math-eval/TAL-SCQ5K huggingface dataset:
Problem: Mary cooked $9$ cakes. She cut two of them into $3$ pieces. How many cakes does she have now? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$15$$ 
Answer: C
Analysis: $9 - 2 + 2 \times 3 = 13$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Camila writes down five positive integers. The unique mode of these integers is 2 greater than their median, and the median is 2 greater than their arithmetic mean. What is the least possible value for the mode? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$13$$ 
Answer: D
Analysis: 1，$$3$$，$$9$$，$$11$$，11 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: If $$a$$@$$b=\frac {a\times b}{a+b}$$ for positive integers $$a$$ and $$b$$ , what is $$5$$@$$10$$? 
Options:
aoVal: A
content: $$\frac 3{10}$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$\frac {10}3$$ 
aoVal: E
content: $$50$$ 
Answer: D
Analysis: Substitute $$a=5$$ and $$b=10$$ into the expression for $$a$$@$$b$$ to get: $$5$$@$$10=\frac {5\times 10}{5+10}=\frac {50}{15}=\frac {10}3$$.  Thus, the answer choice $$\frac {10}3$$ is correct. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly->Ordinary Type
Problem: Is it true that If an odd function has zero in its domain, then it must pass through the origin. (Hint: $0$ is in the domain, we can apply $f(-x) = -f(x)$ at $x=0$) 
Options:
aoVal: A
content: True 
aoVal: B
content: False 
Answer: A
Analysis: By definition of an odd function, $f(-x) = -f(x)$ for all $x$ in the domain. Since $x = 0$ is in the domain, we have $f(-0) = -f(0)$, which means $f(0) = -f(0)$, or $2f(0) = 0$, and thus $f(0) = 0$ (this represents the origin). 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Dividing a certain two-digit number by $$10$$ leaves a remainder of $$9$$, and dividing it by $$9$$ leaves a remainder of $$8$$. The sum of the number\textquotesingle s two digits is． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$17$$ 
Answer: D
Analysis: Dividing a certain two-digit number by $$10$$ leaves a remainder of $$9$$, so it is $$19$$, $$29$$, $$39$$, $$49$$, $$59$$, $$69$$, $$79$$, $$89$$, or $$99$$. The only number listed with remainder $$8$$ when divided by $$9$$ is $$89$$, so the number is $$89$$ and $$8+9=17$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Lucy writes numbers from $1$ to $200$ on the blackboard. Then, she plays a game with her friend Mike. Each time they take turn to delete the first two numbers in the sequence and write their sum by the end of the sequence. For example, Lucy deletes $1$ and $2$, then writes $3$ behind $200$ for the first time. They play this game until there is only one number on the blackboard. What is the number? 
Options:
aoVal: A
content: $$399$$ 
aoVal: B
content: $$402$$ 
aoVal: C
content: $$5050$$ 
aoVal: D
content: $$20100$$ 
aoVal: E
content: $$25050$$ 
Answer: D
Analysis: The result is equal to the sum of all numbers from $1$ to $200$, which is $(1+200)\times200\div2=20100$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: Calculate: $$1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4=$$． 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$98$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$94$$ 
Answer: D
Analysis: $$1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4$$  $$=1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1-3-2-1$$  $$=10\times 10-6$$  $$=100-6$$  $$=94$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences->Pyramid Sequences
Problem: How many types of vitamins are there in the fruits below? There are~\uline{~~~~~~~~~~}~types 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
Answer: C
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: Which of the following expression is a perfect square? 
Options:
aoVal: A
content: $98! \cdot 99!$ 
aoVal: B
content: $98! \cdot 100!$ 
aoVal: C
content: $99! \cdot 100!$ 
aoVal: D
content: $99! \cdot 101!$ 
aoVal: E
content: $100! \cdot 101!$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Lee has $1$ red cube, $1$ yellow cube, $2$ blue cubes and $3$ green cubes. How many ways are there for Lee to arrange his cubes in a row if cubes of the same color is indistinguishable? 
Options:
aoVal: A
content: $$210$$ 
aoVal: B
content: $$420$$ 
aoVal: C
content: $$630$$ 
aoVal: D
content: $$840$$ 
aoVal: E
content: $$1050$$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: How many minutes are there in $$2$$ weeks? 
Options:
aoVal: A
content: $$2\times7\times24$$ 
aoVal: B
content: $$(7+7)\times24\times60$$ 
aoVal: C
content: $$2\times7\times12\times60$$ 
aoVal: D
content: $$2\times24\times60$$ 
aoVal: E
content: $$(7+7)\times12\times60$$ 
Answer: B
Analysis: One week has $$7$$ days. Two weeks has $$7+7$$  One day has $$24$$ hours.  One hour has $$60$$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: One ticket to a mini concert costs $\textbackslash$20$ at full price. Nicole buys $4$ tickets using a coupon that gives her a $25\textbackslash\%$ discount. Bel buys $5$ tickets using a coupon that gives her a $30\textbackslash\%$ discount. How many more dollars does Nicole pay than Bel? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$20$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The students in a class sit in rows. There is the same number of students in each row. There are $2$ rows of students in front of Robert and $1$ row of students behind him. In his row, there are $3$ students on his left and $5$ students on his right. How many students are there in this class? (2022 Math Kangaroo Problem, Level 3-4, Question \#15) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$17$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: There are $3+5+1=9$ students in each row, and there are $2+1+1=4$ rows. Thus, there are $9\times4=36$ students in this class. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
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 Eddie\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: Find the sum of $$\frac{1}{5}$$ and $$\frac{7}{10}$$. 
Options:
aoVal: A
content: $$\frac{8}{10}$$ 
aoVal: B
content: $$\frac{9}{10}$$ 
aoVal: C
content: $$\frac{8}{15}$$ 
aoVal: D
content: $$\frac{9}{20}$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: $$10000\div 200\times$$$$=10000$$. 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$200$$ 
aoVal: C
content: $$1000$$ 
aoVal: D
content: $$2000$$ 
Answer: B
Analysis: $$10000\div 200=50$$; $$50\times \underline{200}=10000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: In a competitive market, a producer is able to sell their good for $\textbackslash$ 10$ per unit, while the cost of producing each unit is $\textbackslash$ 8$. What is the producer surplus in this scenario? 
Options:
aoVal: A
content: $\textbackslash$ 2$ 
aoVal: B
content: $\textbackslash$ 8$ 
aoVal: C
content: $\textbackslash$ 10$ 
aoVal: D
content: $\textbackslash$ 12$ 
Answer: A
Analysis: $\textbackslash$2$, as the producer surplus is calculated as the difference between the price the good is sold for and the cost of production, which is $\textbackslash$10$ - $\textbackslash$8$ = $\textbackslash$2$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The real numbers $c, b, a$ form an arithmetic sequence with $a \geq b \geq c \geq 0$. The quadratic $a x^{2}+b x+c$ has exactly one root. What is this root? 
Options:
aoVal: A
content: $-7-4 \sqrt{3}$ 
aoVal: B
content: $-2-\sqrt{3}$ 
aoVal: C
content: $$-1$$ 
aoVal: D
content: $-2+\sqrt{3}$ 
aoVal: E
content: $-7+4 \sqrt{3}$ 
Answer: D
Analysis: It is given that $a x^{2}+b x+c=0$ has 1 real root, so the discriminant is zero, or $b^{2}=4 a c$. Because $a, b, c$ are in arithmetic progression, $b-a=c-b$, or $b=\frac{a+c}{2}$. We need to find the unique root, or $-\frac{b}{2 a}$ (discriminant is 0 ). From $b^{2}=4 a c$, we can get $-\frac{b}{2 a}=-\frac{2 c}{b}$ Ignoring the negatives(for now), we have $\frac{2 c}{b}=\frac{2 c}{\frac{a+c}{2}}=\frac{4 c}{a+c}=\frac{1}{\frac{1}{\frac{4 c}{a+c}}}=\frac{1}{\frac{a+c}{4 c}}=\frac{1}{\frac{a}{4 c}+\frac{1}{4}}$. Fortunately, finding $\frac{a}{c}$ is not very hard. Plug in $b=\frac{a+c}{2}$ to $b^{2}=4 a c$, we have $a^{2}+2 a c+c^{2}=16 a c$, or $a^{2}-14 a c+c^{2}=0$, and dividing by $c^{2}$ gives $\left(\frac{a}{c}\right)^{2}-14\left(\frac{a}{c}\right)+1=0$, so $\frac{a}{c}=\frac{14 \pm \sqrt{192}}{2}=7 \pm 4 \sqrt{3}$. But $7-4 \sqrt{3}\textless1$, violating the assumption that $a \geq c$. Therefore, $\frac{a}{c}=7+4 \sqrt{3}$. Plugging this in, we have $\frac{1}{\frac{a}{4 c}+\frac{1}{4}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$. But we need the negative of this, so the answer is (D). 