math-eval/TAL-SCQ5K huggingface dataset:
Problem: Among the natural numbers $1-75$: How many are divisible by $3$ or $5$? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$45$$ 
Answer: B
Analysis: All natural numbers divisible by $3$ in the range of $1-75$ are：$75\div3=25$．  All natural numbers divisible by $5$ in the range of $1-75$ are：$75\div5=15$．  The natural numbers divisible by 3 and 5 that are divisible by $15$：$75\div15=5$．  The natural numbers divisible by $3$ or $5$ are：$25+15-5=35$． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Miruna had to multiply two $$2$$-digit numbers together, but she accidentally reversed the digits of both of them before multiplying and reached the answer $$209$$. Which of the following answers should she have got? 
Options:
aoVal: A
content: $$1001$$ 
aoVal: B
content: $$1003$$ 
aoVal: C
content: $$1005$$ 
aoVal: D
content: $$1007$$ 
aoVal: E
content: $$1009$$ 
Answer: A
Analysis: The prime factors of $$209$$ are $$11$$ and $$19$$, so these must have been the reversed numbers that Miruna multiplied. The correct multiplication was $$11 \times 91=1001$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Problem: The product of three consecutive numbers is $$15600$$. What is their sum? 
Options:
aoVal: A
content: $$75$$ 
aoVal: B
content: $$78$$ 
aoVal: C
content: $$81$$ 
aoVal: D
content: $$84$$ 
aoVal: E
content: $$87$$ 
Answer: A
Analysis: Given that the product, $$15600$$, is a multiple of $$25$$, the three numbers between  them must contribute two factors of $$5$$; since they are consecutive and there are only three of them, this can happen only if one number is itself a multiple of $$25$$. It is now worth observing that the product of three consecutive numbers is roughly the same as the cube of the middle of the three. Since $$20^{3}\textless{} 15 600 \textless{} 30^{3}$$, the middle number must lie between $$20$$ and $$30$$, hence one of the numbers must be $$25$$. The numbers can therefore be $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$,~ $$\left\textbackslash{ {24, 25, 26} \right\textbackslash}$$ or $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$. The product $$15600 =13 \times1200$$, so it has factors both $$4$$ and $$13$$. The triple $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$ has no factor of $$4$$, and $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$ no factor of $$13$$. So, by elimination, the numbers are $$24$$, $$25$$ and $$26$$, and their total is $$75$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Problem: A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? ($2011$ AMC $8$ Problem, Question \#$18$) 
Options:
aoVal: A
content: $\dfrac{1}{6}$ 
aoVal: B
content: $\dfrac{5}{12}$ 
aoVal: C
content: $\dfrac{1}{2}$ 
aoVal: D
content: $\dfrac{7}{12}$ 
aoVal: E
content: $\dfrac{5}{6}$ 
Answer: D
Analysis: There are $6\times6=36$ ways to roll a die twice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first time is greater than the second should be the same as the number of rolls where the second time is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\frac{7}{12}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Wendy wrote all the whole numbers from $$13$$ to $$78$$ on her paper. How many times did she write all the digit "$$3$$ on her paper? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$17$$ 
aoVal: E
content: $$18$$ 
Answer: D
Analysis: Ones: $$13, 23, 33, 43, 53, 63, 73$$  Tens: $$30, 31, 32, 33, 34, 35, 36, 37, 38, 39$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: $12 + 8 + 8 + 4 $ is the same as~\uline{~~~~~~~~~~}~$\times 4$. 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$10$$ 
Answer: C
Analysis: $$12+8+8+4=32$$  $$32\div4=8$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: A school has 100 students and 5 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50,20,20,5$, and 5 . Let $l$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$ ? 
