math-eval/TAL-SCQ5K huggingface dataset:
Problem: A meal was priced at $\textbackslash$80$. The shop offered Jane the meal at a $40\textbackslash\%$ discount. After service charges of $10\textbackslash\%$, how much did Jane pay for her meal? 
Options:
aoVal: A
content: $\textbackslash$48$ 
aoVal: B
content: $\textbackslash$88$ 
aoVal: C
content: $\textbackslash$35.20$ 
aoVal: D
content: $\textbackslash$52.80$ 
Answer: D
Analysis: Price after discount $=\textbackslash$80\times(100\textbackslash\%-40\textbackslash\%)=\textbackslash$48$  Price after service charges $=\textbackslash$48\times110\textbackslash\%=\textbackslash$52.80$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: Amy, Bill and Celine are friends with different ages. Only 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. 
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 Bill is the youngest. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Express $$\frac{7}{8}$$ as a decimal. 
Options:
aoVal: A
content: $$0.725$$ 
aoVal: B
content: $$0.785$$ 
aoVal: C
content: $$0.825$$ 
aoVal: D
content: $$0.875$$ 
Answer: D
Analysis: Perform long division.  $7\div8=0.875$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Overseas In-curriculum->Knowledge Point->Knowing Numbers->Decimals->Converting Between Fractions and Decimals->Converting Fractions to Decimals
Problem: How many minutes is it from $$22:45$$ today to $$00:35$$ tomorrow? 
Options:
aoVal: A
content: $$90 $$ 
aoVal: B
content: $$ 100 $$ 
aoVal: C
content: $$ 110 $$ 
aoVal: D
content: $$120 $$ 
aoVal: E
content: $$130$$ 
Answer: C
Analysis: It is $$75$$ minutes from $$22:45$$ to midnight and then another $$35$$ minutes from midnight until $$00:35$$. So the required number of minutes is $$75 + 35 = 110$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many of the following statements are correct?  Statement $1$: The probability of a certain event to happen is $$1$$.  Statement $2$: Indefinite events include impossible events.  Statement $3$: The probability of an impossible event to happen is $$0$$.  Statement $4$: The probability of an indefinite event to happen is between $$0$$ and $$1$$. 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: Impossible events are definite events.  $$\text{Statement 2}$$ is wrong.  $$\text{Statement 1}$$, $$\text{Statement 3}$$, and $$\text{Statement 4}$$ are right.  Thus, the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $$38$$ students in the class. Each student is in either math club or music club or both. Given that among students in this class, $$26$$ students are in the math club, and $$18$$ students are in the music club, there are~\uline{~~~~~~~~~~}~students in the math club only and not in the music club. 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$21$$ 
Answer: C
Analysis: $$26+18-38=6$$, $$26-6=20$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: What is the sum of the smallest $4$-digit number and the largest $1$-digit number? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$109$$ 
aoVal: C
content: $$999$$ 
aoVal: D
content: $$1009$$ 
aoVal: E
content: $$9999$$ 
Answer: D
Analysis: $9+1000=1009$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: There are some pieces of candy on a table. You are challenged by your friend to play the following game: The two players take turns taking some candy. Every turn, you can take away either $$1$$, $$2$$, $$3$$, $$4$$ or $$5$$ pieces from the table. The person who takes away the final piece of candy from the table wins. If you go second, how many pieces of candy should be on the table before the game starts such that you can ensure victory? 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$21$$ 
Answer: C
Analysis: Only $$18$$ is one of the multiples of $$5+1$$, and the second player can ensure victory. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many of the following statements are current?  Statement $1$: The probability of an indefinite event to happen is between $$0$$ and $$1$$.  Statement $2$: The probability of an impossible event to happen is $$0$$.  Statement $3$: The probability of a certain event to happen is $$1$$.  Statement $4$: Indefinite events include impossible events. 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: Impossible events are definite events.  $$\text{Statement 4}$$ is wrong.  $$\text{Statement 1}$$, $$\text{Statement 2}$$, and $$\text{Statement 3}$$ are right.  Thus, the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Between them, the two four-digit integers $$M$$ and $$N$$ contain all ten digits from $$1$$ to $$8$$.  What is the least possible difference between $$M$$ and $$N$$? 