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations
Source: 其它
Problem: What is the $$100\rm th$$ number in the arithmetic sequence $$1$$, $$5$$, $$9$$, $$13$$, $$17$$, $$21$$, $$25$$, $$\cdots$$． 
Options:
aoVal: A
content: $$397$$ 
aoVal: B
content: $$399$$ 
aoVal: C
content: $$401$$ 
aoVal: D
content: $$403$$ 
aoVal: E
content: $$405$$ 
Answer: A
Analysis: $$1+(5-1)\times 99=397$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: The sum of five consecutive natural numbers is equal to $$2005$$. The greatest of these numbers is:~\uline{~~~~~~~~~~}~. (2005 Math Kangaroo Problem, Level 5-6, Question \#17) 
Options:
aoVal: A
content: $$401$$ 
aoVal: B
content: $$403$$ 
aoVal: C
content: $$404$$ 
aoVal: D
content: $$405$$ 
aoVal: E
content: $$2001$$ 
Answer: B
Analysis: Based on the middle term rule, we can find the middle number is $$401$$, and the greatest number is $$401+1+1=403$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: $$10^{5}+10^{6}=10^{5}\times $$． 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$10^{2}$$ 
aoVal: D
content: $$10^{6}$$ 
Answer: A
Analysis: $$10^{5}+10^{6}=1100000=11\times 10^{5}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: In December, Tom-the-cat slept for exactly 3 weeks. Which calculations should we do in order to find how many minutes he stayed awake during this month? 
Options:
aoVal: A
content: (31 - 7) x 3 x 24 x 60 
aoVal: B
content: (31 - 7) x 24 x 60 
aoVal: C
content: (31 -7 x 3) x 24 x 60 
aoVal: D
content: (31 -7 x 3) x 24 x 60 x 60 
aoVal: E
content: (30 -7 x 3) x 24 x 60 
Answer: B
Analysis: The cat slept for exactly 3 weeks for 7 x 3 days.  The cat was awake for 31 - 7 x 3 days.  A day has 24 hours, and an hour has 60 minutes.  Therefore, the cat was awake in (31- 7 x 3) x 24 x 60 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion
Source: 其它
Problem: Calculate: $$894-89-111-95-105-94=$$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$300$$ 
aoVal: B
content: $$400$$ 
aoVal: C
content: $$500$$ 
aoVal: D
content: $$600$$ 
Answer: B
Analysis: $$(894-94)-(89+111)-(95+105)$$  $$=800-200-200$$  $$=400$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Operation Strategy in Addition and Subtraction of Rounding Whole Numbers
Problem: If $$2$$ dogs weigh as much as $$3$$ cats, and $$2$$ cats weigh as much as $$15$$ mice, how many dogs weigh as much as $$45$$ mice? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: In weight, $$45$$ mice $$=3\times (15$$ mice$$)=3\times (2$$ cats$$)= 2\times (3$$ cats$$)=2\times (2$$ dogs$$)= 4$$ dogs. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: What is the simplest form of $$2$$ yards $$:30$$ feet? (Note that $1$ yard is equal to $3$ feet)． 
Options:
aoVal: A
content: $$2:30$$ 
aoVal: B
content: $$1:5$$ 
aoVal: C
content: $$5:1$$ 
aoVal: D
content: $$30:2$$ 
Answer: B
Analysis: We need to make the units same first. $$2$$ yards equal to $$6$$ feet. Now we could remove the same unit, feet. We get $$6:30$$ and simplify it to $$1:5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: The sum of the first $$100$$ positive even whole numbers is $$10100$$.  What is the sum of the first $$101$$ positive even whole numbers? 
Options:
aoVal: A
content: $$10302$$ 
aoVal: B
content: $$10202$$ 
aoVal: C
content: $$10201$$ 
aoVal: D
content: $$10102$$ 
Answer: A
Analysis: The $$100$$ even numbers that add up to $$10100$$ are $$2$$, $$4$$, $$\cdots $$, $$200$$.  The sum we want is $$2+4+ \cdots + 200 + 202 = 10 100 + 202$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: The ones digit of $$106\times107\times108\times109\times110$$ is． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$8$$ 
Answer: A
Analysis: The ones digit is the same as the ones digit of $$6 \times7\times8\times9 \times0$$. ` 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$2009\times 2009-2008\times 2008=$$． 
Options:
aoVal: A
content: $$4017$$ 
aoVal: B
content: $$4007$$ 
aoVal: C
content: $$4027$$ 
aoVal: D
content: $$3017$$ 
Answer: A
Analysis: $$=(2008+1)\times 2009-2008\times 2008$$  $$=2008\times 2009+2009-2008\times 2008$$  $$=2008\times (2009-2008)+2009$$  $$=2008+2009$$  $$=4017$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas-> Difference of Two Squares Formula
Problem: What is the tens digit of $7^{2011}$? (2011 AMC 8 Problems, Question \#22) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$7$$ 
Answer: D
Analysis: Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic.  $$ \begin{aligned} 7 \equiv 07 \&(\bmod 100) \textbackslash\textbackslash{} 7^{2} \equiv 49 \&(\bmod 100) \textbackslash\textbackslash{} 7^{3} \equiv 43 \&(\bmod 100) \textbackslash\textbackslash{} 7^{4} \equiv 01 \&(\bmod 100) \end{aligned} $$  We can write $7^{2011}$ as $\left(7^{4}\right)^{502} \times 7^{3}$. Using this, we can say: $$ 7^{2011} \equiv\left(7^{4}\right)^{502} \times 7^{3} \equiv 7^{3} \equiv 343 \equiv 43 \quad(\bmod 100) . $$  From the above, we can conclude that the last two digits of $7^{2011}$ are $43$. Since they have asked us to find the tens digit, our answer is (D) $4$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: A palindrome is a number that can be read the same forward and backward. For example, the numbers 99, 252 and 4884 are palindromes. How many 3. digit palindrome numbers are there? 
Options:
aoVal: A
content: $$81$$ 
aoVal: B
content: $$900$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: From 100-199: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191 (10 numbers)  From 200 - 299: 202, 212, 222, 232, 242, 252, 262, 272, 282, 292 (10 numbers)  From 300 to 399: 10 numbers  From 400 to 499: 10 numbers  $$\cdots $$  From 900 to 999: 10 numbers.  Hence, there are 10 x 9 = 90 three-digit palindromes. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: Given the equation $3x+4y=5$, find the slope and $y$-intercept. 
Options:
aoVal: A
content: $m=-3, b=4$ 
aoVal: B
content: $m=-3, b=5$ 
aoVal: C
content: $m=\frac{3}{4}, b=\frac{5}{4}$ 
aoVal: D
content: $m=-\frac{3}{4}, b=\frac{5}{4}$ 
Answer: D
Analysis: $3x+4y=5$,  $y=-\frac{3}{4}x+\frac{5}{4}$,  Its slope is $-\frac{3}{4}$ and intercept is $\frac{5}{4}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable
Source: 其它
Problem: Store $A$ is selling watermelon at the price of $32$ dollars per pound. Store $B$ is selling the same watermelon at the price of $30$ dollars per $16$ ounces. Which store has a better buy? 
Options:
aoVal: A
content: $A$ 
aoVal: B
content: $B$ 
Answer: B
Analysis: $16$ ounce equals $1$ pound 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion
Source: 其它
Problem: The number of hours in $$10$$ days $$=$$ the number of minutes inhours. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
Answer: C
Analysis: The number of hours in $$10$$ days is $$240$$; $$240$$ minutes is $$240\div60 =4$$ hours. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Time
Problem: $$4:14=14:$$． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$114$$ 
Answer: C
Analysis: Divide through by $$2$$ then multiply by $$7$$ to get $$4:14=2:7=14:49$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: What is the product of $763$ and $5$? 
