math-eval/TAL-SCQ5K huggingface dataset:
Problem: Real numbers $x$ and $y$ satisfy $x+y=4$ and $x\cdot y=-2$. What is the value of $x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y$? (2020 AMC 10A Problem, Question \#14) 
Options:
aoVal: A
content: $$360$$ 
aoVal: B
content: $$400$$ 
aoVal: C
content: $$420$$ 
aoVal: D
content: $$440$$ 
aoVal: E
content: $$480$$ 
Answer: D
Analysis: $x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y=x+\frac{y^{3}}{x^{2}}+y+ \frac{x^{3}}{y^{2}}=\frac{x^{3}}{x^{2}}+\frac{y^{3}}{x^{2}}+\frac{y^{3}}{y^{2}}+\frac{x^{3}}{y^{2}}$  Continuing to combine  $\frac{x^{3}+y^{3}}{x^{2}}+\frac{x^{3}+y^{3}}{y^{2}}=\frac{\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)}{x^{2} y^{2}}=\frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}}$  From the givens, it can be concluded that $x^{2}y^{2}=4$. Also,  $(x+y)^{2}=x^{2}+2 x y+y^{2}=16$  This means that $x^{2}+y^{2}=20$. Substituting this information into $\frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}}$,  we have $\frac{20\times 4\times22}{4}=440$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Given that $$a\Phi b=2\times a-b$$, for example, $$2\Phi 1 = 2\times2 -1$$, what is $$3\Phi4$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$12$$ 
Answer: A
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition
Source: 其它
Problem: For how many integers $x$ is the number $x^{4}-51 x^{2}+50$ negative? (  2014 AMC 10B Problems, Question \#20) 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: C
Analysis: First, note that $50+1=51$, which motivates us to factor the polynomial as $\left(x^{2}-50\right)\left(x^{2}-1\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-50\textless0\textless x^{2}-1$. Solving this inequality, we find $1\textless x^{2}\textless50$. There are exactly $12$ integers $x$ that satisfy this inequality, $\pm\textbackslash{2,3,4,5,6,7\textbackslash}$. Thus our answer is $(\mathbf{C}) 12$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: How many of the following options are equations?  1. $x = y$  2. $x \textgreater{} 1$  3. $x \geq x-1$  4. $ 1 = 2$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: 1 and 4 are equations. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: How many sections will Linda get if she cuts a piece of wood $4$ times? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: It depends on how long the piece of wood is. 
Answer: B
Analysis: $4 + 1 = 5$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the product of $$628$$ and $$6$$?（) 
Options:
aoVal: A
content: $$3628$$ 
aoVal: B
content: $$3668$$ 
aoVal: C
content: $$3728$$ 
aoVal: D
content: $$3768$$ 
Answer: D
Analysis: Stack the two numbers as shown below ,~ lining up the unit digits  Remember that the 2 in 628 stands for 2 tens(20) ,the 6 in the 628 stands for 6 hundreds(600)  First, multiply the ones~ $6\times8=48$~, regroup the 4 tens to the tens column  Write 8 in the ones place.  Then,~ Multiply and add the tens .~$2\times6+4=16$  Write 6 in the tens place and regroup the 1 hundred.  Last, multiply and add the hundreds.~$6\times6+1=37$  Write 7 in the hundreds place and write 3 in the thousands place. 
Knowledge points: Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Multiplication of Whole Numbers->Multiplication of Multi-Digit Numbers and 1-Digit Numbers->Multiplication of 3-Digit and 1-Digit (with regrouping for more than once)
Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: If $x+2y=3$, what is $2^{x}\cdot 4^{}y$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$16$$ 
Answer: C
Analysis: $2^{}x \cdot 4^{}y=2^{}x\cdot 2^{2y}=2^{x+2y}=2^{3}=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: In the animal school, some sheep are taking lessons. The teacher cow finds out that the sheep have $$24$$ legs altogether. How many sheep are there? (Adapted from 2012 Math Kangaroo Problem, Level 3-4, Question \#14) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$2$$ 
Answer: A
Analysis: $$24 \div 4 = 6$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Source: 其它
