math-eval/TAL-SCQ5K huggingface dataset:
Problem: Amy, Bill, and Celine are friends with different ages. Exactly one of the following statements is true.  $$\rm I$$. Bill is the oldest.  $$\rm II$$. Amy is not the oldest.  $$\rm III$$. Celine is not the youngest.  Rank the friends from oldest to youngest. ($$2004$$ AMC $$8$$ Problem, Question \#$$13$$) 
Options:
aoVal: A
content: Bill, Amy, Celine 
aoVal: B
content: Amy, Bill, Celine 
aoVal: C
content: Celine, Amy, Bill 
aoVal: D
content: Celine, Bill, Amy 
aoVal: E
content: Amy, Celine, Bill 
Answer: E
Analysis: If Bill is the oldest, then Amy is not the oldest, and both statements $$\rm I$$ and $$\rm II$$ are true, so statement $$\rm I$$ is not the true one.  If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements $$\rm II$$ and $$\rm III$$ are true, so statement $$\rm II$$ is not the true one.  Therefore, statement $$\rm III$$ is the true statement, and both $$\rm I$$ and $$\rm II$$ are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is $$\rm E$$: Amy, Celine, Bill. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: There is a three-digit number \textquotesingle$$502$$\textquotesingle. John writes a digit before this number to make it a new four-digit number. If he writes a~\uline{~~~~~~~~~~}~, the new four-digit number would be as small as possible. 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$0$$ 
Answer: C
Analysis: Note that the digit \textquotesingle$$0$$\textquotesingle{} could not be the highest digit of a four-digit number. Therefore, the smallest number is $$1502$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value in Enumeration Problems
Problem: The average of the \emph{different~}prime factors of $$2009$$ is． 
Options:
aoVal: A
content: $$2009$$ 
aoVal: B
content: $$147$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$24$$ 
Answer: D
Analysis: $$2009=7^{2}\times 41$$, and the average of $$7$$ and $$41$$ is $$(7+41)\div 2=24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: In the expression $$1□2□3□4$$ each $$□$$ is to be replaced by either $$+$$ or $$\times$$. What is the largest value of all the expressions that can be obtained in this way? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$25$$ 
Answer: E
Analysis: If $$m$$ and $$n$$ are positive integers, then $$mn{}\textgreater m+n$$ unless at least one of $$m$$ or $$n$$ is equal to $$1$$, or $$m=n=2$$. So, to maximise the expression, we need to place multiplication signs between $$2$$ and $$3$$ and between $$3$$ and $$4$$. However, we need to place an addition sign between $$1$$ and $$2$$ because $$1+2\times3\times4=25$$, whereas $$1\times2\times3\times4=24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Two tiles numbered $3$ and $4$ are turned face down, respectively. One tile is turned up at random, and throw a die to get a number from $1$ to $6$. What is the probability that the product of the numbers on the tile and the die is greater than or equal to $20$? 
Options:
aoVal: A
content: $\frac {1}{2}$ 
aoVal: B
content: $\frac {1}{6}$ 
aoVal: C
content: $\frac {1}{3}$ 
aoVal: D
content: $\frac {1}{4}$ 
aoVal: E
content: $\frac {1}{12}$ 
Answer: B
Analysis: There are $12$ different combinations. The product of two numbers is greater than $20$ will be $4\times5$ and $4\times6$. Thus, the probability is $\frac 2{12}$ = $\frac 16$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A cat divides $$24$$ fish into $$4$$ groups, and each group has at least $$1$$ fish. There are  fish in the group that has the largest number of fish. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$21$$ 
Answer: D
Analysis: $$ 1 + 1+1 + 21 = 24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In how many ways can the letters in $CPCCKBY$ be rearranged so that two or more $C$s do not appear together? 
Options:
aoVal: A
content: $$240$$ 
aoVal: B
content: $$180$$ 
aoVal: C
content: $$64$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$140$$ 
Answer: A
Analysis: There are $3$ $C$s in total now with other $4$ letters remaining.  There are $\_4P\_4$ ways for us to arrange the $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4\times \_5C\_3 =240$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: The digits $1$, $2$, and $3$ can make~\uline{~~~~~~~~~~}~three-digit numbers. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$30$$ 
Answer: C
Analysis: $$3\times 3\times 3=27$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In my suitcase I have $$5$$ sweaters and $$6$$ pairs of pants. If I make an outfit of a sweater and a pair of pants, how many different outfits can I select? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$30$$ 
Answer: D
Analysis: I have $$5$$ sweaters and $$6$$ pairs of pants. For each sweater, there are $$6$$ pairs of pants with which that sweater can be paired. There are $$5$$ sweaters, so there are $$5\times6=30$$ different possible outfits. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: There were $20$ ducks, pigs, and sheep in total in Sam\textquotesingle s farm. After Sam bought some new sheep, the number of sheep has doubled. There are $27$ ducks, pigs, and sheep in total. Originally, how many ducks and pigs were there? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: A
Analysis: $27-20=7$  $7+7=14$  $20-14=6$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Two tiles numbered $5$ and $6$ are turned face down, respectively. One tile is turned up at random, and throw a die to get a number from $1$ to $6$. What is the probability that the product of the numbers on the tile and the die is smaller than or equal to $8$? 
