math-eval/TAL-SCQ5K huggingface dataset:
Problem: Half of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of 4 cups. What percent of the total capacity of the pitcher did each cup receive?~ (adapted from 2020 AMC 8, Question \#5) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12.5$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$25$$ 
Answer: C
Analysis: The pitcher is half full, i.e. $50 \textbackslash\%$ full. Therefore each cup receives $\frac{50}{4}=(\mathbf{C}) 12.5$ percent of the total capacity. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: \textbf{The distribution of the number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and standard deviation 7. Volunteers in the top 20 percent of hours worked will receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90?} 
Options:
aoVal: A
content: $$0.077$$ 
aoVal: B
content: $$0.123$$ 
aoVal: C
content: $$0.134$$ 
aoVal: D
content: $$0.618$$ 
aoVal: E
content: $$0.923$$ 
Answer: B
Analysis: \textbf{A: receive a merit B: hour \textless{} 90}  \textbf{P(B) = P(hour \textless{} 90) = P(Z\textless(90-80)/7) = 0.9236}  \textbf{P(A) = 0.2}  \textbf{P(hour \textgreater{} x) = 0.2}  \textbf{P(hour ≤ x) = 0.8}  \textbf{P(Z ≤ (x-80)/7) =0.8}  \textbf{P(A∩B) = P(x\textless hour\textless90) = P($$\frac{x-80}{7} \textless{} Z \textless{} \frac{90-80}{7}$$) = P($$Z \textless{} \frac{90-80}{7}$$) - P($$Z \textless{} \frac{x-80}{7}$$) = 0.9236-0.8 = 0.1236}  \textbf{P(A\textbar B) = $$\frac{P(A∩B)}{P(B)}$$ = $$\frac{0.1236}{0.9236}$$=0.134} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $x=?$ 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: A 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: In a fruit market, $$1\textasciitilde\text{kg}$$ of apples is priced at $$\textbackslash$a$$. $$1\textasciitilde\text{kg}$$ of strawberries costs $$\textbackslash$b$$ more than $$2$$ times $$a$$. How much will $$3\textasciitilde\text{kg}$$ of strawberries cost? 
Options:
aoVal: A
content: $$3a+6b$$ 
aoVal: B
content: $$2a+b$$ 
aoVal: C
content: $$6a+3b$$ 
aoVal: D
content: $$6a+b$$ 
Answer: C
Analysis: $$1\textasciitilde\text{kg}$$ of strawberries costs $$2a+b$$,  so $$3\textasciitilde\text{kg}$$ of strawberries costs $$3\times (2a+b)=6a+3b$$.  So, the answer is C. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: What is the median of the following list of $4042$ number?  $1$, $2$, $3$, $\cdots $, $2021$, $1^{2}$, $2^{2}$, $3^{2}$, $\cdots $, $2021^{2}$ 
Options:
aoVal: A
content: $$1974.5$$ 
aoVal: B
content: $$1975.5$$ 
aoVal: C
content: $$1976.5$$ 
aoVal: D
content: $$1977.5$$ 
aoVal: E
content: $$1978.5$$ 
Answer: D
Analysis: We want to know the $2021^{th}$ term and the $2022^{th}$ term to get the median. We know that $44^{2}=1936\textless2021$, and $45^{2}=2025\textgreater2021$.  So, the number $1^{2}$, $2^{2}$, $3^{2}$, $\cdots $, $44^{2}$ are between $1$ to $1936$. $1936+44=1980$, which mean that $1936$ is the $1980^{th}$ number.  Thus, the $2021^{th}$ term will be $1936+41=1977$, and similarly the $2021^{th}$ term will be $1978$.  So, the answer is $1977.5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: If the degree measures of the angles of a triangle are in the ratio $3: 3: 4$, what is the degree measure of the largest angle of the triangle? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$72$$ 
Answer: D
Analysis: The sum of the ratios is 10 . Since the sum of the angles of a triangle is $180^{\circ}$, the ratio can be scaled up to $54: 54: 72(3 \cdot 18: 3 \cdot 18: 4 \cdot 18)$. The numbers in the ratio $54: 54: 72$ represent the angles of the triangle. The question asks for the largest, so the answer is 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: The GPAs (grade point averages) of students who take the AP Statistics exam are approximately normally distributed with a mean of 3.4 and a standard deviation of 0.3. What is the probability that a student selected at random from this group has a GPA lower than 3.0? 