Options:
aoVal: A
content: $$123$$ 
aoVal: B
content: $$247$$ 
aoVal: C
content: $$427$$ 
aoVal: D
content: $$472$$ 
aoVal: E
content: $$742$$ 
Answer: B
Analysis: The first digits of the two numbers will need to be as close as possible to each other. Since they cannot be equal, they will have to differ by $$1$$; say they are $$n$$ and $$n + 1$$. The difference between the two numbers will then be minimised by making the four digits after $$n + 1$$ as small as possible and the four digits after $$n$$ as large as possible. The smallest four-digit number available is $$123$$ and the largest is $$876$$. So we need to make $$n= 4$$ and then the required diference is $$5123 - 4876 = 247$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value in Enumeration Problems
Problem: Eddie reads at least one chapter of a book with $$16$$ chapters each day. If he is asked to read a different number of chapters each day, this book can be finished reading for at most~\uline{~~~~~~~~~~}~days. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$16$$ 
Answer: B
Analysis: $$16=1+2+3+4+6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
Problem: A chess singles tournament had $10$ players. Each player played with every other player only once. $2$ points are earned for winning a game, $0$ points are earned for losing a game, and $1$ point is earned by each player in a tie. After the tournament, the judge finds that the sum of these $10$ players\textquotesingle{} points is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$180$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$90$$ 
Answer: E
Analysis: There are $$10$$ players, so there are $$10\times9\div2=45$$ games in total. The sum of the points of two players in each game must be $2$. Thus, the sum of all points earned by $10$ players in $45$ games is $2\times45=90$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Hilla can make $9$ burgers every hour. In a day, she makes burgers for $6$ hours. How many burgers can she make? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$72$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$54$$ 
Answer: D
Analysis: $6\times9=54$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Only $$1$$ of the $$3$$ boys Abel, Ben and Charles can swim. Abel says, "I can swim." Ben says, "I cannot swim." Charles says, "Abel cannot swim." Only $$1$$ boy is telling the truth. Who can swim? 
Options:
aoVal: A
content: Abel 
aoVal: B
content: Ben 
aoVal: C
content: Charles 
aoVal: D
content: more information needed 
Answer: B
Analysis: Only $$1$$ of the $$3$$ boys can swim. Only $$1$$ of the $$3$$ boys is telling the truth!  Since Abel and Cain contradict each other, there must be one who is telling the truth!  If Abel is true, Ben is lying and Ben can swim. Then we have $$2$$ boys (Abel and Ben)  who can swim. Contradiction.  Hence, Cain is true. Abel is lying and so is Ben. Then, Ben can swim. Abel cannot swim and we are not sure if Cain can swim. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Abby, Bret, Carl, and Dana are sitting in a row of four seats numbered $$1$$ to $$4$$. Joe looks at them and says:  "Bret is next to Carl."  "Abby is between Bret and Carl."  However, all of Joe\textquotesingle s statements are false. Bret is actually sitting in the seat $$3$$. Who is sitting in seat the $$2$$? ． 
Options:
aoVal: A
content: Abby 
aoVal: B
content: Bret 
aoVal: C
content: Carl 
aoVal: D
content: Dana 
Answer: D
Analysis: We know that Carl does not sit next to Bret, so he must sit in seat $$1$$. Since Abby is not between Bret and Carl, she must sit in seat $$4$$. Finally, Dana has to take the last seat available, which is $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In numbers 30 to 90, there are some two-digit numbers in which the sum of the tens digit and the ones digit is 13. How many of such numbers are there?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$5$$ 
Answer: D
Analysis: 49, 58, 67, 76, 85 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn? 
Options:
aoVal: A
content: $$\frac{3}{10}$$ 
aoVal: B
content: $$\frac{2}{5}$$ 
aoVal: C
content: $$\frac{1}{2}$$ 
aoVal: D
content: $$\frac{3}{5}$$ 
aoVal: E
content: $$\frac{7}{10}$$ 
Answer: B
Analysis: There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$ red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the $10$ ways to end the game, $4$ of them ends with a green chip. The answer is $\frac{4}{10}=\frac{2}{5}$, or $(\mathbf{B}) \frac{2}{5}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability
Source: 其它
Problem: Ali, Barb, and Cal were all born on April $$1$$, in different years.  This coming April $$1$$, if I add all their ages together, I\textquotesingle ll get $$9$$.  On that day, Ali\textquotesingle s age could not be． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$1$$ 
Answer: A
Analysis: Ali, Barb, and Cal \emph{were} all born on April $$1$$, in \emph{different} years.  This coming Apr $$1$$, if I add all their ages, I\textquotesingle ll get $$9$$. Since the youngest possible ages are $$1$$ and $$2$$, the oldest possible age is $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: There are $$10$$ boys in Pat\textquotesingle s math class. If there are twice as many girls as boys in the class, how many girls are there in the class? 
