Difference between revisions of "2021 April MIMC 10"

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{{AMC10 Problems|year=2021|ab=B}}
 
==Problem 1==
 
==Problem 1==
 
What is the sum of <math>2^{3}-(-3^{4})-3^{4}+1</math>?
 
What is the sum of <math>2^{3}-(-3^{4})-3^{4}+1</math>?

Revision as of 23:12, 20 April 2021

2021 AMC 10B (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 SAT 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 sum of $2^{3}-(-3^{4})-3^{4}+1$?

$\textbf{(A)} ~-155 \qquad\textbf{(B)} ~-153 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~9 \qquad\textbf{(E)} ~171$

Solution

Problem 2

Okestima is reading a $150$ page book. He reads a page every $\frac{2}{3}$ minutes, and he pauses $3$ minutes when he reaches the end of page 90 to take a break. He does not read at all during the break. After, he comes back with food and this slows down his reading speed. He reads one page in $2$ minutes. If he starts to read at $2:30$, when does he finish the book?

$\textbf{(A)} ~4:33 \qquad\textbf{(B)} ~5:30 \qquad\textbf{(C)} ~5:33 \qquad\textbf{(D)} ~6:30 \qquad\textbf{(E)} ~7:33$

Solution

Problem 3

Find the number of real solutions that satisfy the equation $(x^2+2x+2)^{3x+2}=1$.

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4$

Solution

Problem 4

Stiskwey wrote all the possible permutations of the letters $AABBCCCD$ ($AABBCCCD$ is different from $AABBCCDC$). How many such permutations are there?

$\textbf{(A)} ~420 \qquad\textbf{(B)} ~630 \qquad\textbf{(C)} ~840 \qquad\textbf{(D)} ~1680 \qquad\textbf{(E)} ~5040$

Solution

Problem 5

5. Given $x:y=5:3, z:w=3:2, y:z=2:1$, Find $x:w$.

$\textbf{(A)} ~3:1 \qquad\textbf{(B)} ~10:3 \qquad\textbf{(C)} ~5:1 \qquad\textbf{(D)} ~20:3 \qquad\textbf{(E)} ~10:1$

Solution

Problem 6

A worker cuts a piece of wire into two pieces. The two pieces, $A$ and $B$, enclose an equilateral triangle and a square with equal area, respectively. The ratio of the length of $B$ to the length of $A$ can be expressed as $a\sqrt[b]{c}:d$ in the simplest form. Find $a+b+c+d$.

$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~12 \qquad\textbf{(D)} ~14 \qquad\textbf{(E)} ~15$

Solution

Problem 7

Find the least integer $k$ such that $838_k=238_k+1536$ where $a_k$ denotes $a$ in base-$k$.

$\textbf{(A)} ~12 \qquad\textbf{(B)} ~13 \qquad\textbf{(C)} ~14 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

Solution

Problem 8

In the morning, Mr.Gavin always uses his alarm to wake him up. The alarm is special. It always rings in a cycle of ten rings. The first ring lasts $1$ second, and each ring after lasts twice the time than the previous ring. Given that Mr.Gavin has an equal probability of waking up at any time, what is the probability that Mr.Gavin wakes up and end the alarm during the tenth ring?

$\textbf{(A)} ~\frac{511}{1023} \qquad\textbf{(B)} ~\frac{1}{2} \qquad\textbf{(C)} ~\frac{512}{1023} \qquad\textbf{(D)} ~\frac{257}{512} \qquad\textbf{(E)} ~\frac{129}{256}$

Solution

Problem 9

Find the largest number in the choices that divides $11^{11}+13^2+126$.

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

Solution

Problem 10

If $x+\frac{1}{x}=-2$ and $y=\frac{1}{x^{2}}$, find $\frac{1}{x^{4}}+\frac{1}{y^{4}}$.

$\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\textbf{(C)} ~0 \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution