math-eval/TAL-SCQ5K huggingface dataset:
Problem: Evaluate $$\frac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}$$ 
Options:
aoVal: A
content: $\dfrac{1}{3}$ 
aoVal: B
content: $\dfrac{2}{5}$ 
aoVal: C
content: $\dfrac{2}{9}$ 
aoVal: D
content: $\dfrac{5}{12}$ 
Answer: D
Analysis: $$\frac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}=\frac{1}{2+ \dfrac{1}{ \dfrac{5}{2}}}= \frac{1}{2+ \dfrac{2}{5}}= \frac{1}{ \dfrac{12}{5}}= \frac{5}{12}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions
Problem: Find the value of $$ 6\times \frac{5}{3}$$． 
Options:
aoVal: A
content: $5$ 
aoVal: B
content: $\frac{33}{3}$ 
aoVal: C
content: $10$ 
aoVal: D
content: $6\frac{12}{3}$ 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: The $$40^{}\text{th}$$ number in the sequence $$2$$, $$6$$, $$10$$, $$14$$, $$\cdots \cdots $$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$171$$ 
aoVal: B
content: $$158$$ 
aoVal: C
content: $$164$$ 
aoVal: D
content: $$160$$ 
Answer: B
Analysis: 4$\times$40-2=158 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: Teacher Nicole bought some sweets and divided it equally among $9$ children. If everyone got $6$ sweets, there would still be some sweets remaining. What is the most number of sweets Teacher Nicole could have bought? What is the least number of sweets Teacher Nicole could have bought? 
Options:
aoVal: A
content: $$63$$，$$54$$ 
aoVal: B
content: $$63$$，$$55$$ 
aoVal: C
content: $$62$$，$$54$$ 
aoVal: D
content: $$62$$，$$55$$ 
Answer: D
Analysis: ~\uline{~~~~~~~~~~}~$\div 9=6$ $\text{R}$~\uline{~~~~~~~~~~}~  Greatest possible remainder is $8$ while smallest possible remainder is $1$.  Greatest possible number of sweets is $$9\times 6+8=62$$, while the least possible number of sweets is $$9\times 6+1=55$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: What is the correct ordering of the three numbers $$\dfrac{5}{19}$$, $$\dfrac{7}{21}$$, and $$\dfrac{9}{23}$$, in increasing order? ($$2012$$ AMC $$8$$ Problem, Question \#$$4$$) 
Options:
aoVal: A
content: $$\dfrac{9}{23}\textless{} \dfrac{7}{21}\textless\dfrac{5}{19}$$ 
aoVal: B
content: $$\dfrac{5}{19}\textless{} \dfrac{7}{21}\textless{} \dfrac{9}{23}$$ 
aoVal: C
content: $$\dfrac{9}{23}\textless{} \dfrac{5}{19}\textless{} \dfrac{7}{21}$$ 
aoVal: D
content: $$\dfrac{5}{19}\textless{} \dfrac{9}{23}\textless{} \dfrac{7}{21}$$ 
aoVal: E
content: $$\dfrac{7}{21}\textless{} \dfrac{5}{19}\textless{} \dfrac{9}{23}$$ 
Answer: B
Analysis: Method $$1$$:  The value of $$\dfrac{7}{21}$$ is $$\dfrac{1}{3}$$. Now we give all the fractions a common denominator.  $$\dfrac{5}{19} \Rightarrow \dfrac{345}{1311}$$,  $$\dfrac{1}{3} \Rightarrow \dfrac{437}{1311}$$,  $$\dfrac{9}{23} \Rightarrow \dfrac{513}{1311}$$.  Ordering the fractions from least to greatest, we find that they are in the order  listed. Therefore, $$\frac{5}{19}\textless{} \frac{7}{21}\textless{} \frac{9}{23}$$.  Method $$2$$:  Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus,  $$1- \dfrac{5}{19}= \dfrac{14}{19}$$,  $$1- \dfrac{7}{21}= \dfrac{14}{21}$$,  $$1- \dfrac{9}{23}= \dfrac{14}{23}$$.  All three fraction have the common numerator $$14$$. Now the order of the fractions is obvious. $$\dfrac{14}{19}\textgreater\dfrac{14}{21}\textgreater\dfrac{14}{23}\Rightarrow\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: Which of the following is a linear equation written in function form? 