Options:
aoVal: A
content: $\frac {1}{2}$ 
aoVal: B
content: $\frac {1}{3}$ 
aoVal: C
content: $\frac {1}{4}$ 
aoVal: D
content: $\frac {1}{5}$ 
aoVal: E
content: $\frac {1}{6}$ 
Answer: E
Analysis: There are $2\times6=12$ different combinations. The product of two numbers is smaller than $8$ will be $5\times1$ and $6\times1$. Thus, the probability is $\frac 2{12}$ = $\frac 16$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: In how many ways can the letters in $AAABCDA$ be rearranged so that two or more $A$s do not appear together? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$18$$ 
Answer: B
Analysis: There are $4$ $A$s in total, which have $3$ intervals leaving for the other $3$ letters. Thus, the answer is $\_3P\_3=6$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: If Keyue fished every day from July $$17$$ through July $$31$$ (within the same year), she fished fordays． 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$24$$ 
Answer: B
Analysis: Total number of days $$=31-17 +1= 15$$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In a toy store, cars are available in $5$ different colours: blue, white, yellow, black and red. A car has either $2$ or $4$ doors. How many different version of the car are available? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: $$5 \times 2 = 10$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Two distinct numbers from $$1$$ to $$100$$ inclusive will form a pair if the sum of these two is a multiple of $$5$$. How many different pairs are there? 
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$150$$ 
aoVal: C
content: $$800$$ 
aoVal: D
content: $$990$$ 
aoVal: E
content: $$1200$$ 
Answer: D
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Today is Amy\textquotesingle s birthday! She takes photos with her four close friends. All of them stand in a line and they make Amy stand in the middle. There are~\uline{~~~~~~~~~~}~different ways for them to form the line. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$36$$ 
Answer: D
Analysis: $$4 \times 3 \times 1 \times 2 \times 1 = 24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: There are $$24$$ four-digit numbers which is formed using each of the digits $$3$$, $$5$$, $$6$$ and $$9$$ once only. When all of these $$24$$ four-digit numbers are put in order from smallest to largest, which one is in the eighth position? 
Options:
aoVal: A
content: $$3569$$ 
aoVal: B
content: $$5369$$ 
aoVal: C
content: $$5396$$ 
aoVal: D
content: $$5639$$ 
aoVal: E
content: $$5936$$ 
Answer: C
Analysis: When put in order, the numbers are: $$3569$$, $$3596$$, $$3659$$, $$3695$$, $$3956$$, $$3965$$, $$5369$$, $$5396$$, $$5639$$, $$5693$$, $$\ldots $$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: The maximum number of intersection points of $$4$$ different circles is． 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: Each pair has $$2$$ intersection points. The $$6$$ pairs have $$12$$ such points. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Dotted Line Arrangement
Problem: In how many ways can the digits in $3433256337$ be rearranged so that two or more $3$s do not appear together? 
Options:
aoVal: A
content: $$1800$$ 
aoVal: B
content: $$1200$$ 
aoVal: C
content: $$1000$$ 
aoVal: D
content: $$720$$ 
aoVal: E
content: $$120$$ 
Answer: D
Analysis: There are $5$ $3$s in total now with other $5$ digits remaining.  There are $\_5P\_5$ ways for us to arrange the $5$ letters\textquotesingle{} positions. Then, we can put the $5$ $3$s in the $6$ intervals.  So the answer is $\_5P\_5 \times \_6C\_5=720$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Elvis is packing shirts for a trip. He just randomly grabs $3$ shirts from his closet without observing them. The closet contains $10$ shirts: $5$ striped, $3$ plaid, and $2$ pure-colored ones. What is the probability that he chooses $2$ striped shirts and $1$ pure-colored shirt? 
