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Difference between revisions of "Mock AMC 10B Problems"

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[[2019 Mock AMC 10B  Problems/Problem 24|Solution]]
 
[[2019 Mock AMC 10B  Problems/Problem 24|Solution]]
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===Solution 1===
 +
 +
Let's label the people from three schools with A,B,C. We can consider the case in which A is in seat number 1. Now, we consider to space between two successive A's. There are totally 8 spaces, which can be broken into sum of four positive integers:
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<math>8 = 1 + 1 + 1 + 5 = 1 + 1 + 2 + 4 = 1 + 2 +2 + 3 = 1 + 1 + 3 + 3 = 2 + 2 + 2 + 2</math>.
 +
 +
For case of <math>1+1+1+5</math>, there are four possible orders. For each order, we can arrange 5 people in the space with 5 seats as <math>BCBCB</math> or <math>CBCBC</math>, and then arrange two B's or two C's in two of three remaining spaces, making a total of <math>2\times 3\times = 24</math>.
 +
 +
For case of <math>1+1+2+4</math>, there are twelve possible orders, <math>1124</math>, <math>1241</math>, <math>2411</math>, <math>4112</math>, <math>1142</math>, <math>1421</math>, <math>4211</math>, <math>2114</math>, <math>1214</math>, <math>2141</math>, <math>1412</math>, <math>4121</math>. For each order, we can arrange 4 people in the space with 4 seats as <math>BCBC</math> or <math>CBCB</math>, and then arrange the space with two seats as <math>BC</math> or <math>CB</math>, and then choose 1 space for <math>B</math> and one space for <math>C</math>, making a total of <math>2\times 2\times 2\times 12 = 96</math>.
 +
 +
For case of <math>1+2+2+3</math>, there are twelve possible orders, similar to the case above. For each order, we can arrange 3 people in the space with 3 seats as <math>BCB</math> or <math>CBC</math>, and then arrange the space with two seats as <math>BC</math> or <math>CB</math>, and then arrange the last people in the space with 1 seat, making a total of <math>2\times 2\times 2\times 12= 96</math>.
 +
 +
For case of <math>1+1+3+3</math>, there are six possible orders, <math>1133</math>, <math>1331</math>, <math>3311</math>, <math>3113</math>, <math>1313</math>, <math>3131</math>. For each order, we can arrange 3 people in the two spaces with 3 seats as <math>BCB/BCB</math> or <math>BCB/CBC</math> or <math>CBC/BCB</math> or <math>CBC/CBC</math>, and then arrange the space with one seats for the people left, so there six arrangements in total: <math>BCB/BCB/C/C</math>, <math>BCB/CBC/B/C</math>, <math>BCB/CBC/C/B</math>, <math>CBC/BCB/B/C</math>, <math>CBC/BCB/C/B</math>, <math>CBC/CBC/B/B</math>, making a total of <math>6\times 6 = 36</math>.
 +
 +
For case of <math>2+2+2+2</math>, there is only one order, and each space can be arranged as <math>BC</math> or <math>CB</math>, making a total of 16.
 +
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Considering the three possibilities for seat 1, the final result is
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<cmath>
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3\times(24+96+96+36+16) = \boxed{804}.
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</cmath>
  
 
===Problem 25===
 
===Problem 25===

Revision as of 01:59, 21 January 2021

Problem 1

What is the difference between $6+7+8+9+10$ and $1+2+3+4+5$?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$

Solution

Problem 2

Al, Bob, Clayton, Derek, Ethan, and Frank are six Boy Scouts that will be split up into two groups of three Boy Scouts for a boating trip. How many ways are there to split up the six boys if the two groups are indistinguishable?


$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 35$

Solution

Problem 3

Which of these numbers is a rational number?


$\textbf{(A) }(\sqrt[3]{3})^{2018} \qquad \textbf{(B) }(\sqrt{3})^{2019} \qquad \textbf{(C) }(3+\sqrt{2})^2 \qquad \textbf{(D) }(2\pi)^2 \qquad \textbf{(E) }(3+\sqrt{2})(3-\sqrt{2}) \qquad$

Solution

Problem 4

In the diagram below, $ABC$ is an isosceles right triangle with a right angle at $B$ and with a hypotenuse of $40\sqrt{2}$ units. Find the greatest integer less than or equal to the value of the radius of the quarter circle inscribed inside $ABC$.

[asy] label("$B$", (8.5, -0.5), S); label("$A$", (8.5, (9sqrt(2)+0.5)), S); label("$C$", ((9.5+9sqrt(2)), -0.5), S); draw((9,0)--((9+9sqrt(2)),0)); draw((9,0)--(9,9sqrt(2))); draw(((9+9sqrt(2)),0)--(9,9sqrt(2))); draw(arc((9,0),9,0,90)); [/asy]

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

Solution

Problem 5

The three medians of the unit equilateral triangle $ABC$ intersect at point $P$. Find $PA + PB + PC$.

$\textbf{(A)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ 2$

Solution

Problem 6

Mark rolled two standard dice. Given that he rolled two distinct values, find the probability that he rolled two primes.

$\textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{7}\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{5}$

Solution

Problem 7

What is the sum of the solutions to $n^2=x^2-8x+96$?, where $n$ is a positive integer?

$\textbf{(A) }8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) }12$

Solution

Problem 8

In the following diagram, Bob starts at the origin and makes a certain number of moves. A move is defined as him starting at $(x,y)$ and moves to $(x,y+1)$, $(x+1,y)$, $(x,y-1)$, and $(x-1,y)$ with equal probability. The probability that Bob will eventually reach the point $(4,3)$ is $N$. Find the number of distinct points, including $(4, 3)$, that satisfy that the probability that he will eventually reach that point is $N$.

$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 12$

Solution

Problem 9

Consider the line segment $OA_0$, which has two endpoints $O = (0, 0)$ and $A = (5, 0)$. $OA_n$ is constructed by rotating $OA_0$ about the point $O$ clockwise $\frac{360n}{\mu}$ degrees, where $\mu$ is a positive integer greater than 2 and $n < \mu$. After this operation, the line segments $A_0A_1$, $A_1A_2$, $A_2A_3$, $...$, $A_{n-2}A_{n-1}$, $A_{n-1}A_0$ are drawn. Let $S$ be the sum of the areas of the Triangles $OA_0A_1, OA_1A_2, OA_2A_3, ..., OA_{n-2}A_{n-1}, OA_{n-1}A_0$. As $n$ approaches infinity, $S$ approaches a constant $p$. Find $\lfloor p \rfloor$.


$\textbf{(A)}\ 77\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 79\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 81$

Solution

Problem 10

A certain period of time $P$ starts at exactly 6:09PM on a Tuesday and ends at exactly 6:09AM on a Thursday. Which of these numbers listed in the choices here is a possible length in days for $P$?

$\mathrm{(A) \ } 100.5\qquad \mathrm{(B) \ } 1000.5\qquad \mathrm{(C) \ } 10,000.5\qquad \mathrm{(D) \ } 100,000.5\qquad \mathrm{(E) \ } 1,000,000.5$

Solution

Problem 11

Consider Square $ABCD$, a square with side length $10$. Let Points $E$, $F$, $G$, $H$ be the midpoints of sides $AB$, $BC$, $CD$, and $DA$, respectively. Find the area of the square formed by the four line segments $AG$, $BH$, $CE$, and $DF$.

[asy] draw((0,0)--(10,0)); draw((0,0)--(0, 10)); draw((10,0)--(10, 10)); draw((10,10)--(0, 10)); draw((0,10)--(5, 0)); draw((0,0)--(10, 5)); draw((10,0)--(5, 10)); draw((10,10)--(0, 5)); [/asy]

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$

Solution

Problem 12

[asy] draw((0,0)--(10*sqrt(3),0)); draw((0,0)--(0, 10)); draw((10*sqrt(3),0)--(0, 10)); draw(arc((5*sqrt(3),0),5*sqrt(3),0,180)); label("$A$",(3,4)); [/asy]

In the figure shown here, the triangle has two legs of length $10$ and $10\sqrt{3}$, and the semicircle has diameter $10\sqrt{3}$. The area of Region $A$ can be expressed as $\frac{a\pi+b\sqrt{c}}{d}$, where $a, b, c, d$ are positive integers, $c$ is square-free, $\text{ gcd }(a, d) = 1$, and $\text{ gcd }(b, d) = 1$. Find $a+b+c+d$.

$\textbf{(A)}\ 130 \qquad\textbf{(B)}\ 131 \qquad\textbf{(C)}\ 132 \qquad\textbf{(D)}\ 133 \qquad\textbf{(E)}\ 134$

Solution

Problem 13

Kevin has a friend named Anna. The two of them are both in the same class, BC Calculus, which is a class that has $32$ students. To split the class up into partners that work on a group project involving integrals, the teacher, Mrs. Jannesen, randomly partitions the class into groups of two. If he is assigned to be partners with his friend, he will be happy. What is the probability that Kevin is assigned to be with Anna?

$\mathrm{(A) \ } \frac{1}{30}\qquad \mathrm{(B) \ } \frac{1}{31}\qquad \mathrm{(C) \ } \frac{1}{32}\qquad \mathrm{(D) \ } \frac{1}{33}\qquad \mathrm{(E) \ } \frac{1}{34}$

Solution

Problem 14

Let $S$ be the number of distinct triangles that can be formed from $5$ coplanar points. Find the sum of all possible values of $S$.

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 33$

Solution

Problem 15

In the figure below, a square of area $108$ is inscribed inside a square of area $144$. There are two segments, labeled $m$ and $n$. The value of $m$ can be expressed as $a + b \sqrt{c}$, where $a, b, c$ are positive integers and $c$ is square-free. Find $a+b+c$.

[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$n$",(-0.1,0.15)); label("$m$",(-0.1,1.15));[/asy]

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

Solution

Problem 16

For a particular positive integer $n$, the number of ordered sextuples of positive integers $(a, b, c, d, e, f)$ that satisfy $a+b+c+d+e+f \leq n$ is exactly $3003$. Find $n$.

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

Solution

Problem 17

Let $S$ be a regular octagon. How many distinct quadrilaterals can be formed from the vertices of $S$ given that two quadrilaterals are not distinct if the latter can be obtained by a rotation of the former?

[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 35 \qquad\textbf{(E)}\ 70$

Solution

Problem 18

Two logs of length 10 are laying on the ground touching each other. Their radii are 3 and 1, and the smaller log is fastened to the ground. The bigger log rolls over the smaller log without slipping, and stops as soon as it touches the ground again. What is the volume of the set of points swept out by the larger log as it rolls over the smaller one?

[asy] draw(Circle((0,0),1)); draw(Circle((-2*sqrt(3),2),3)); [/asy]

$\textbf{(A) } 250\pi \qquad \textbf{(B) } 260\pi \qquad \textbf{(C) } 270\pi \qquad \textbf{(D) } 280\pi \qquad \textbf{(E) } 290\pi$

Solution

Problem 19

What is the largest power of $2$ that divides $3^{2016}-1$?

$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 64 \qquad\textbf{(D)}\ 128 \qquad\textbf{(E)}\ 256$

Solution

Problem 20

Define a permutation $a_1a_2a_3a_4a_5a_6$ of the set $1, 2, 3, 4, 5, 6$ to be $\text{ factor-hating }$ if $\text{ gcd }(a_k, a_{k+1}) = 1$ for all $1 \leq k \leq 5$. Find the number of $\text{ factor-hating }$ permutations.

$\textbf{(A) }36 \qquad \textbf{(B) }48 \qquad \textbf{(C) }56 \qquad \textbf{(D) }64 \qquad \textbf{(E) }72 \qquad$

Solution

Problem 21

There are $N$ distinct $4\times4$ arrays of integers that satisfy: 1. Each integer in the array is a $1, 2, 3$ or $4$. 2. Every row and column contains all the integers $1, 2, 3$ and $4$. 3. No row or column contains two of the same number. Find $N$.

$\textbf{(A)}\ 432 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 864 \qquad\textbf{(D)}\ 1,152 \qquad\textbf{(E)}\ 1,296$

Solution

Problem 22

Let $S = \{r_1, r_2, r_3, ..., r_{\mu}\}$ be the set of all possible remainders when $15^{n} - 7^{n}$ is divided by $256$, where $n$ is a positive integer and $\mu$ is the number of elements in $S$. The sum $r_1 + r_2 + r_3 + ... + r_{\mu}$ can be expressed as \[p^qr,\]where $p, q, r$ are positive integers and $p$ and $r$ are as small as possible. Find $p+q+r$.

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 41\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 43\qquad\textbf{(E)}\ 44$

Solution

Problem 23

Four real numbers $x_1, x_2, x_3, x_4$ are randomly and independently selected from the range $[0, 9]$. Let the sets $S_1$, $S_2$, $S_3$, $S_4$ contain all of the real numbers in the range $[x_1, x_1+1], [x_2, x_2+1], [x_3, x_3+1],$ and $[x_4, x_4+1]$, respectively. The probability that the four aforementioned sets are disjoint can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

$\textbf{(A)}\ 95\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 97\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 99$

Solution

Problem 24

Four elementary schoolers, four middle schoolers, and four high schoolers sit around a round table with $12$ seats. There is a rule that no two people of the same school may sit adjacent to each other. Let $N$ be the number of distinct seating arrangements following the rule. Find $\frac{N}{(4!)^3}$.

$\textbf{(A)}\ 804\qquad\textbf{(B)}\ 876\qquad\textbf{(C)}\ 948\qquad\textbf{(D)}\ 984 \qquad\textbf{(E)}\ 1,020$

Solution


Solution 1

Let's label the people from three schools with A,B,C. We can consider the case in which A is in seat number 1. Now, we consider to space between two successive A's. There are totally 8 spaces, which can be broken into sum of four positive integers:

$8 = 1 + 1 + 1 + 5 = 1 + 1 + 2 + 4 = 1 + 2 +2 + 3 = 1 + 1 + 3 + 3 = 2 + 2 + 2 + 2$.

For case of $1+1+1+5$, there are four possible orders. For each order, we can arrange 5 people in the space with 5 seats as $BCBCB$ or $CBCBC$, and then arrange two B's or two C's in two of three remaining spaces, making a total of $2\times 3\times = 24$.

For case of $1+1+2+4$, there are twelve possible orders, $1124$, $1241$, $2411$, $4112$, $1142$, $1421$, $4211$, $2114$, $1214$, $2141$, $1412$, $4121$. For each order, we can arrange 4 people in the space with 4 seats as $BCBC$ or $CBCB$, and then arrange the space with two seats as $BC$ or $CB$, and then choose 1 space for $B$ and one space for $C$, making a total of $2\times 2\times 2\times 12 = 96$.

For case of $1+2+2+3$, there are twelve possible orders, similar to the case above. For each order, we can arrange 3 people in the space with 3 seats as $BCB$ or $CBC$, and then arrange the space with two seats as $BC$ or $CB$, and then arrange the last people in the space with 1 seat, making a total of $2\times 2\times 2\times 12= 96$.

For case of $1+1+3+3$, there are six possible orders, $1133$, $1331$, $3311$, $3113$, $1313$, $3131$. For each order, we can arrange 3 people in the two spaces with 3 seats as $BCB/BCB$ or $BCB/CBC$ or $CBC/BCB$ or $CBC/CBC$, and then arrange the space with one seats for the people left, so there six arrangements in total: $BCB/BCB/C/C$, $BCB/CBC/B/C$, $BCB/CBC/C/B$, $CBC/BCB/B/C$, $CBC/BCB/C/B$, $CBC/CBC/B/B$, making a total of $6\times 6 = 36$.

For case of $2+2+2+2$, there is only one order, and each space can be arranged as $BC$ or $CB$, making a total of 16.

Considering the three possibilities for seat 1, the final result is \[3\times(24+96+96+36+16) = \boxed{804}.\]

Problem 25

Let $S_{n, k} = \sum_{a=0}^{n} \dbinom{a}{k}\dbinom{n-a}{k}$. Find the remainder when $\sum_{n=0}^{200} \sum_{k=0}^{200} S_{n, k}$ is divided by $1000$.


$\textbf{(A)}\ 374 \qquad\textbf{(B)}\ 375 \qquad\textbf{(C)}\ 503 \qquad\textbf{(D)}\ 750 \qquad\textbf{(E)}\ 751$

Solution

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