Options:
aoVal: A
content: $x=15$ 
aoVal: B
content: $y=5x+7b-120c$ 
aoVal: C
content: $y^{2}=4$ 
aoVal: D
content: $y=-3x^{2}-1$ 
Answer: B
Analysis: An equation is in function form when it is solved for $y$.  A linear equation is also an equation in which the highest power of the variable is always $1$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable
Source: 其它
Problem: The school has $$11200$$ books. The school puts them into 70 equal piles. How many books are in each pile? 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$160$$ 
aoVal: C
content: $$1600$$ 
Answer: B
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division within the Multiplication Tables
Problem: \textbf{The height and age of each child in a random sample of children was recorded. The value of the correlation coefficient between height and age for the children in the sample was 0.8. Based on the least-squares regression line created from the data to predict the height of a child based on age, which of the following is a correct statement?} 
Options:
aoVal: A
content: \textbf{~On average, the height of a child is 80\% of the age of the child.} 
aoVal: B
content: \textbf{The least-squares regression line of height versus age will have a slope of 0.8.} 
aoVal: C
content: \textbf{The proportion of the variation in height that is explained by a regression on age is 0.64.~} 
aoVal: D
content: \textbf{The least-squares regression line will correctly predict height based on age 80\% of the time.} 
aoVal: E
content: \textbf{The least-squares regression line will correctly predict height based on age 64\% of the time.} 
Answer: C
Analysis: $$R^{2} = r^{2} = 0.8^{2} = 0.64$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Karen works part-time at a local convenience store and earns 10 dollars per hour. She wants to spend next Saturday afternoon attending a music concert. The full price of a concert ticket is 75 dollars, but Karen was able to get a discounted price of 50 dollars from a friend who purchased the ticket but has become unable to attend. If Karen took 4 hours off from her job to attend the concert, what was her opportunity cost of attending the concert? 
Options:
aoVal: A
content: $40 
aoVal: B
content: $50 
aoVal: C
content: $75 
aoVal: D
content: $90 
aoVal: E
content: $115 
Answer: D
Analysis: Economic costs are the sum of explicit and implicit costs. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The measures of angles $A, B$ and $C$ are $32^{\circ}, 68^{\circ},90^{\circ}$. The triangle formed by the three angles will be~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: right triangle 
aoVal: B
content: acute triangle 
aoVal: C
content: obtuse triangle 
aoVal: D
content: no triangle 
Answer: D
Analysis: $32^{\circ} + 68^{\circ} + 90^{\circ}=190^{\circ}$  The sum of a triangle\textquotesingle s interior angles is $180^{\circ}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns of Number Tables->Triangle Number Table
Source: 其它
Problem: The zeroes of the function $f(x)=x^{2}-a x+a$ are integers. What is the sum of the possible values of $a$? (Adapted From 2015 AMC 10A Problems, Question \#23) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: By Vieta\textquotesingle s Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.  Because the zeros are integral, the discriminant of the function, $a^{2}-4a$, is a perfect square, say $k^{2}$. Then adding $4$ to both sides and completing the square yields $$ (a-2)^{2}=k^{2}+4$$.  Therefore, $(a-2)^{2}-k^{2}=4$ and $$((a-2)-k)((a-2)+k)=4$$.  Let $(a-2)-k=u$ and $(a-2)+k=v$; then, $a-2=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+2$.  Listing all possible $(u, v)$ pairs such that $a$ is integral: $(2,2), (-2,-2)$ (not counting transpositions because this does not affect $u+v$). Then, $a=4,0$.  These $a$ sum to $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Consider these two operations:  $a ~♦ ~b = a^{2}~−~b^{2}$  $a✭b = (a~−~b)^{2}$  What is the value of $(5♦3)✭6$? 