Options:
aoVal: A
content: $\frac12$ 
aoVal: B
content: $\frac13$ 
aoVal: C
content: $\frac1{12}$ 
aoVal: D
content: $\dfrac{1}{6}$ 
aoVal: E
content: $\frac14$ 
Answer: D
Analysis: $\dfrac{\_5C\_2 \times~~\_2C\_1}{\_{10}C\_3}=\dfrac{10\times2}{120}=\dfrac{2}{12}=\dfrac{1}{6}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: We can formdifferent two-digit numbers with the numbers $$1$$, $$3$$ and $$0$$.（without using the same number two times like $$33$$ ） 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $$13$$、$$10$$、$$31$$、$$30$$， notice that $$0$$ cannot be in the first place, so only $$4$$ numbers can be formed．  So the answer is $$\text{B}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Dictionary Ordering
Problem: How many four-digit numbers can be made using the digits $1-8$ without repeating digits? 
Options:
aoVal: A
content: $$1680$$ 
aoVal: B
content: $$4096$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$256$$ 
Answer: A
Analysis: $$8\times 7\times 6\times 5=1680$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Eight cards are numbered from $$1$$ to $$8$$. The cards are placed in two boxes $$P$$ and $$Q$$ so that the sum of the numbers on the three cards in box $$P$$ is equal to the sum of the numbers on the five cards in box $$Q$$. Which of the following statements must be true? 
Options:
aoVal: A
content: The card numbered $$1$$ is not in box $$Q$$ 
aoVal: B
content: Four cards in box $$Q$$ have even numbers on 
aoVal: C
content: The card numbered $$5$$ is in box $$Q$$ 
aoVal: D
content: The card numbered $$2$$ is in box $$Q$$ 
aoVal: E
content: Exactly three cards in box $$Q$$ have odd numbers on 
Answer: D
Analysis: Note first that the sum of the numbers on the eight cards is $$36$$. Therefore the sum of the numbers on the cards in each of the boxes is $$18$$. There are only three cards in box $$P$$ and hence the possible combinations for the numbers on the cards in box $$P$$ are $$\left( 8,7,3 \right)$$, $$\left( 8,6,4 \right)$$ and $$\left( 7,6,5 \right)$$ with the corresponding combinations for box $$Q$$ being $$\left( 6,5,4,2,1 \right)$$, $$\left( 7,5,3,2,1 \right)$$ and $$\left( 8,4,3,2,1 \right)$$. The only statement which is true for all three possible combinations for box $$Q$$ is that the card numbered $$2$$ is in box $$Q$$. Hence the only statement which must be true is statement $$\rm D$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: Amy has $$7$$ gray balls, $$4$$ white balls and $$3$$ black balls in a bag. What is the least number of balls she has to take out of the bag with her eyes closed to be sure that she takes at least two balls of each color? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $$7+4+2=13$$． 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems
Problem: Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (2013 AMC 8 Problem, Question \#14) 
Options:
aoVal: A
content: $\frac{1}{4}$ 
aoVal: B
content: $\frac{1}{3}$ 
aoVal: C
content: $\frac{3}{8}$ 
aoVal: D
content: $\frac{1}{2}$ 
aoVal: E
content: $\frac{2}{3}$ 
Answer: C
Analysis: The probability that both show a green bean is $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\left(\right.$ C) $\frac{3}{8}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A dinner set in a restaurant is free to choose a main course, a salad and a drink. Today, the restaurant offers five main courses, three salads, and six drinks for customers to choose. How many different sets can be matched? ． 
Options:
aoVal: A
content: $14$ 
aoVal: B
content: $30$ 
aoVal: C
content: $48$ 
aoVal: D
content: $80$ 
aoVal: E
content: $90$ 
Answer: E
Analysis: $5\times3\times6=90$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Throw two dice of the same quality and size. The six sides of each die are marked with number of dots from $$1$$ to $$6$$. Among the following options,~\uline{~~~~~~~~~~}~is an impossible event. 
Options:
aoVal: A
content: The sum of dots is $$12$$. 
aoVal: B
content: The sum of dots is smaller than $$3$$. 
aoVal: C
content: The sum of dots is larger than $$4$$ but smaller than $$8$$. 
aoVal: D
content: The sum of dots is $$13$$. 
Answer: D
Analysis: The maximum sum is $$6+6=12$$, so "the sum of dots is $$13$$" is an impossible event; so $$\text{D}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $13$ kids in line. They are numbered following the pattern: $1$, $2$, $3$, $4$, $5$, $1$, $2$, $3$, $4$, $5$, and so on. How many kids are numbered with odd numbers? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$13$$ 
Answer: C
Analysis: $13\div5=2R3$  There are $3$ odd numbers in each cycle.  $3\times2+2=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 2020年希望杯二年级竞赛模拟第30题
其它
Problem: A palindromic number is a number that reads the same when the order of its digits is reversed. What is the difference between the largest and smallest five-digit palindromic numbers that are both multiples of $$45$$? 
