== | ||
− | + | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length <math>2</math> so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles? | |
+ | |||
+ | <asy> | ||
+ | size(140); | ||
+ | fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); | ||
+ | fill(arc((2,0),1,180,0)--(2,0)--cycle,white); | ||
+ | fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); | ||
+ | fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); | ||
+ | fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white); | ||
+ | fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); | ||
+ | fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); | ||
+ | draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); | ||
+ | draw(arc((2,0),1,180,0)--(2,0)--cycle); | ||
+ | draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); | ||
+ | draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); | ||
+ | draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); | ||
+ | draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); | ||
+ | draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); | ||
+ | label("$2$",(3.5,3sqrt(3)/2),NE); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ 6\sqrt3-3\pi \qquad\textbf{(B)}\ \frac{9\sqrt3}{2}-2\pi \qquad\textbf{(C)}\ \frac{3\sqrt3}{2}-\frac{\pi}{3} \qquad\textbf{(D)}\ 3\sqrt3-\pi \\ \qquad\textbf{(E)}\ \frac{9\sqrt3}{2}-\pi</math> | ||
[[2020 AMC 12B Problems/Problem 11|Solution]] | [[2020 AMC 12B Problems/Problem 11|Solution]] | ||
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==Problem 12== | ==Problem 12== | ||
− | + | Let <math>\overline{AB}</math> be a diameter in a circle of radius <math>5\sqrt2.</math> Let <math>\overline{CD}</math> be a chord in the circle that intersects <math>\overline{AB}</math> at a point <math>E</math> such that <math>BE=2\sqrt5</math> and <math>\angle AEC = 45^{\circ}.</math> What is <math>CE^2+DE^2?</math> | |
+ | |||
+ | <math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | ||
[[2020 AMC 12B Problems/Problem 12|Solution]] | [[2020 AMC 12B Problems/Problem 12|Solution]] | ||
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==Problem 13== | ==Problem 13== | ||
− | These problems will not be available until the 2020 AMC 12B contest is released on | + | These problems will not be available until the 2020 AMC 12B contest is released on Which of the following is the value of <math>\sqrt{\log_2{6}+\log_3{6}}?</math> |
+ | |||
+ | <math>\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}</math> | ||
[[2020 AMC 12B Problems/Problem 13|Solution]] | [[2020 AMC 12B Problems/Problem 13|Solution]] |
Revision as of 19:01, 7 February 2020
2020 AMC 12B (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
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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 |
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 See also
Problem 1
What is the value in simplest form of the following expression?
Problem 2
What is the value of the following expression?
Problem 3
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to ?
Problem 4
The acute angles of a right triangle are and , where and both and are prime numbers. What is the least possible value of ?
Problem 5
Teams and are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team has won of its games and team has won of its games. Also, team has won more games and lost more games than team How many games has team played?
Problem 6
For all integers the value of is always which of the following?
Problem 7
Two nonhorizontal, non vertical lines in the -coordinate plane intersect to form a angle. One line has slope equal to times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
Problem 8
How many ordered pairs of integers satisfy the equation
Problem 9
A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?
Problem 10
In unit square the inscribed circle intersects at and intersects at a point different from What is
Problem 11
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?
Problem 12
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is
Problem 13
These problems will not be available until the 2020 AMC 12B contest is released on Which of the following is the value of
Problem 14
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 15
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 16
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 17
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 18
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 19
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 20
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
See also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by 2020 AMC 12A Problems |
Followed by 2021 AMC 12A Problems |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.