Problem: The $1985^{}\text{th}$ digit at the right of the decimal point in the decimal expression of $\dfrac{1}{7}$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: $$\frac{1}{7}=0.\overline{142857}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$\textsuperscript{th}~digit is $7$. The $1986$\textsuperscript{th} digit is $7$, so the $1985$\textsuperscript{th} digit is $5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: $$6.98-4.53+10.02-5.27=$$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$6.9$$ 
aoVal: B
content: $$7.0$$ 
aoVal: C
content: $$7.1$$ 
aoVal: D
content: $$7.2$$ 
aoVal: E
content: $$7.3$$ 
Answer: D
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde6.98-4.53+10.02-5.27$$  $$=(6.98+10.02)-(4.53+5.27)$$  $$=17-9.8$$  $$=7.2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals
Problem: How many different three-digit numbers can we make using $6,7,8,$ and $9$? The digits can be repeated. 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$64$$ 
aoVal: E
content: $$72$$ 
Answer: D
Analysis: $4\times4\times4=64$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Addition and Subtraction of Decimals
Source: 其它
Problem: For how many values of $a$ is it true that the line $y=x+a^{2}-6$ passes through the vertex of the parabola $y=x^{2}-4x+a^{2}$? (Adapted From 2005 AMC 12B Problem, Question \#8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: infinitely many 
Answer: E
Analysis: We see that the vertex of the quadratic function $y=x^{2}-4x+a^{2}$ is $\left(2, a^{2}-4\right)$. If $\left(2, a^{2}-4\right)$ will be on the line $y=x+a^{2}-6$, $a^{2} -4=2+a^{2}-6$. Solve for $a$, any value of $a$ will satisfy this equation. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Line $l\_1$ has equation $2x-y=3$ and goes through $A=(1,-1)$. Line $l\_2$ has equation $y=1$ and meets line $l\_1$ at point $B$. Line $l\_3$ has negative slope, goes through point $A$, and meets $l\_2$ at point $C$. The area of $\triangle A B C$ is $3$. What is the slope of $l\_3$? (Adapted From 2013 AMC 12B Problems, Question \#8) 
Options:
aoVal: A
content: $$-\frac23$$ 
aoVal: B
content: $$-\frac34$$ 
aoVal: C
content: $$-1$$ 
aoVal: D
content: $$-\frac43$$ 
aoVal: E
content: $$-2$$ 
Answer: C
Analysis: Line $l\_1$ has the equation $y=2x-3$ when rearranged. Substituting $1$ for $y$, we find that line $l\_2$ will meet this line at point $(2,1)$, which is point $B$. We call $\overline{B C}$ the base and the altitude from $A$ to the line connecting $B$ and $C, y=1$, the height. The altitude has length $\textbar-1-1\textbar=2$. The area of $\triangle A B C=3$. Since $A=\frac{bh}{2}, b=3$. Points that are on the line $y= 1$ and has a distance of $3$ from $B$ are $(5,1)$ and $(-1,1)$. Since $l\_3$ has negative slope, point $C$ is $(-1,1)$. $l\_3$ passes through $(-1,1)$ and $(1,-1)$, and thus has slope $\frac{1-(-1)}{-1-1}=(\mathbf{C}) -1$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the median of the following distribution: 6, 2, 9, 4, 7, 3? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$5.5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$6.5$$ 
Answer: B
Analysis: The median of the distribution is 5. The problem is easier if vou put the scores in order: 2, 3, 4, 6, 7, 9. Since the distribution has an even number of scores, there is no middle score and you must average the two middle scores, 4 and 6. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: 26-18=． 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$11$$ 
Answer: A
Analysis: $$26-18=8$$ 
Knowledge points: Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Addition and Subtraction of Whole Numbers->Adding and Subtracting within 10000->Subtraction of 3-digit Numbers
Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Alicia and Emily agreed to meet at the cinema at $3.55\rm{pm}$. Emily left her house at $1.47\rm{pm}$ but arrived at the cinema $17$ minutes late. How long was Emily\textquotesingle s journey from her house to the cinema? 
Options:
aoVal: A
content: $$189$$ minutes 
aoVal: B
content: $$172$$ minutes 
aoVal: C
content: $$216$$ minutes 
aoVal: D
content: $$206$$ minutes 
aoVal: E
content: None of the above. 
Answer: E
Analysis: 3: 55pm - 1: 47om + 15 minutes = 145 minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion
Source: 其它
Problem: Calculate~ ~ ~$\dfrac{2014}{2013-\dfrac{2012}{2011-\dfrac{2010}{5-\dfrac{4}{3-\dfrac{2}{1}}}}}$  After calculating, Chuan says that answer is D. Is he right or wrong? If not, choose the right option. 