Options:
aoVal: A
content: $$9180$$ 
aoVal: B
content: $$9090$$ 
aoVal: C
content: $$9000$$ 
aoVal: D
content: $$8910$$ 
aoVal: E
content: $$8190$$ 
Answer: B
Analysis: For a number to be a multiple of $$45$$ it must be a multiple of $$5$$ and also of $$9$$. In order to be a multiple of $$5$$, a number\textquotesingle s units digit must be $$0$$ or $$5$$. However, the units digit of a palindromic number cannot be $$0$$, so it may be deduced that any palindromic number which is a multiple of $$45$$ both starts and ends in the digit $$5$$. In order to make the desired number as large as possible, its second digit should be $$9$$ and for it to be as small as possible its second digit should be $$0$$. So, if possible, the numbers required are of the form \textquotesingle$$59x95$$\textquotesingle{} and \textquotesingle$$50y05$$\textquotesingle{} . In addition, both numbers are to be multiples of $$9$$ which means the sum of the digits of both must be a multiple of $$9$$. For this to be the case, $$x=8$$ and $$y=8$$ , giving digit sums of $$36$$ and $$18$$ respectively. So the two required palindromic numbers are $$59895$$ and $$50805$$. Their difference is $$9090$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value in Enumeration Problems
Problem: There are 70 beads.~ $$A$$ and $$B$$ take turns taking beads from the pile ($$A$$ goes first). Each person can take between 1 and 4 beads per turn. The person who takes the last bead loses. If $$A$$ wants to win, how many beads should he take on the first turn? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$A$$ can\textquotesingle t win 
Answer: A
Analysis: $$(70-1)\div (1+4)=17\cdots 1$$  $$A$$ takes one bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Math Game->Sudoku
Problem: What is $$6$$ ones and $$15$$ thousandths? 
Options:
aoVal: A
content: $$0.615$$ 
aoVal: B
content: $$6.0015$$ 
aoVal: C
content: $$6.015$$ 
aoVal: D
content: $$6.105$$ 
aoVal: E
content: $$6.15$$ 
Answer: C
Analysis: $$6$$ ones $=6$  $$15$$ thousandths $=$ $1$ hundredth and $5$ thousandths $=0.015$  $\textasciitilde$  Total $=6.015$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: There are $$20$$ students in a class standing in a line for recess. Mike is the $$7$$\textsuperscript{th} counting from front to back. How many students are behind Mike? What is his position counting backwards? 
Options:
aoVal: A
content: $$12$$, $$13$$ 
aoVal: B
content: $$13$$, $$14$$ 
aoVal: C
content: $$13$$, $$12$$ 
aoVal: D
content: $$14$$, $$13$$ 
Answer: B
Analysis: The number of students behind Mike plus the position of Mike equals the total number of students in the class. So there are $$13$$ students behind Mike: $$20-7=13$$. However Mike is not included in these $$13$$ students, which means Mike is the $$14$$\textsuperscript{th} student counting from back to front. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: Sophia's average score on six tests is $$82$$. Her average scores on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. Sophia then took another test and the average score of all $9$ tests is $87$? 
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->Counting Modules->Statistics and Probability
Problem: In an opaque bag, there are $5$ red balls, $5$ white balls, and $5$ yellow balls. The balls are the same except for their colours. Take out one ball from the bag. What is the probability that you take out a yellow ball? 
Options:
aoVal: A
content: $\dfrac{1}{2}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{1}{4}$ 
Answer: B
Analysis: $\dfrac{5}{15}=\dfrac{1}{3}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Maria had $$28$$ dreams last month. If $$16$$ of them involved monkeys, $$15$$ involved squirrels, and $$4$$ involved no animals, then at least how many dreams involved both monkeys and squirrels? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
Answer: B
Analysis: Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: $\frac{7}{9}$, $\frac{5}{4}$  Subtract the sum from the difference of the above two fractions. 
Options:
aoVal: A
content: $2\frac{1}{2}$ 
aoVal: B
content: $\frac{17}{36}$ 
aoVal: C
content: $\frac{73}{36}$ 
aoVal: D
content: $\frac{14}{9}$ 
Answer: D
Analysis: Sum $=\frac{5}{4}+\frac{7}{9}=\frac{73}{36}$  Difference $=\frac{5}{4}-\frac{7}{9}=\frac{17}{36}$  Difference (Sum and difference) $=\frac{73}{36}-\frac{17}{36}=\frac{56}{36}=\frac{14}{9}=1\frac{5}{9}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它