Options:
aoVal: A
content: $$3505$$ 
aoVal: B
content: $$3815$$ 
aoVal: C
content: $$3515$$ 
aoVal: D
content: $$3805$$ 
Answer: B
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: If $ a◆b$ means$(a\times b)+b$ , then $2◆3$ has the value． 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: If $a◆b$~ represents$(a\times b)+b$ , $2◆3=(2\times3)+3=9$ .  So the answer is $\rm B$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: There are $$30$$ students in a classroom. They sit at desks in pairs in such a way that every boy sits with girl, and only half of girls sit with a boy. How many boys are in the classroom? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$22$$ 
Answer: A
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Problem: Calculate：$$\frac{1}{2}\times \frac{22}{7}\div \frac{11}{5}$$． 
Options:
aoVal: A
content: $$\frac{5}{7}$$ 
aoVal: B
content: $$\frac{4}{7}$$ 
aoVal: C
content: $$\frac{6}{7}$$ 
aoVal: D
content: $$\frac{3}{7}$$ 
Answer: A
Analysis: $$\frac{1}{2}\times \frac{22}{7}\div \frac{11}{5}=\frac{1}{2}\times \frac{22}{7}\times \frac{5}{11}=\frac{5}{7}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: The product of any whole number and $$2$$ is always． 
Options:
aoVal: A
content: 　prime 
aoVal: B
content: 　composite 
aoVal: C
content: 　odd 
aoVal: D
content: 　even 
Answer: D
Analysis: The product of a whole number and an even number must be even. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner $A$, added to the number on Spinner $B$. What is the probability that $N$ is a perfect square number? (2022 AMC 8 Problems, Question \#12) 
Options:
aoVal: A
content: $\frac{1}{16}$ 
aoVal: B
content: $\frac{1}{8}$ 
aoVal: C
content: $\frac{1}{4}$ 
aoVal: D
content: $\frac{3}{8}$ 
aoVal: E
content: $\frac{1}{2}$ 
Answer: B
Analysis: First, we calculate that there are a total of $4 \cdot 4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=$ (B) $\frac{1}{8}$ probability that the number formed by the two spinners is a perfect square. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $16^{-2}=\left(\frac{1}{4}\right)^{}x$, what is the value of $x$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: Convert both sides into powers with the same bases:  $4^{-4}=4^{-x}$  $x=4$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The 2000 Census identified the ethnic breakdown of the state of California to be approximately as follows: White:46％, Latino:32\%, Asian:11\%, Blcak:7\%,and Other:4\%.~Assuming that these are mutually exclusive categories (this is not a realistic assumption), what is the probability that a randomly selected person from the state of California is of Asian or Latino descent? 
Options:
aoVal: A
content: 46\% 
aoVal: B
content: 32\% 
aoVal: C
content: 11\% 
aoVal: D
content: 43\% 
aoVal: E
content: 3.5\% 
Answer: D
Analysis: The correct answer is(d). Because ethnic group categories are assumed to be mutually exclusive, P(Asian or Latino)=P(Asian)+P(Latino)=32\%+11\%=43\% 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: Kitty writes down a sequence of five integers. The rule she uses is, "after the first two terms, each term is the sum of the two previous terms." She sequence is ---, ---, ---, ~$$18$$, $$29$$.  What is her first term? ． 
Options:
aoVal: A
content: $$0 $$ 
aoVal: B
content: $$ 3 $$ 
aoVal: C
content: $$ 4 $$ 
aoVal: D
content: $$ 5 $$ 
aoVal: E
content: $$ 7$$ 
Answer: C
Analysis: Let the first six terms of Kitty\textquotesingle s sequence be $$a$$, $$b$$, $$c$$, $$18$$ and $$29$$ respectively.  Then $$c+ 18= 29$$, so $$c= 11$$. Hence $$b+11= 18$$, so $$b=7$$.  Therefore, $$a+7=11$$, so $$a=4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: How many multiples of $$7$$ are between $$20$$ and $$100$$? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$14$$ 
Answer: C
Analysis: $$3\times7=21$$, start from the $$3^{}\rm{rd}$$  $$14\times7=98$$, end at the $$14^{}\rm{th}$$  Thus, the number of term remain: $$14-2=12$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它