Options:
aoVal: A
content: $$-18.5$$ 
aoVal: B
content: $$-13.5$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$13.5$$ 
aoVal: E
content: $$18.5$$ 
Answer: B
Analysis: The formula for expected values is $$ \text { Expected Value }=\sum(\text { Outcome } \cdot \text { Probability }) . $$ We have $$ \begin{aligned} t \& =50 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+5 \cdot \frac{1}{5}+5 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =(50+20+20+5+5) \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =100 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =20, \textbackslash\textbackslash{} s \& =50 \cdot \frac{50}{100}+20 \cdot \frac{20}{100}+20 \cdot \frac{20}{100}+5 \cdot \frac{5}{100}+5 \cdot \frac{5}{100} \textbackslash\textbackslash{} \& =25+4+4+0.25+0.25 \textbackslash\textbackslash{} \& =33.5 . \end{aligned} $$ Therefore, the answer is $t-s=(\mathbf{B})-13.5$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Julia, Kasia, Susanna, and Helena have their birthdays on March $$1^{\rm st}$$, May $$17^{\rm th}$$, July $$20^{\rm th}$$, and March $$20^{\rm th}$$. Kasia and Susanna were born in the same month. Julia and Susanna were born on the same day of a month. Which of the girls was born on May $$17^{\rm th}$$? ． 
Options:
aoVal: A
content: Julia 
aoVal: B
content: Kasia 
aoVal: C
content: Susanna 
aoVal: D
content: Helena 
Answer: D
Analysis: Susanna: March 20\textsuperscript{th}  Kasia: March 1\textsuperscript{st}  Julia: July 20\textsuperscript{th}  Helena: May 17\textsuperscript{th} 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: $$7\times 11\times 13=$$ ． 
Options:
aoVal: A
content: $$1001$$ 
aoVal: B
content: $$1111$$ 
aoVal: C
content: $$1221$$ 
aoVal: D
content: $$1101$$ 
Answer: A
Analysis: 7x(10+1)x13 =(70+7)x13 =77x(10+3) =77x10+77x3 =770+231 =1001 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: Two different numbers are randomly selected from the set $-5, -3, -1, 3, 5$~and multiplied together. What is the probability that the product is a negative number? (adapted from $2016$ AMC $8$ Problem, Question \#$13$) 
Options:
aoVal: A
content: $\frac{5}{6}$ 
aoVal: B
content: $\frac{2}{5}$ 
aoVal: C
content: $\frac{3}{4}$ 
aoVal: D
content: $\frac{1}{2}$ 
aoVal: E
content: $\frac{3}{5}$ 
Answer: E
Analysis: $\frac25\times\frac34+\frac35\times\frac24=\frac{3}{5}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Given that $x+2y=3$, $3^{}x\cdot 9^{}y=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$81$$ 
aoVal: E
content: $$243$$ 
Answer: B
Analysis: $3^{}x \cdot 9^{}y=3^{x+2y}=3^{3}=27$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: The total score of $8$ students is an even number. Each of $3$ of them gets an odd number score, and each of $4$ of them gets an even number score. Which of the following would be the possible score that the last student gets? 
Options:
aoVal: A
content: $$66$$ 
aoVal: B
content: $$70$$ 
aoVal: C
content: $$74$$ 
aoVal: D
content: $$77$$ 
aoVal: E
content: $$80$$ 
Answer: D
Analysis: The sum of $3$ odd numbers is an odd number.  The sum of $4$ even numbers is an even number.  The sum of an odd number and an even number is an odd number, so the total score of the $7$ students is an odd number.  The total score of $8$ students is an even number, and the sum of $2$ odd numbers is an even number.  Therefore, the score of the last one should be an odd number. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: A deck of cards contains a white card and two red cards. Take one card out randomly, record its colour, and put it back in the deck. Then, take another card out randomly.  Compare the probability of the following situations.  I. Both cards are of the same colour.  II. Both cards are red.  III. Two cards are of different colours.  Which one has the highest probability? 
Options:
aoVal: A
content: $$I$$ 
aoVal: B
content: $$II$$ 
aoVal: C
content: $$III$$ 
Answer: A
Analysis: I. Both cards are in the same colour: $$ \dfrac{5}{9}$$;  II. Both cards are red: $$\dfrac{4}{9}$$;  III. Two cards are in different colours: $$\dfrac{4}{9}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Each of the 5 sides and the 5 diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color? 
Options:
aoVal: A
content: $\frac{2}{3}$ 
aoVal: B
content: $\frac{105}{128}$ 
aoVal: C
content: $\frac{125}{128}$ 
aoVal: D
content: $\frac{253}{256}$ 
aoVal: E
content: $$1$$ 
Answer: D
Analysis: Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.  Case 1: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.  Case $2: 3$ are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.  Case $3: 2$ are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are $\left(\begin{array}{l}4 \textbackslash\textbackslash{} 2\end{array}\right)$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6 \cdot 2=12$ ways to color the pentagon so that no such triangle has the same color for all of its sides. These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024}=1-\frac{3}{256}=\frac{253}{256}=D$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: How many different positive integers at most can add up to $$80$$? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$13$$ 
Answer: D
Analysis: $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10+11+12 = 78$$, $$80 -- 78 = 2$$. Since all the numbers should be different, the remaining \textquotesingle$$2$$\textquotesingle{} cannot make up a new positive integer. It can only be added to the number(s) before. Therefore, there are at most $$12$$ different positive integers that can add up to $$80$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Which of the following statements is not correct? 
