Difference between revisions of "2020 AMC 12B Problems"

Line 125: Line 125:
 
==Problem 13==
 
==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 <math>\sqrt{\log_2{6}+\log_3{6}}?</math>
+
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>
 
<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>
Line 133: Line 133:
 
==Problem 14==
 
==Problem 14==
  
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
+
Bela and Jenn play the following game on the closed interval <math>[0, n]</math> of the real number line, where <math>n</math> is a fixed integer greater than <math>4</math>. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval <math>[0, n]</math>. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?
 +
 
 +
<math>\textbf{(A) } \text{Bela will always win.}</math>
 +
<math>\textbf{(B) } \text{Jenn will always win.} </math>
 +
<math>\textbf{(C) } \text{Bela will win if and only if }n \text{ is odd.}</math>
 +
<math>\textbf{(D) } \text{Jenn will win if and only if }n \text{ is odd.} </math>
 +
<math>\textbf{(E) } \text{Jenn will win if and only if }n > 8.</math>
  
 
[[2020 AMC 12B Problems/Problem 14|Solution]]
 
[[2020 AMC 12B Problems/Problem 14|Solution]]
Line 139: Line 145:
 
==Problem 15==
 
==Problem 15==
  
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
+
There are 10 people standing equally spaced around a circle. Each person knows exactly 3 of the other 9 people: the 2 people standing next to her or him, as well as the person directly across the circle. How many ways are there for the 10 people to split up into 5 pairs so that the members of each pair know each other?
 +
 
 +
<math>\textbf{(A) } 11 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15</math>
  
 
[[2020 AMC 12B Problems/Problem 15|Solution]]
 
[[2020 AMC 12B Problems/Problem 15|Solution]]
Line 145: Line 153:
 
==Problem 16==
 
==Problem 16==
  
These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.
+
An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
 +
 
 +
<math>\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12</math>
  
 
[[2020 AMC 12B Problems/Problem 16|Solution]]
 
[[2020 AMC 12B Problems/Problem 16|Solution]]

Revision as of 19:03, 7 February 2020

2020 AMC 12B (Answer Key)
Printable versions: WikiAoPS ResourcesPDF

Instructions

  1. 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.
  2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
  4. Figures are not necessarily drawn to scale.
  5. You will have 75 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

What is the value in simplest form of the following expression?\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]

$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

Solution

Problem 2

What is the value of the following expression? \[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$

Solution

Problem 3

The ratio of $w$ to $x$ is $4 : 3$, the ratio of $y$ to $z$ is $3 : 2$, and the ratio of $z$ to $x$ is $1 : 6$. What is the ratio of $w$ to $y$?

$\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3$

Solution

Problem 4

The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$, where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad\textbf{(E) }11$

Solution

Problem 5

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?

$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution

Problem 6

For all integers $n \geq 9,$ the value of \[\frac{(n+2)!-(n+1)!}{n!}\]is always which of the following?

$\textbf{(A) } \text{a multiple of }4 \qquad \textbf{(B) } \text{a multiple of }10 \qquad \textbf{(C) } \text{a prime number} \\ \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}$

Solution

Problem 7

Two nonhorizontal, non vertical lines in the $xy$-coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution

Problem 8

How many ordered pairs of integers $(x, y)$ satisfy the equation\[x^{2020}+y^2=2y?\] $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution

Problem 9

A three-quarter sector of a circle of radius $4$ 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]

$\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7$

Solution

Problem 10

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$

Solution

Problem 11

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length $2$ 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]

$\textbf{(A)}\ 6\sqrt3-3\pi \qquad\textbf{(B)}\ \frac{9\sqrt3}{2}-2\pi \qquad\textbf{(C)}\ \frac{3\sqrt3}{2}-\frac{\pi}{3} \qquad\textbf{(D)}\ 3\sqrt3-\pi \\ \qquad\textbf{(E)}\ \frac{9\sqrt3}{2}-\pi$

Solution

Problem 12

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$

$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution

Problem 13

Which of the following is the value of $\sqrt{\log_2{6}+\log_3{6}}?$

$\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}$

Solution

Problem 14

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4$. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n]$. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

$\textbf{(A) } \text{Bela will always win.}$ $\textbf{(B) } \text{Jenn will always win.}$ $\textbf{(C) } \text{Bela will win if and only if }n \text{ is odd.}$ $\textbf{(D) } \text{Jenn will win if and only if }n \text{ is odd.}$ $\textbf{(E) } \text{Jenn will win if and only if }n > 8.$

Solution

Problem 15

There are 10 people standing equally spaced around a circle. Each person knows exactly 3 of the other 9 people: the 2 people standing next to her or him, as well as the person directly across the circle. How many ways are there for the 10 people to split up into 5 pairs so that the members of each pair know each other?

$\textbf{(A) } 11 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

Solution

Problem 16

An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?

$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$

Solution

Problem 17

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 18

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 19

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 20

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

2020 AMC 12B (ProblemsAnswer KeyResources)
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. AMC logo.png