Options:
aoVal: A
content: $\dfrac{1}{2014}$ 
aoVal: B
content: $\dfrac{1}{2013}$ 
aoVal: C
content: $$2014$$ 
aoVal: D
content: $$2013$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: $$3- \frac{2}{1}=1$$  $$5- \frac{4}{1}=1$$  $$\cdots \cdots$$  $$2013- \frac{2012}{1}=1$$  $$\frac{2014}{1}=2014$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions
Problem: Which number has to be subtracted from $-17$ in order to obtain $-33$? (2017 Math Kangaroo Problem, Level 7-8, Question \#3) 
Options:
aoVal: A
content: $$-50$$ 
aoVal: B
content: $$-16$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$40$$ 
aoVal: E
content: $$50$$ 
Answer: C
Analysis: $(-17)-16=-33$, so the answer is $C$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Source: 其它
Problem: If $$\left\textbar{} 5x+3 \right\textbar=8$$, $x=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$-1$$ or $$\frac{11}{5}$$ 
aoVal: C
content: $$1$$ or $$\frac{1}{5}$$ 
aoVal: D
content: $$1$$ or $$-\frac{11}{5}$$ 
Answer: D
Analysis: $$5x+3=\pm 8$$  $$x=1$$ or $$x=-\frac{11}{5}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities->Solving Inequalities
Source: 其它
Problem: The operator $$\bigtriangleup$$ acts on two numbers to give the following outcomes:  $$3 \bigtriangleup 2 = 12$$  $$4 \bigtriangleup 5 = 40$$  $$5 \bigtriangleup 9 = 90$$  $$6 \bigtriangleup 1 = 12$$  What is $$2 \bigtriangleup 7$$ equal to? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: The pattern of the operation is $$a \bigtriangleup b = a\times b\times2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition
Source: 其它
Problem: Find the integer part of the following fractional expression:  $$ \frac{1}{\frac{1}{50}+\frac{1}{51} +\frac{1}{52} +\frac{1}{53} +\frac{1}{54}} $$ 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$50$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Three football teams participate in a sport tournament. Each team plays the other two teams exactly once. In each game, the winner gets three points and the loser doesn\textquotesingle t get any points. If the game ends in a tie, each team gets $1$ point. At the end of tournament, Which number of points is it impossible for any team to have? (2022 Math Kangaroo Problem, Level 3-4, Question \#18) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: Each team plays $2$ games.  $1$ points: a lost and a tie.  $2$ points: two ties.  $3$ points: a lost and a win.  $4$ points: a win and a tie.  $6$ points: two wins. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Suppose $5\times8=\triangle$, $4\times7=\angle$, $3\times9=\square$. Arrange the three shapes according to their values from smallest to largest. 
Options:
aoVal: A
content: $\angle$ $\triangle$ $\square$ 
aoVal: B
content: $\square$ $\angle$ $\triangle$ 
aoVal: C
content: $\triangle$ $\angle$ $\square$ 
aoVal: D
content: $\triangle$ $\square$ $\angle$ 
aoVal: E
content: $\square$ $\triangle$ $\angle$ 
Answer: B
Analysis: $\square=27$  $\angle=28$  $\triangle=40$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Source: 其它
Problem: The students in Mr. Neatkin\textquotesingle s class took a penmaship test. Two-thirds of the boys and~$\dfrac{3}{4}$~of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class? (2008 AMC 8, 20) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$17$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$36$$ 
Answer: B
Analysis: Let~$b$~be the number of boys and~$g$~be the number of girls.  $\dfrac{2}{3}b=\dfrac{3}{4}g\textbackslash{} \Rightarrow\textbackslash{} b=\dfrac{9}{8}g$  For~$g$~and~$b$~to be integers,~~must cancel out with the denominator, and the smallest possible value is . This yields~~boys. The minimum number of students is~$8+9=\boxed{\left( B\right)17}$  Solution 2  We know that~$\dfrac{2}{3}B=\dfrac{3}{4}G\textbackslash{} or\textbackslash{} \dfrac{6}{9}B=\dfrac{6}{8}G.$~So, the ratio of the number of boys to girls is~$9:8$. The So, the ratio of the number of boys to girls is~$8+9=\boxed{\left( B\right)17}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: In how many different ways can Chloe select two digits out of $$ 0$$, $$1$$, $$ 2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ and put into the two boxes to make the below equality correct?  $$ 20- \square =22-\square $$ 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Problem: I multiply a whole number by itself, then multiply that product by itself. The ones digit of my final product cannot be． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: The ones digit of the product is the same as the ones digit of $$0^{4}$$, $$1^{4}$$, $$2^{4}$$, $$3^{4}$$, $$4^{4}$$,  $$5^{4}$$, $$6^{4}$$, $$7^{4}$$, $$8^{4}$$, or $$9^{4}$$. The ones digit can be $$0$$, $$1$$, $$5$$, or $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: The sum of five consecutive whole numbers is $$280$$. What is the sum of the next five consecutive whole numbers? 
Options:
aoVal: A
content: $$285$$ 
aoVal: B
content: $$305$$ 
aoVal: C
content: $$405$$ 
aoVal: D
content: $$425$$ 
Answer: B
Analysis: Each whole number in the second sequence is $$5$$ more than the corresponding number in the first sequence. We have $$280 +5 \times5 = 305$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: What is the range of the number in tens place? 