Options:
aoVal: A
content: $$-20$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$100$$ 
aoVal: E
content: $$220$$ 
Answer: D
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $\frac{1}{4}$ 
aoVal: C
content: $\frac{1}{2}$ 
aoVal: D
content: $$1$$ 
aoVal: E
content: $$2$$ 
Answer: D
Analysis: We expand the original expression, then factor the result by grouping: $$ \begin{aligned} (x y-1)^{2}+(x+y)^{2} \&=\left(x^{2} y^{2}-2 x y+1\right)+\left(x^{2}+2 x y+y^{2}\right) \textbackslash\textbackslash{} \&=x^{2} y^{2}+x^{2}+y^{2}+1 \textbackslash\textbackslash{} \&=x^{2}\left(y^{2}+1\right)+\left(y^{2}+1\right) \textbackslash\textbackslash{} \&=\left(x^{2}+1\right)\left(y^{2}+1\right) . \end{aligned} $$ Clearly, both factors are positive. By the Trivial Inequality, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \geq(0+1)(0+1)= 1 . $$ Note that the least possible value of $(x y-1)^{2}+(x+y)^{2}$ occurs at $x=y=0$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations
Source: 其它
Problem: How many even numbers are there?  1, 3, 4, 6, 7, 9, 5, 8. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
Answer: A
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: Find the value of $$26+(12-9\div3)\times4-2$$. 
Options:
aoVal: A
content: $$28$$ 
aoVal: B
content: $$54$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$138$$ 
Answer: C
Analysis: $$26+(12-9\div3)\times4-2$$  $$=26+9\times4-2$$  $$=26+36-2$$  $$=62-2$$  $$=60$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: Avril has a card. The number on the card is a neighbouring number of 10, but is not a neighbouring number of 12. What is the number on Avril\textquotesingle s card?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: If $$☆-10=△-5$$, then $$☆$$~\uline{~~~~~~~~~~}~$$△$$. 
Options:
aoVal: A
content: $$\textgreater$$ 
aoVal: B
content: $$\textless$$ 
aoVal: C
content: $$=$$ 
Answer: A
Analysis: If $$☆=10$$, then $$△=5$$，$$☆\textgreater△$$．  So, the answer is $$\text{A}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Problem: \textbf{In a population of university students, 20\% of the students have experienced feelings of math anxiety. If we select 8 students, what is the probability that exactly three have experienced math anxiety?} 
Options:
aoVal: A
content: $${8 \choose 3} (0.20)^{3} (0.80)^{8}$$ 
aoVal: B
content: $${8 \choose 3} (0.20)^{3} (0.80)^{5}$$ 
aoVal: C
content: $${5 \choose 3} (0.20)^{3} (0.80)^{5}$$ 
aoVal: D
content: $$(0.20)^{3} (0.80)^{5}$$ 
aoVal: E
content: $$(0.20)^{5} (0.80)^{3}$$ 
Answer: B
Analysis: \textbf{Binomial distribution with parameters n = 8 and p = 0.20.}  \textbf{$$P(X = 3) ={8 \choose 3}(0.20)^{3}(0.80)^{5}$$} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is ．(1996 AJHSME, Question 8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$17$$ 
Answer: B
Analysis: If $A B=10$ and $B C=4$, then $(10-4) \leq A C \leq(10+4)$ by the triangle inequality. In the triangle inequality, the equality is only reached when the "triangle" $A B C$ is really a degenerate triangle, and $A B C$ are collinear. Simplifying, this means the smallest value $A C$ can be is 6 . Applying the triangle inequality on $A C D$ with $A C=6$ and $C D=3$, we know that $6-3 \leq A D \leq 6+3$ when $A C$ is minimized. If $A C$ were larger, then $A D$ could be larger, but we want the smallest $A D$ possible, and not the largest. Thus, $A D$ must be at least 3 , but cannot be smaller than 3 . Therefore, $B$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: There are $20$ balls of the same size in a box.  Lucas says: "Half of them are red."  Peter says: "The number of red balls is $5$ times that of green balls."  Claire says: "There are three colors of balls in the box: red, light blue, and green."  How many dark blue balls are there in the box? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$10$$ 
Answer: A
Analysis: There is no dark blue ball in the box. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: $$9.25\times 0.8+9\frac{1}{4}\times 0.2=$$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$9.25$$ 
aoVal: B
content: $$92.5$$ 
aoVal: C
content: $$925$$ 
Answer: A
Analysis: $$9.25\times 0.8+9\frac{1}{4}\times 0.2$$  $$=9.25\times 0.8+9.25\times 0.2$$  $$=9.25\times (0.8+0.2)$$  $$=9.25\times 1$$  $$=9.25$$  So, $$\text{A}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{x+y}{x}=2$, what is the value of $\frac{x}{y}$? 