Options:
aoVal: A
content: $$0.0918$$ 
aoVal: B
content: $$0.4082$$ 
aoVal: C
content: $$0.9082$$ 
aoVal: D
content: $$-0.0918$$ 
aoVal: E
content: $$0$$ 
Answer: A
Analysis: P(X\textless3.0) = $P(z\textless\frac{3-3.4}{0.3}=-1.33)$ = 0.0918  or  normalcdf(-100, 3, 3.4, 0.3) = 0.0912 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Let $a$ and $b$ be two consecutive odd integers. If $a$ is three times $b$, what is their sum $a+b$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: A 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which is the smallest fraction in $$\frac{2020}{2021}$$, $$\frac{2021}{2022}$$, $$\frac{2022}{2023}$$ and $$\frac{2023}{2024}$$? 
Options:
aoVal: A
content: $$\frac{2020}{2021}$$ 
aoVal: B
content: $$\frac{2021}{2022}$$ 
aoVal: C
content: $$\frac{2022}{2023}$$ 
aoVal: D
content: $$\frac{2023}{2024}$$ 
Answer: A
Analysis: Sugar water theory. 1 gram of sugar added each time, and the sugar water gets sweeter. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating
Problem: What is the smallest whole number larger than the perimeter of any triangle with a side of length $12$ and a side of length $13$? (adapted from 2015 AMC8, Question 8) 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$51$$ 
aoVal: D
content: $$49$$ 
aoVal: E
content: $$33$$ 
Answer: B
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s\textless12+13$. Adding $12+13$ to both sides of the inequality, we get $s+12+13\textless50$, and because $s+12+13$ is the perimeter of our triangle, (B) 50 is our answer. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: Which of the following groups has equivalent ratios? 
Options:
aoVal: A
content: $1$ to $5$, $\frac{0.5}{1}$ 
aoVal: B
content: $$\frac{1}{5}$$, $1:5$ 
aoVal: C
content: $\frac{1}{5}$, $5$ to $10$ 
aoVal: D
content: $1:5$, $\frac{5}{10}$ 
Answer: B
Analysis: $$A$$, $1$ to $5$ $$=\frac{1}{5}$$, $$\frac{0.5}{1}=\frac{5}{10}=\frac{1}{2}$$, so wrong.  $$B$$, $$\frac{1}{5}=1:5$$, so true.  $$C$$, $5$ to $10$ $$=\frac{5}{10}=\frac{1}{2}$$, so wrong.  $$D$$, $$1:5=\frac{1}{5}$$, $$\frac{5}{10}=\frac{1}{2}$$, so wrong. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Source: 其它
Problem: Which of the following numbers\textquotesingle{} value does not change after removing all "$$0$$" s ． 
Options:
aoVal: A
content: $$120.221$$ 
aoVal: B
content: $$4.401$$ 
aoVal: C
content: $$2424.390$$ 
Answer: C
Analysis: $$2424.390=2424.39$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Source: 其它
Problem: Match the linear equation, $6x-3y=2$, with its corresponding $y$-intercept. 