Options:
aoVal: A
content: $$0-5$$ 
aoVal: B
content: $$1-5$$ 
aoVal: C
content: $$0-9$$ 
aoVal: D
content: $$1-9$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: For $\triangle ABC$, all its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $25$ and a side length of $18$ is at least ． 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$52$$ 
aoVal: D
content: $$51$$ 
aoVal: E
content: $$50$$ 
Answer: D
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+18\textgreater25$. Therefore, $P\textgreater25+25$. The least integer value of $P$ is $51$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $a$ and $$b$$ are reciprocals， $$\frac{2}{a}\div \frac{b}{12}=$$． 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$\frac{1}{24}$$ 
aoVal: C
content: $$\frac{b}{6a}$$ 
aoVal: D
content: $$\frac{a}{6b}$$ 
Answer: A
Analysis: $$\frac{2}{a}\div \frac{b}{12}= \frac{2}{a}\times \frac{12}{b}= \frac{24}{ab}=24 $$， 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering->Comparing and Ordering Fraction by Comparing Its Reciprocal
Source: 其它
Problem: If $$x=120$$, what is the value of $$\frac{x^{2}}{12^{2}}$$? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$36$$ 
Answer: B
Analysis: $$\frac{x^{2}}{12^{2}}=\left(\frac{x}{12}\right)^{2}=\left(\frac{120}{12}\right)^{2}=\left(10\right)^{2}=100$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: Let $Z$ be a $6$-digit positive integer, such as $247247$, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$? (adapted from 2017 AMC 8 Problem, Question \#$7$) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$1001$$ 
aoVal: D
content: $$111$$ 
aoVal: E
content: $$1111$$ 
Answer: C
Analysis: chair number 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Source: 其它
Problem: Which of the following is neither a positive nor a negative number? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$-23$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$-1$$ 
Answer: C
Analysis: $$\text{A}$$ is a positive number, $$\text{B}$$ and $$\text{D}$$ are two negative numbers, $$\text{C}$$ is zero, which is neither positive nor negative. We choose $$\text{C}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Knowing the Number Lines
Problem: $$(x+5)-(x+4)+(x+3)-(x+2)+(x+1)-(x+0)=$$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$-3$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$3x+3$$ 
aoVal: D
content: $$6x+3$$ 
Answer: B
Analysis: We can observe that when we remove the parentheses, all the $$x$$s are offset. $$5-4+3-2+1-0= 3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: Simplify the following expression: $$4a-3(a-b)$$． 
Options:
aoVal: A
content: $$a-b$$ 
aoVal: B
content: $$a+b$$ 
aoVal: C
content: $$a+3b$$ 
aoVal: D
content: $$a-3b$$ 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Using a Letter to Represent an Unknown Number
Problem: There were $$9$$ pieces of paper. Some of them were cut into three pieces. As a result, there are now $$15$$ pieces of paper now. How many pieces of paper were cut? ($$2005$$ Math Kangaroo Problem, Level $$3-4$$, Question \#$$17$$) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $15-9\times1=6$  $6\div(3-1)=3$ pieces of paper were cut. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: For how many values of $a$ is it true that the line $y=x+a$ passes through the vertex of the parabola $y=x^{2}+a^{2}$? (2005 AMC 12B Problem, Question \#8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: infinitely many 
Answer: C
Analysis: We see that the vertex of the quadratic function $y=x^{2}+a^{2}$ is $\left(0, a^{2}\right)$. The $y$-intercept of the line $y=x+a$ is $(0, a)$. We want to find the values (if any) such that $a=a^{2}$. Solving for $a$, the only values that satisfy this are 0 and 1, so the answer is $(\text{C}) 2$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Values for $A, B, C$, and $D$ are to be selected from $\textbackslash{1,2,3,4,5,6\textbackslash}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=A x^{2}+B$ and $y=C x^{2}+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $ A=4, B=1, C=3, D=2 $.) 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$180$$ 
aoVal: E
content: $$360$$ 
Answer: C
Analysis: Setting $y=A x^{2}+B=C x^{2}+D$, we find that $A x^{2}-C x^{2}=x^{2}(A-C)=D-B$, so $x^{2}=\frac{D-B}{A-C} \geq 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are 2 ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by 2 at the end, however, since the 2 curves aren\textquotesingle t considered distinct. Calculating, we get $$ \frac{1}{2} \cdot\left(\begin{array}{l} 6 \textbackslash\textbackslash{} 2 \end{array}\right)\left(\begin{array}{l} 4 \textbackslash\textbackslash{} 2 \end{array}\right) \cdot 2=(\text { C) } 90 $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: For how many integers $x$ is the number $x^{4}-7x^{2}+12$ negative? (  2014 AMC 10B Problems, Question \#20) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$1$$ 
Answer: C
Analysis: Factor the polynomial as $\left(x^{2}-4\right)\left(x^{2}-3\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-4\textless0\textless x^{2}-3$. Solving this inequality, we find $3\textless x^{2}\textless4$. There is no integer $x$ that satisfies this inequality. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它