Options:
aoVal: A
content: $$\frac{1}{3}$$ 
aoVal: B
content: $$-1$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: Rearranging, we find $x+y=2x, x=y, \frac xy = 1$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Cathy wants to cut a wooden stick. In how many places does she need to break a wooden stick in order to get $7$ pieces? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: $7 - 1 = 6$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $a◆b=a\times 2+b$, then $2◆3=$． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$10$$ 
Answer: A
Analysis: $2◆3=2\times2+3=7$  So the answer is $\rm A$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have? ( 2017 AMC8, Questions \#17) 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$63$$ 
aoVal: E
content: $$81$$ 
Answer: C
Analysis: We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$ \begin{gathered} 9 c-18=g \textbackslash\textbackslash{} 6 c+3=g \end{gathered} $$ We do this because for $9$ chests there are $2$ empty and if $9$ were in each $9$ multiplied by $2$ is $18$ left.  Therefore, $6 c+3=9 c-18$. This implies that $c=7$. We therefore have $g=45$. So, our answer is (C) 45 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Source: 其它
Problem: \textbf{A stationery store owner calculated the mean and median price of all types of pen: 5 and 3, respectively. He plans to have a 20\% off sale. What are the new mean and median of pens in that store?} 
Options:
aoVal: A
content: \textbf{mean: 4; median: 2.4} 
aoVal: B
content: \textbf{mean: 5; median: 2.4} 
aoVal: C
content: \textbf{~mean: 4; median: 3} 
aoVal: D
content: \textbf{~mean: 5; median: 3} 
aoVal: E
content: \textbf{mean: 2.4; median: 4} 
Answer: A
Analysis: \textbf{If you times some number to the data set,~ the mean or median needs to be times with the same number. So for here, everything times with 0.8.} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which of the following is the same as~$44\times25$? (~\uline{~~~~~~~~~~}~) 
Options:
aoVal: A
content: $$4\times 25+4\times 25$$ 
aoVal: B
content: $$44\times 2+44\times 5$$ 
aoVal: C
content: $$4\times 2+4\times 5$$ 
aoVal: D
content: $$40\times 25+4\times 25$$ 
Answer: D
Analysis: $44\times25=\left( 40+4\right)\times25=40\times25+4\times25$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Distributive Law of Whole Numbers
Problem: $$50\textbackslash\%$$ of $$30\textbackslash\%=15\textbackslash\%$$of ． 
Options:
aoVal: A
content: $$10\textbackslash\%$$ 
aoVal: B
content: $$25\textbackslash\%$$ 
aoVal: C
content: $$100\textbackslash\%$$ 
aoVal: D
content: $$150\textbackslash\%$$ 
Answer: C
Analysis: $$50\textbackslash\%\times 30\textbackslash\%=$$ half of $$30\textbackslash\%=15\textbackslash\%=15\textbackslash\%$$ of $$100\textbackslash\%$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: $$2001 + (2000 - 1999 + 1998 - 1997 + 1996 - \cdots + 2 - 1) =$$． 
Options:
aoVal: A
content: $$2001$$ 
aoVal: B
content: $$3001$$ 
aoVal: C
content: $$4001$$ 
aoVal: D
content: $$4002$$ 
Answer: B
Analysis: $$2001+(2000-1999)+(1998-1997)+\cdots +(2-1)=2001+(1000$$ones$$)$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: How many polynomials are there in the following expressions:  $$-4{{x}^{2}}+2$$，$$-\frac{1}{3}mn$$，$$ \pi $$，$$\frac{{{(2x-y)}^{2}}}{3}$$，$$32\frac{1}{4}$$~ ~． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: $$-4{{x}^{2}}+2$$，$$\frac{{{(2x-y)}^{2}}}{3}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: What value does the $4$ represent in the number $55.431$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$0.4$$ 
aoVal: C
content: $$0.04$$ 
aoVal: D
content: $$0.004$$ 
Answer: B
Analysis: The number $$4$$ is located on the tenth place, thus representing $$0.04$$.  Check Lesson 4 Concept 1 on textbook 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Source: 其它
Problem: Assuming $a \neq 3, b \neq 4$, and $c \neq 5$, what is the value in simplest form of the following expression? (Adapted From 2020 AMC 10A Problems, Question \#3)  $$ \frac{2a-6}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{2c-10}{4-b} $$ 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$-4$$ 
aoVal: C
content: $\frac{a b c}{15}$ 
aoVal: D
content: $\frac{4}{a b c}-\frac{1}{15}$ 
aoVal: E
content: $\frac{1}{15}-\frac{1}{a b c}$ 
Answer: B
Analysis: If $x \neq y$, then $\frac{x-y}{y-x}=-1$. We use this fact to simplify the original expression: $$ \frac{2a-6}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{2c-10}{4-b} = -2\times 2= 4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The zeroes of the function $f(x)=x^{2}-a x+2 a$ are integers. What is the sum of the possible values of $a$? (2015 AMC 10A Problems, Question \#23) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$18$$ 
Answer: C
Analysis: By Vieta\textquotesingle s Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.  Because the zeros are integral, the discriminant of the function, $a^{2}-8 a$, is a perfect square, say $k^{2}$. Then adding $16$ to both sides and completing the square yields $$ (a-4)^{2}=k^{2}+16 $$.  Therefore, $(a-4)^{2}-k^{2}=16$ and $$((a-4)-k)((a-4)+k)=16$$.  Let $(a-4)-k=u$ and $(a-4)+k=v$; then, $a-4=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+4$.  Listing all possible $(u, v)$ pairs such that $a$ is integral: $(2,8),(4,4),(-2,-8),(-4,-4)$ (not counting transpositions because this does not affect $u+v$). Then, $a=9,8,-1,0$.  These $a$ sum to $16$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Alice, Tom, Chloe and Susan ate chicken nuggets together. Everyone ate at least 2 pieces. The person who ate the least ate 4 pieces less than the person who ate the most. How many pieces of chicken nuggets did they each eat? (adapted from $$2021$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$8$$） 
Options:
aoVal: A
content: $8, 2, 4, 3$ 
aoVal: B
content: $4, 1, 5,9$ 
aoVal: C
content: $6,5,0,10$ 
aoVal: D
content: $2,7,4,1$ 
aoVal: E
content: $2,6,4,8$ 
Answer: A
Analysis: In B, C, D, the smallest number less than $2$. In the E, the difference between the largest number and the smallest number is $6$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Subtraction of Whole Numbers->Subtraction in Horizontal Form
Problem: Which of the following quotients is $$1$$ more than $$162\div18$$? 
Options:
aoVal: A
content: $$128\div 16$$ 
aoVal: B
content: $$120\div 15$$ 
aoVal: C
content: $$132\div 12$$ 
aoVal: D
content: $$110\div 11$$ 
Answer: D
Analysis: Since $$162 \div18 =9$$, we want a quotient of $$10$$. That\textquotesingle s $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: A box contains five cards, numbered $1,2,3,4$, and $5$. Three cards are selected randomly without replacement from the box. What is the probability that $4$ is the largest value selected? (2017 AMC 8 Problems, Question \#10) 
Options:
aoVal: A
content: $\frac{1}{10}$ 
aoVal: B
content: $\frac{1}{5}$ 
aoVal: C
content: $\frac{3}{10}$ 
aoVal: D
content: $\frac{2}{5}$ 
aoVal: E
content: $\frac{1}{2}$ 
Answer: C
Analysis: There are $\_5C\_3 = 10$ possible groups of cards that can be selected. If $4$ is the largest card selected, then the other two cards must be either $1$,$2$, or $3$, for a total $\_3C\_2 = 3$ groups of cards. Then, the probability is just $(\text{C}) \frac{3}{10}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The distance between A and B is $$350$$ $\text{km}$. Kin and Mary drive away from A and B respectively at $8$ a.m. and go towards each other at same time. Kin drives $$40$$ $\text{km/h}$, and Mary drives $$50$$ $\text{km/h}$. Mary rested for $$2$$ hours on her way and then continues driving. They will meet at~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$10$$ a.m. 
aoVal: B
content: $$11$$ a.m. 
aoVal: C
content: $$12$$ p.m. 
aoVal: D
content: $$1$$ p.m. 
Answer: D
Analysis: $$(350-80)$$$\div$$$(40+50)=3$$hr  $$8+2+3=13$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它