Options:
aoVal: A
content: $\frac{2}{3}$ 
aoVal: B
content: $-\frac{2}{3}$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $6x-3y=2$,  $y=2x-\frac{2}{3}$,  Its $y$-intercept is $-\frac{2}{3}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable
Source: 其它
Problem: Given that $$a\Psi b=a\times b+2$$, for example, $$3\Psi 1 = 3\times1 +2$$, what is $$3\Psi4$$? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
Answer: D
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition
Source: 其它
Problem: Real numbers $x$ and $y$ satisfy $x+y=4$ and $x \cdot y=-2$. What is the value of $$ x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y? $$ 
Options:
aoVal: A
content: $$360$$ 
aoVal: B
content: $$400$$ 
aoVal: C
content: $$420$$ 
aoVal: D
content: $$440$$ 
aoVal: E
content: $$480$$ 
Answer: D
Analysis: $$ x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y=x+\frac{x^{3}}{y^{2}}+y+\frac{y^{3}}{x^{2}}=\frac{x^{3}}{x^{2}}+\frac{y^{3}}{x^{2}}+\frac{y^{3}}{y^{2}}+\frac{x^{3}}{y^{2}} $$ 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 $\frac{(20)(4)(22)}{4}=20 \cdot 22=$ 440. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
Source: 其它
Problem: The ratio of $A$ to $B$ is $3:4$. If we multiply $A$ by $3$, $B$ should~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: multiply by $4$ 
aoVal: B
content: divide by $3$ 
aoVal: C
content: add by $8$ 
Answer: C
Analysis: $\frac{3}{4}=\frac{9}{12}$  $12-4=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Fill in the blanks with ``$$+$$'' or ``$$-$$'' to make the equation true．  $$6$$~~~~ $$6$$~~~~ $$6$$~~~~ $$6$$~~~~ $$6=6$$ 
Options:
aoVal: A
content: $$+++-$$ 
aoVal: B
content: $$++++$$ 
aoVal: C
content: $$++-\/-$$ 
Answer: C
Analysis: $$6+6+6-6-6=6$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Problem: One apple, one banana, and two peaches together weigh $12$ lbs. One apple and one peach together weigh $5$ lbs. One banana and $2$ peaches together weigh $5$ lbs more than one apple and one peach weigh together. Each peach weighs the same. How many pounds does one banana weigh? 
Options:
aoVal: A
content: $3$ lbs 
aoVal: B
content: $4$ lbs 
aoVal: C
content: $5$ lbs 
aoVal: D
content: $6$ lbs 
Answer: B
Analysis: We can write their relationships as the equations below:  $A+B+P+P=12$  $A+P=5$  $B+P+P=A+P+5$  So, $B+P+P=5+5=10$, $A=12-10=2$, $P=5-2=3$, $B=12-2-3-3=4$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution
Source: 其它
Problem: Calculate: $$\sqrt{\frac12 }+\sqrt{12.5}-\frac12\sqrt{200}+\sqrt{60\frac12}$$=． 
Options:
aoVal: A
content: $$2\sqrt{2}$$ 
aoVal: B
content: $$\frac{5}{2} \sqrt{2}$$ 
aoVal: C
content: $$3\sqrt{2}$$ 
aoVal: D
content: $$\frac{7}{2} \sqrt{2}$$ 
Answer: D
Analysis: n/a. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: Let $a, b$, and $c$ be positive integers with $a \geq b \geq c$ such that $a^{2}-b^{2}-c^{2}+a b=2011$ and $a^{2}+3 b^{2}+3 c^{2}-3 a b-2 a c-2 b c=-1997$. What is $a$? 
Options:
aoVal: A
content: $$249$$ 
aoVal: B
content: $$250$$ 
aoVal: C
content: $$251$$ 
aoVal: D
content: $$252$$ 
aoVal: E
content: $$253$$ 
Answer: E
Analysis: Add the two equations. $$ 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c=14 . $$ Now, this can be rearranged and factored.  $$ \begin{aligned} \&\left(a^{2}-2 a b+b^{2}\right)+\left(a^{2}-2 a c+c^{2}\right)+\left(b^{2}-2 b c+c^{2}\right)=14 \textbackslash\textbackslash{} \&(a-b)^{2}+(a-c)^{2}+(b-c)^{2}=14 \end{aligned} $$.  $a, b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14=9+4+1$ $(a-c)^{2}=9 \Rightarrow a-c=3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b=1$ or 2 , or whether $b-c=1$ or 2 . We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c)=(a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$. $a^{2}-(a-1)^{2}-(a-3)^{2}+a(a-1)=2011$, after some algebra, simplifies to $7 a=2021$. 2021 is not divisible by 7 , so $a$ is not an integer. The other case gives $a^{2}-(a-2)^{2}-(a-3)^{2}+a(a-2)=2011$, which simplifies to $8 a=2024$. Thus, $a=253$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Addition and Subtraction of Equations
Source: 其它
Problem: There are four soccer teams that are competing in a round-robin format. In the event of a draw, $$1$$ point would be awarded to both teams. $$3$$ points would be given to the team that wins and $$0$$ points would be given to the team that loses. The scores of Team $$A$$, $$B$$, $$C$$ and $$D$$ are $$7$$, $$4$$, $$4$$ and $$1$$ point respectively. How many matches ended in a draw? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
Answer: A
Analysis: We can first assume that all games ended in a $win-lose$ scenario.  Number of games played $=3+2+1=6$  Maximum total score $=6\times3=18$  Everytime a $win-lose$ scenario changes to a $draw-draw$ scenario, the total score decreases by $3-2=1$  Difference in score $=18-16=2$  Number of matches that ended in a draw $=2\div1=2$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns
Problem: The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$? 