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$20$$ 
Answer: D
Analysis: girls = $10\times2=20$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: There are three boxes labeled $$A$$, $$B$$, and $$C$$. Box $$A$$ contains $$1$$ red ball and $$4$$ white balls. Box $$B$$ contains $$2$$ red balls and $$3$$ white balls. Box $$C$$ contains $$3$$ red balls. All the balls are the same except for the color. Oscar randomly selects one box and then randomly selects one ball from the selected box. What is the probability that he selects a red ball? 
Options:
aoVal: A
content: $$\frac{6}{13}$$ 
aoVal: B
content: $$\frac{2}{5}$$ 
aoVal: C
content: $$\frac{8}{13}$$ 
aoVal: D
content: $$\frac{8}{15}$$ 
aoVal: E
content: $$\frac{11}{15}$$ 
Answer: D
Analysis: $$\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{2}{5}+\frac{1}{3}=\frac{8}{15}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Matt had to deliver flyers to all houses numbered from 25 to 57. How many houses got th flyers? 
Options:
aoVal: A
content: $$31$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$33$$ 
aoVal: D
content: $$34$$ 
aoVal: E
content: $$35$$ 
Answer: C
Analysis: L - F +1 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Which of the following expression doesn\textquotesingle t mean the sum of 6 + 6 + 6 + 6 + 6? 
Options:
aoVal: A
content: $$6+6+6+6+6$$ 
aoVal: B
content: $$5+6$$ 
aoVal: C
content: $$5\times 6$$ 
Answer: B
Analysis: By the meaning of multiplication, $$5\times 6$$ means the sum of five $$6$$\textquotesingle s. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: How many thousands of seconds are there in $$365$$ days? 
Options:
aoVal: A
content: $$31536$$ 
aoVal: B
content: $$525600$$ 
aoVal: C
content: $$1892160$$ 
aoVal: D
content: $$3536000$$ 
Answer: A
Analysis: $$365$$ days have $$365\times24\times60\times60\div1000=31536$$ thousand seconds． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: There is a box contains $7$ chips numbered $1$, $2$, $3$, $4$, $5$, $6$, and $7$. A chip is drawn randomly from the box. What is the probability that the number on the chip is an even number? (adapted from 2015 AMC 8 Problem, Question \#7) 
Options:
aoVal: A
content: $\frac17$ 
aoVal: B
content: $\frac37$ 
aoVal: C
content: $\frac57$ 
aoVal: D
content: $\frac47$ 
aoVal: E
content: $\frac67$ 
Answer: B
Analysis: $2$, $4$, and $6$ are even numbers. Thus, the probability is $\frac37$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely? 
Options:
aoVal: A
content: all $4$ are boys 
aoVal: B
content: all $4$ are girls 
aoVal: C
content: $2$ are girls and $2$ are boys 
aoVal: D
content: $3$ are of one gender and $1$ is of the other gender 
aoVal: E
content: all of these outcomes are equally likely 
Answer: D
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Which of the following statements is a certain event? 
Options:
aoVal: A
content: $3$ is a factor of $98765$. 
aoVal: B
content: The sum of the interior angles of a triangle is $180$°. 
aoVal: C
content: The age difference between Michael and Candy will increase in $10$ years. 
aoVal: D
content: Choose a $3$-digit number and it can be divided by $9$ without any remainder. 
Answer: B
Analysis: $A$ is an impossible event. $C$ is an impossible event. $D$ is a random event. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Eddie reads at least one chapter of a $$15-$$chapter book each day. If he is asked to read a different number of chapters each day, this book can be read for at most~\uline{~~~~~~~~~~}~days. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
Answer: B
Analysis: $$15=1+2+3+4+5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
Problem: Seven girls and three boys are standing in a line randomly. What is the probability that three boys are next to each other? 