Options:
aoVal: A
content: Indefinite events include impossible events. 
aoVal: B
content: The probability of an impossible event to happen is $$0$$. 
aoVal: C
content: The probability of an indefinite event to happen is between $$0$$ and $$1$$. 
aoVal: D
content: The probability of a certain event to happen is $$1$$. 
Answer: D
Analysis: Impossible events are definite events.  $$\text{B}$$, $$\text{C}$$, and $$\text{D}$$ are right.  Thus, the answer is $$\text{A}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: A bag contains four pieces of paper, each labeled with one of the digits $$1$$, $$2$$, $$3$$ or $$4$$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $$3$$? ($$2007$$ AMC $$8$$ Problem, Question \#$$24$$) 
Options:
aoVal: A
content: $$\frac{1}{4}$$ 
aoVal: B
content: $$\frac{1}{3}$$ 
aoVal: C
content: $$\frac{1}{2}$$ 
aoVal: D
content: $$\frac{2}{3}$$ 
aoVal: E
content: $$\frac{3}{4}$$ 
Answer: C
Analysis: The combinations of digits that give multiples of $$3$$ are $$(1$$, $$2$$, $$3)$$ and $$(2$$, $$3$$, $$4)$$. For each of them, there are $3\times2\times1=6$ possibilities. Thus, there are $6+6=12$ possibilities in total. The number of ways to choose three digits out of four is $$4\times3\times2=24$$. Therefore, the probability is $$\frac{12}{24}=\frac12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: How many two-digit numbers are there where the ones digit is greater than the tens digit?． 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: $$8+7+6+5+4+3+2+1=36$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Problem: There is a ball in a box. Three kids are guessing the colour of the ball.  Val says: "The ball is white."  John says: "The ball is blue."  Elvis says: "I agree with Val."  Then, they open the box and find only one of them guessed right.  The ball is  . 
Options:
aoVal: A
content: white 
aoVal: B
content: Blue 
aoVal: C
content: Uncertain 
Answer: B
Analysis: Val\textquotesingle s point and Elvis\textquotesingle{} points are identical, so both of them guessed incorrectly.  Therefore, John guessed correctly. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: A number is formed by writing $2022$ consecutively for $3$ times. Which of the following statements is correct? 
Options:
aoVal: A
content: The number is a multiple of $3$, $6$ and $9$. 
aoVal: B
content: The number is a multiple of $3$ and $6$ but not a multiple of $9$. 
aoVal: C
content: The number is a multiple of $3$ and $9$ but not a multiple of $6$. 
aoVal: D
content: The number is a multiple of $3$ but not a multiple of $6$. 
aoVal: E
content: None of the above 
Answer: A
Analysis: The sum of the digits of 202220222022 is 2 x 9 = 18. Hence this number is divisible by 3 and 9. As its last digit is 2, it is also divisible by 6. The answer is \textbf{Option A.} 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: The average height of June, Ali and Linda is $$160\text{cm}$$. Ali is $$166 \text{cm}$$ tall. June and Linda are as tall as each other. What is Linda\textquotesingle s height? 
Options:
aoVal: A
content: $$\rm~~154 cm$$ 
aoVal: B
content: $$\rm~~157 cm$$ 
aoVal: C
content: $$\rm~~162 cm$$ 
aoVal: D
content: $$\rm~~164 cm$$ 
Answer: B
Analysis: The total height is $160\times 3 = 480\text{cm}$ and hence the height of both June and Linda is $480-166=314\text{cm}$.  Therefore the height of Linda is $314 \div 2 = 157\text{cm}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: SASMO 2014 P2 Q8  Find the number A such that the following statement is true:  7 x A = 3 x 8 + 4 x 8 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: 7A = 24 + 32  7A = 56  A = 8 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: What is the probability of choosing a composite number from $0\sim10$? 
Options:
aoVal: A
content: $\frac12$ 
aoVal: B
content: $\frac5{11}$ 
aoVal: C
content: $\frac3{10}$ 
aoVal: D
content: $\frac25$ 
aoVal: E
content: $\frac{6}{11}$ 
Answer: B
Analysis: There are $11$ numbers in total. Among them, $4,6, 8, 9,$ and $10$ are composite numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: There are $$30$$ pupils in my class. $$20$$ pupils like Maths and $$18$$ pupils like English. Twice as many pupils like both subjects as like neither of them. How many pupils like only Maths? 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$4$$ 
Answer: E
Analysis: Let the number of pupils who like neither subject be $$x$$. Hence the number who like both subjects is $$2x$$. Therefore the number of pupils who like only Maths is $$20−2x$$ and the number who like only English is $$18−2x$$. Since there are $$30$$ pupils in my class, we have $$\left( 20-2x \right)+2x+\left( 18-2x \right)+x=30$$ and hence $$38−x = 30$$. This has solution $$x = 8$$ and hence the number of pupils who like only Maths is $$20-2\times 8=4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Amos is taller than Eugene.  Leo is shorter than James but taller than Eugene.  James is shorter than Amos.  is the tallest andis the shortest. 
Options:
aoVal: A
content: Amos, James 
aoVal: B
content: James, Eugene 
aoVal: C
content: James, Leo 
aoVal: D
content: Amos, Eugene 
Answer: D
Analysis: From tallest to shortest: \textbf{Amos}, James, Leo, \textbf{Eugene}. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? (2019 AMC 8 Problems, Question \#18) 
Options:
aoVal: A
content: $\dfrac{4}{9}$ 
aoVal: B
content: $\dfrac{1}{2}$ 
aoVal: C
content: $\dfrac{5}{9}$ 
aoVal: D
content: $\dfrac{3}{5}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: We have a $2$ dice with $2$ evens and $4$ odds on both dice. For the sum to be even, the $2$ rolls can be $2$ odds or $2$ evens.  Ways to roll $2$ odds: The total number of ways to obtain $2$ odds on $2$ rolls is $4 * 4=16$, as there are $4$ possible odds on the first roll and $4$ possible odds on the second roll.  Ways to roll $2$ evens: Similarly, we have $2 * 2=4$ ways to obtain $2$ evens.  The probability is $\frac{20}{36}=\frac{5}{9}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Source: 其它
Problem: How many digits are there from $5$ to $118$? 
Options:
aoVal: A
content: $$114$$ 
aoVal: B
content: $$204$$ 
aoVal: C
content: $$239$$ 
aoVal: D
content: $$242$$ 
Answer: D
Analysis: $5 - 9: 9 - 5 + 1 = 5$  $10 - 99: 99 - 10 + 1 = 90$  $100 - 118: 118 - 100 + 1 = 19$  $5\times1 + 90\times2 + 19\times3 = 5 + 180 + 57 = 242$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Calculate the value of $$7 + 16 + 34 + 45 + 50 - 6 - 15 - 4 - 7$$. 
Options:
aoVal: A
content: $$70$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$127$$ 
aoVal: D
content: $$124$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: $$7-7 + 16-6 + 34-4 + 45-15 +50$$  $$= 0+10+30+30+50$$  $$= 120$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: In the UK it is now $$11$$ am. The time in San Francisco is $$8$$ hours behind the UK. What time do the clocks now show in San Francisco? 
Options:
aoVal: A
content: $$3$$ am 
aoVal: B
content: $$4$$ am 
aoVal: C
content: $$5$$ am 
aoVal: D
content: $$11$$ am 
aoVal: E
content: $$7$$ pm 
Answer: A
Analysis: Eight hours behind $$11$$ am is $$3$$ am. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Nick wants to bake a cake that consists of only $$1$$ flavour and $$1$$ topping. He gets to choose from $$3$$ different flavours and $$3$$ different toppings. How many different kinds of cake can he make? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: $$3\times3=9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: Two different numbers are randomly selected from the set $-2, -1, 0, 3, 4, 5$~and multiplied together. What is the probability that the product is $0$? ($2016$ AMC $8$ Problem, Question \#$13$) 
Options:
aoVal: A
content: $\frac{1}{6}$ 
aoVal: B
content: $\frac{1}{5}$ 
aoVal: C
content: $\frac{1}{4}$ 
aoVal: D
content: $\frac{1}{3}$ 
aoVal: E
content: $\frac{1}{2}$ 
Answer: D
Analysis: The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are 15 ways of choosing $2$ numbers randomly. Thus $\frac{5}{15} = \frac{1}{3}$. The answer is $D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Sophia's average score on six tests is $$82$$. Her average score on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ test is $$98$$. What is her average score on all eight tests? 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$94$$ 
Answer: A
Analysis: Sophia\textquotesingle s total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: How many distinguishable arrangements are there of $1$ blue tile, $2$ green tiles, and $3$ yellow tiles in row from left to right? (Tiles of the same color are indistinguishable) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$60$$ 
Answer: E
Analysis: $\frac{6!}{3!\times2!}=60$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它