Options:
aoVal: A
content: $$255$$ 
aoVal: B
content: $$256$$ 
aoVal: C
content: $$257$$ 
aoVal: D
content: $$258$$ 
aoVal: E
content: $$259$$ 
Answer: A
Analysis: The sum of the first $m$ odd integers is given by $m^{2}$. The sum of the first $n$ even integers is given by $n(n+1)$. Thus, $m^{2}=n^{2}+n+212$. Since we want to solve for $n$, rearrange as a quadratic equation: $n^{2}+n+\left(212-m^{2}\right)=0$. Use the quadratic formula: $n=\frac{-1+\sqrt{1-4\left(212-m^{2}\right)}}{2}$. Since $n$ is clearly an integer, $1-4\left(212-m^{2}\right)=4 m^{2}-847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.  Let $x=\sqrt{4 m^{2}-847}$; note that this means $n=\frac{-1+x}{2}$. It can be rewritten as $x^{2}=4 m^{2}-847$, so $4 m^{2}-x^{2}=847$. Factoring the left side by using the difference of squares, we get $(2 m+x)(2 m-x)=847=7 \cdot 11^{2}$. Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2 m+x)-(2 m-x)=2 m+x-2 m+x=2 x$. We have three pairs of factors, $847 \cdot 1,121 \cdot 7$, and $77 \cdot 11$. The differences between these factors are 846,114 , and 66 - those are all possible values for $2 x$. Thus the possibilities for $x$ are $423$, $57$, and $33$. Now plug in these values into the equation $n=\frac{-1+x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $255$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations
Source: 其它
Problem: Adding five of the six fractions $$\frac{1}{2}$$, $$\frac{2}{3}$$, $$\frac{1}{4}$$, $$\frac{1}{6}$$, $$\frac{1}{9}$$ and $$\frac{1}{18}$$ gives a total of $$1.5$$.  Which of the fractions is not used? 
Options:
aoVal: A
content: $$\frac{1}{3}$$ 
aoVal: B
content: $$\frac{1}{4}$$ 
aoVal: C
content: $$\frac{1}{6}$$ 
aoVal: D
content: $$\frac{1}{9}$$ 
aoVal: E
content: $$\frac{1}{18}$$ 
Answer: B
Analysis: The sum of the five given fractions is $$\frac{1}{2}+\frac{2}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=\frac{18+24+9+6+4+2}{36}$$.  $$\frac{63}{36}= \frac{7}{4}=1 \frac{3}{4}$$.  So the fraction which is not used is $$1 \frac{3}{4}-1 \frac{1}{2}=\frac{1}{4}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: One tour minibus can seat no more than $$8$$ people. What is the smallest number of minibuses needed to take $$75$$ people? (Adapted from 2000 Math Kangaroo Problem, Level 3-4, Question \#6) 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: $75\div8=9R3$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division with Remainders
Problem: Let $A$ and $B$ be positive whole numbers. $A$ is a $2$-digit number and $B$ is a $4$-digit number. If $A+B$ and $A^{2}+B^{2}$ are both multiples of $7$, find the largest possible value of $B-A$. 
Options:
aoVal: A
content: $$9968$$ 
aoVal: B
content: $$9982$$ 
aoVal: C
content: $$9989$$ 
aoVal: D
content: $$9996$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Source: 其它
Problem: How many millimetres are there in $$0.08$$ kilometres? 
Options:
aoVal: A
content: $$\rm 80 mm$$ 
aoVal: B
content: $$\rm 800 mm$$ 
aoVal: C
content: $$\rm 8000 mm$$ 
aoVal: D
content: $$\rm 80000 mm$$ 
Answer: D
Analysis: 0.08km=80m；80m=80000mm 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length
Problem: Two integers are inserted into the list $3,3,8,11,28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers? 