Options:
aoVal: A
content: $\dfrac{1}{15}$ 
aoVal: B
content: $\dfrac{2}{15}$ 
aoVal: C
content: $\dfrac{1}{5}$ 
aoVal: D
content: $\dfrac{4}{15}$ 
aoVal: E
content: $\dfrac{1}{3}$ 
Answer: A
Analysis: $\dfrac{A\_8^{8}\times A\_3^{3}}{A\_{10}^{10}}=\dfrac{6}{90}=\dfrac{1}{15}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: What is the value of the following sum?  $$902+804+700+609+508+403+307+201+106$$ 
Options:
aoVal: A
content: $$4450$$ 
aoVal: B
content: $$4540$$ 
aoVal: C
content: $$4500$$ 
aoVal: D
content: $$4505$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: simple math calculation 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: A candy store sells some assorted candies. Each bag of assorted candies is made up of $2$ kilograms of toffee and $3$ kilograms of fruit drops. The price of toffee is $\textbackslash$6$ per kg, and the price of fruit drops is $\textbackslash$1$ per kg. How much is the cost of the assorted candies per kilogram? 
Options:
aoVal: A
content: $$\textbackslash$2$$ 
aoVal: B
content: $$\textbackslash$3$$ 
aoVal: C
content: $$\textbackslash$5$$ 
aoVal: D
content: $$\textbackslash$7$$ 
Answer: B
Analysis: The assorted candy weighs $2+3=5$ kg and is priced at $2\times6+3\times1=\textbackslash$15$ in total. Therefore, each kilogram of the assorted candy is priced at $15\div5=3$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: How many positive integers from $1$ to $100$ do not have $2,3$ or $5$ as its factors? 
Options:
aoVal: A
content: $$23$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: $$30$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: Elson finished reading a storybook last week. He read an average of $19$ pages per day for the first four days of the week. He read an average of $25$ pages per day from Friday to Saturday. He did not read on Sunday. How many pages on average did Elson read per day throughout the entire week? 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$17$$ 
Answer: D
Analysis: $(19\times4+25\times2)\div7=18$ pages. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $6$ more red fish than yellow fish in Joann\textquotesingle s aquarium. Joann buys $8$ new red fish and $3$ new yellow fish. How many more red fish than yellow fish are in the aquarium now? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$14$$ 
Answer: C
Analysis: $8-3=5$  $6+5=11$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Three children have $$10$$ balloons in total, and each of them has a different number of balloons. Joann has the most balloons, and at least how many balloons should she have? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $10=1+2+7=1+3+6=1+4+5=2+3+5$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Source: 其它
Problem: Molly, Dolly, Sally, Elly and Kelly are sitting on a park bench. Molly is not sitting on the far right and Dolly is not sitting on the far left. Sally is not sitting at either end. Kelly is not sitting next to Sally and Sally is not sitting next to Dolly. Elly is sitting to the right of Dolly but not necessarily next to her. Who is sitting at the far right end? 
Options:
aoVal: A
content: Molly　 
aoVal: B
content: Dolly　 
aoVal: C
content: Sally　 
aoVal: D
content: Kelly　 
aoVal: E
content: Elly　 
Answer: E
Analysis: The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions $$2$$, $$3$$ and $$4$$ from the left-hand end. Were Sally to sit in place $$2$$, neither Dolly nor Kelly could sit in places $$1$$ or $$3$$ as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places $$4$$ and $$5$$ which is impossible. Similarly, were Sally to sit in place $$3$$, Dolly could not sit in place $$2$$ or $$4$$ and the question also tells us she cannot sit in place $$1$$ so Dolly would have to sit in place $$5$$ making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place $$4$$, Dolly could sit in place $$2$$, Kelly in place $$1$$, Molly (who cannot sit in place $$5$$) in place $$3$$ leaving Elly to sit in place $$5$$ at the right-hand end. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: There are $$20$$ red balls, $$2$$ black balls, and 1 white ball in a cloth bag. They are of identical shape, size and quality except for color. Take out 1 ball at random. Among the following statements, which one is true? 
Options:
aoVal: A
content: The probability of taking out a black ball is the smallest. 
aoVal: B
content: It\textquotesingle s impossible to take out a white ball. 
aoVal: C
content: The probability of taking out a red ball is larger. 
aoVal: D
content: A red ball is surely to be taken out. 
Answer: C
Analysis: The number of red balls in the bag is the most compared with other colored balls. If we take out one ball at random, the probability of taking out a red ball is larger. So $$\text{C}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability