Difference between revisions of "2021 MECC Mock AMC 10"

(Problem 5)
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<math>\textbf{(A)} ~31 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~33 \qquad\textbf{(D)} ~34 \qquad\textbf{(E)} ~35 </math>
 
<math>\textbf{(A)} ~31 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~33 \qquad\textbf{(D)} ~34 \qquad\textbf{(E)} ~35 </math>
 
[[2021 April MIMC 10 Problems/Problem 1|Solution]]
 
  
 
==Problem 2==  
 
==Problem 2==  

Revision as of 22:52, 20 April 2021

Problem 1

Compute $|2^{2}+2^{1}+2^{0}-3^{1}-3^{2}-3^{3}|$

$\textbf{(A)} ~31 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~33 \qquad\textbf{(D)} ~34 \qquad\textbf{(E)} ~35$

Problem 2

Define a binary operation $a\%b=a^{2}+4ab+4b^{2}$. Find the number of possible ordered pair of positive integers $(a,b)$ such that $a\%b=25$.

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

Solution

Problem 3

$\sqrt{8+4\sqrt{3}}$ can be expressed as $\sqrt{a}+\sqrt{b}$. Find $a+b$.

$\textbf{(A)} ~6 \qquad\textbf{(B)} ~8 \qquad\textbf{(C)} ~10 \qquad\textbf{(D)} ~12 \qquad\textbf{(E)} ~14$

Solution

Problem 4

Compute the number of ways to arrange 2 distinguishable apples and five indistinguishable books.

$\textbf{(A)} ~21 \qquad\textbf{(B)} ~42 \qquad\textbf{(C)} ~63 \qquad\textbf{(D)} ~84 \qquad\textbf{(E)} ~126$

Solution

Problem 5

Galieo, Neton, Timiel, Fidgety and Jay are participants of a game in soccer. Their coach, Mr.Tom, will allocate them into two INDISTINGUISHABLE groups for practice purpose(People in the teams are interchangable). Given that the coach will not put Galieo and Timiel into the same team because they just had a fight. Find the number of ways the coach can put them into two such groups.

$\textbf{(A)} ~24 \qquad\textbf{(B)} ~36 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~72 \qquad\textbf{(E)} ~144$

Solution

Problem 6

Let $a_n$ be a sequence of positive integers with $a_0=1$ and $a_1=2$ and $a_n=a_{n-1}\cdot a_{n+1}$ for all integers $n$ such that $n\geq 1$. Find $a_{2021}+a_{2023}+a_{2025}$.

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~\frac{7}{2} \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~\frac{9}{2} \qquad\textbf{(E)} ~5$

Solution