Options:
aoVal: A
content: $$56$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$58$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$61$$ 
Answer: A
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The hundreds digit of the product $$2014\times400$$ is． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: $$2014\times400 = 805600$$; the hundreds digit is $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Function $f$ is defined for the set of all natural number as follows:  $$f(x)=\begin{cases}\dfrac{x}{2},(\text{when }x\text{ is even}) \textbackslash\textbackslash{} x+1,(\text{when }x\text{ is odd}) \textbackslash\textbackslash{} \end{cases}$$.  For example: $$f(3)=3+1=4$$, $$f(2)=2\div 2=1$$.  What is the value of $$f(f(f(f(f(17)))))$$? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: E
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde f\left( f\left( f\left( f\left( f\left( 17 \right) \right) \right) \right) \right)$$  $$=f\left( f\left( f\left( f\left( 18 \right) \right) \right) \right)$$  $$=f\left( f\left( f\left( 9 \right) \right) \right)$$  $$=f\left( f\left( 10 \right) \right)$$  $$=f\left( 5 \right)$$  $$=6$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Number Machine
Problem: $$0.1\times 0.2\times 0.3=$$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$0.0006$$ 
aoVal: B
content: $$0.006$$ 
aoVal: C
content: $$0.06$$ 
aoVal: D
content: $$0.6$$ 
Answer: B
Analysis: $$0.1\times 0.2\times 0.3=(0.1\times 0.2)\times 0.3=0.02\times 0.3=0.006$$. Therefore, the answer is $$\rm B$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals
Problem: Preview Question:  Which of the following is not an expression? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$a$$ 
aoVal: C
content: $$a+b=a+b$$ 
aoVal: D
content: $$b-3$$ 
Answer: C
Analysis: Equations are not expressions. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: Which one of the following fractions is larger than $\dfrac{1}{4}$? 
Options:
aoVal: A
content: $$\frac{1}{5}$$ 
aoVal: B
content: $$\frac{5}{8}$$ 
aoVal: C
content: $$\frac{2}{9}$$ 
aoVal: D
content: $$\frac{4}{17}$$ 
Answer: B
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$16$$ 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{Ten percent of all trucks undergoing a certain inspection will fail the inspection. Assume that trucks are independently undergoing this inspection one at a time. The expected number of trucks inspected before a truck fails inspection is} 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$20$$ 
Answer: D
Analysis: \textbf{This follows a geometric distribution.}  $$\mu = \frac{1}{p} = \frac{1}{0.10} = 10$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the product of $409$ and $6$? 
Options:
aoVal: A
content: $$2454$$ 
aoVal: B
content: $$2404$$ 
aoVal: C
content: $$2444$$ 
aoVal: D
content: $$2464$$ 
Answer: A
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$3+6+9+\cdots +66+69+72=$$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$600$$ 
aoVal: B
content: $$900$$ 
aoVal: C
content: $$1200$$ 
aoVal: D
content: $$1800$$ 
Answer: B
Analysis: $3\times（1+24）\times24\div 2=3\times300=900$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences->Sum of Terms in Arithmetic Sequences
Problem: 3, 6, , 12, 15.~~Which number should be filled in the bracket? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$11$$ 
Answer: C
Analysis: $$6+3=9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: \textbf{Which of the following events are independent?}  \textbf{P(A) = 0.7, P(B) = 0.2, P(C) = 0.1, P(B\textbar A) = 0.5, P(B\textbar C) = 0.2, P(A∩C) = 0.0} 
Options:
aoVal: A
content: \textbf{A and B only} 
aoVal: B
content: \textbf{A and C only} 
aoVal: C
content: \textbf{A, B, and C} 
aoVal: D
content: \textbf{B and C only} 
aoVal: E
content: \textbf{None are independent} 
Answer: D
Analysis: \textbf{P(B\textbar A)\ne P(B) → A and B not independent}  \textbf{P(B\textbar C) =P(B) → B and C independent}  \textbf{P(A\textbar C)=P(A∩B)/P(C)=0.05/0.1=0.5\ne P(A)→ A and C not independent} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它