math-eval/TAL-SCQ5K huggingface dataset:
Problem: Mia has a date tomorrow. She has $2$ hats, $3$ dresses, and $3$ pairs of shoes. Now she wants to choose one dress and one pair of shoes. How many options does she have? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$18$$ 
Answer: C
Analysis: She does not need to choose a hat. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Amy picks some numbers as shown below:  $13, 20, 14, 15, 19, 20, 20, 19, 19, 15, 19, 19, 20, 13, 15$.  What is the difference between their mode and median? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$19$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: She picks two $13$s, one $14$, three $15$s, five $19$s, and four $20$s.  The mode is $19.$  The median is $19.$  Thus, their difference should be $0.$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: How many multiples of 6 are there between 14 and 100? 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$13$$ 
Answer: C
Analysis: 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
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 those that 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->Permutations and Combinations->Combinations
Problem: Betty and Abby are playing a game. They take turns writing numbers from $$1$$ to $$52 $$ on a blackboard. Each person can only write $$1$$, $$2$$, $$3$$ or $$4$$ numbers at a time, and each number can only be written once. The person who has no more numbers to write loses. Should Betty go first or second in order to win? 
Options:
aoVal: A
content: First 
aoVal: B
content: Second 
Answer: A
Analysis: $$52$$ is not a multiple of $$4+1$$, so the first player will win the game. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Ashley and Elvis are playing a game that requires them to drink a total of $$12$$ cups of coffee. They take turns drinking and each can drink either $$1$$ or $$2$$ cups at a time. The person who finishes the last cup of coffee wins this game. Should Elvis go first or second to ensure victory? 
Options:
aoVal: A
content: Go first 
aoVal: B
content: Go second 
Answer: B
Analysis: $$12$$ is a multiple of $$2+1$$, so, the second player should make the total number for each round to be $$3$$ to ensure victory. Therefore, Elvis should go second to ensure his vitory. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy
Problem: What is the smallest possible sum of two positive integers whose product is $$240$$? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$34$$ 
aoVal: E
content: $$38$$ 
Answer: B
Analysis: Since the product of the two positive integers is $$240$$, the possible pairs of integers are $$\left( 1,240 \right)$$, $$\left( 2,120 \right)$$, $$\left( 3,80 \right)$$, $$\left( 4,60 \right)$$, $$\left( 5,48 \right)$$, $$\left( 6,40 \right)$$, $$\left( 8,30 \right)$$, $$\left( 10,24 \right)$$, $$\left( 12,20 \right)$$ and $$\left( 15,16 \right)$$. The respective sums of these pairs are $$241$$, $$122$$, $$83$$, $$64$$, $$53$$, $$46$$, $$38$$, $$34$$, $$32$$ and $$31$$. Of these, the smallest value is $$31$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: SASMO 2014 P2 Q9  Two $1 coins and ten 50c coins are randomly distributed among 4 children such that each child receives the same number of coins. What is the difference between the biggest amount and the smallest amount a child can receive? 
Options:
aoVal: A
content: $$50c$$ 
aoVal: B
content: $1 
aoVal: C
content: $1.50 
aoVal: D
content: $2 
aoVal: E
content: None of the above 
Answer: B
Analysis: There are a total of 2+10 = 12 coins. so each child receives 12/4 = 3 coins.  Largest possible is $1 + $1 +50c= $2.50.  smallest possible is 50c + 50c + 50c= $1.50  $2.50 - $1.50 = $1 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
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: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$11$$ 
Answer: C
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->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: $6\times6$ =~\uline{~~~~~~~~~~}~groups of $4$. 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$36$$ 
Answer: B
Analysis: 6x6=36;  $$36\div4=9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
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. 
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 are $$17$$ balls in a bag. Each ball has a number from $$1$$ to $$17$$ on it. We randomly pick a ball from the bag. What is the smallest number of balls we have to pick in order to be sure that we have at least one pair of balls with a sum equal to $$18$$? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$17$$ 
Answer: C
Analysis: Among these numbers, there are $$8$$ pairs of numbers can get the sum of $$18$$($$1+17=2+16=3+15=4+14=5+13=6+12=7+11=$$$$8+10$$), and $$9$$ is useless. So in the worst case, after we choose $$9$$, we need $$8+1=9$$ more numbers to make sure a pair appears.  Thus, the answer is $$1+8+1=10$$.  Copyrighted material used with permission from Math Kangaroo in USA, NFP Inc.  ($$2005$$ Math Kangaroo Problem, Level $$9-10$$, Question $$ \textbackslash\# $$$$15$$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems
Problem: Pip multiplies all the digits in the number $$145$$ to get $$20$$. What's the biggest three digit number whose digits multiply together to give $$30$$?~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$999$$ 
aoVal: B
content: $$922$$ 
aoVal: C
content: $$651$$ 
aoVal: D
content: $$532$$ 
Answer: C
Analysis: $$651$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations->Basic Operations of Combinations
Problem: Vicky is ordering lunch at a fast food restaurant that has sandwiches and burgers on the lunch menu, along with coffee, milk, and tea as drink options. If Vicky chooses one food item and one drink item from the lunch menu, she has~\uline{~~~~~~~~~~}~different ways to order lunch. 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: $$2\times 3=6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In the following events, the certain event is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: Ame will get $$100$$ points in the final math exam. 
aoVal: B
content: Claire will eat breakfast tomorrow. 
aoVal: C
content: The news is on when you open the TV. 
aoVal: D
content: There are two red balls and one white ball in a bag. If two balls are taken out, one must be red. 
Answer: D
Analysis: $$\text{A}$$. "Ame will get $100$ points in the final math exam" is a random event.  $$\text{B}$$. "Claire will eat breakfast tomorrow." is a random event.  $$\text{C}$$. "The news is on when you open the TV" is a random event.  $$\text{D}$$. There are two red balls and one white ball in a bag. If two balls are taken out, one must be red. It is a certain event.  So $$\text{D}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: In a two-digit number, the digit in the ones place is 3. The digit in the tens place is 3 more than the digit in the ones place. What is the number? 
Options:
aoVal: A
content: $$63$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$33$$ 
aoVal: D
content: $$66$$ 
Answer: A
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: Two tiles numbered $1$ and $2$ 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 $10$? 
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 or equal to $10$ will be $2\times5$ and $2\times6$. Thus, the probability is $\frac 2{12}$ = $\frac 16$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: The Dragonvale Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: A
Analysis: Boy, Girl, Girl, Girl, Boy.There are $$2\times 1=2$$ ways of arranging two boys; there are $$3\times 2\times 1=6\textasciitilde$$ways of arranging $3$ girls. Therefore, there are $$2\times 6=12$$ ways of arranging all students. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Martina and Linda had $$68$$ seashells.  After Martina gave $$8$$ seashells to Linda, they had an equal amount of seashells.  How many seashells did Martina have at first? ． 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$42$$ 
Answer: D
Analysis: $$ 68\div 2=34$$, $$34+8=42$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $$60$$ pupils in Arnold\textquotesingle s class. $$35$$ pupils like Maths and $$38$$ pupils like English. Twice as many pupils like both subjects as those that like neither of them. How many pupils like only Maths? 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$9$$ 
Answer: D
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 $$35−2x$$ and the number who like only English is $$38−2x$$.  Since there are $$60$$ pupils in Arnold\textquotesingle s class, we have:  $$\left( 35-2x \right)+2x+\left( 38-2x \right)+x=60$$ and hence $$x = 13$$ and hence the number of pupils who like only Maths is $$35-(2\times 13)=9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable
Problem: Julia, Kasia, Zuzanna, 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 Zuzanna were born in the same month. Julia and Zuzanna 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: Zuzanna 
aoVal: D
content: Helena 
Answer: D
Analysis: Given that Kasia and Zuzanna were born in the same month, their birth month must be March. Given that Julia and Zuzanna were born on the same day of a month, they must be born on the $$20^{\rm th}$$. Hence, Zuzanna was born on March $$20$$; Kasia was born on March $$1$$; and Julia was born on July $$20$$. Helena is therefore the one born on May $$17$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: A box contains blue marbles. Another two boxes contain only white marbles.  Label on Box $$\rm A$$: white marbles  Label on Box $$\rm B$$: blue marbles  Label on Box $$\rm C$$: Box $$\rm B$$ contains blue marbles  Which box contains blue marbles if two of the above labels are wrong?． 
Options:
aoVal: A
content: Box A 
aoVal: B
content: Box B 
aoVal: C
content: Box C 
Answer: C
Analysis: If two of the labels are wrong, \textbf{one label is correct.}  If the label on Box A is correct, Box A contains white marbles.  The label on Box B is wrong, so Box B contains white marbles.  The label on Box C is wrong, so Box B contains white marbles.  Therefore, Box A and Box B contain white marbles, while \textbf{Box C} contains blue marbles. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: Three kids line up to play games. How many different ways can they form the line? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: There are six different ways for three kids to line up. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: An elementary school is arranging students\textquotesingle{} schedules. For the first class in the morning, it has to be Chinese or Maths or English. Students won\textquotesingle t start the day with the same class for two days in a row. If Wednesday starts with Chinese, then Thursday can start with~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: Chinese or Maths 
aoVal: B
content: Maths or English 
aoVal: C
content: English or Chinese 
Answer: B
Analysis: Students won\textquotesingle t start the day with the same class for two days in a row. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Tree Diagrams
Problem: Granny has $$10$$ grandchildren. Alice is the oldest. One day, Granny notices that her grandchildren all have different ages. If the sum of her grandchildren\textquotesingle s ages is $$180$$, what is the youngest age that Alice can be． 
Options:
aoVal: A
content: $$19$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$21$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$23$$ 
Answer: E
Analysis: Try to consider the problem of the youngest Alice in such way: make each grandchild has a similar age as possible, and the age of each grandchild should be different, that is, $$1 + 2 + 3 +\cdots 9 + 10 = 55$$; then $$180 - 55 = 125$$, $$125 \div 10 = 12\cdots\cdots5$$, and $$5$$ is left. If every child\textquotesingle s age is added by $$12$$, then $$5$$ is left. Give the extra year to each of the $$5$$ children who are the oldest. In doing so, the minimum number of Alice\textquotesingle s age is $$10+12+1=23$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
Problem: $$\dfrac{1}{2}$$ of the pens in a box are red, $$\dfrac{1}{4}$$ of the remainder are blue and the rest are green. What percentage of the pens in the box are green? 
Options:
aoVal: A
content: $$12.5\textbackslash\%$$ 
aoVal: B
content: $$25\textbackslash\%$$ 
aoVal: C
content: $$37.5\textbackslash\%$$ 
aoVal: D
content: $$75\textbackslash\%$$ 
Answer: C
Analysis: Remainder $=1-\frac{1}{2}=\frac{1}{2}$  Green $=1-\frac{1}{4}=\frac{3}{4}$ of remainder  $=\frac{3}{4}\times\frac{1}{2}$ of total  $=\frac{3}{8}$ of total  $=\frac{3}{8}\times100\textbackslash\%=37.5\textbackslash\%$ of total 
Knowledge points: Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding the Percentage Given a Part and a Whole
Overseas Competition->Knowledge Point->Counting Modules
Problem: Using the digits $$1$$, $$3$$, and $$9$$, we can form~\uline{~~~~~~~~~~}~different 3-digit numbers without repeating digits. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: D
Analysis: Three-digit:$$139$$, $$193$$, $$319$$, $$391$$ , $$913$$, $$931$$ for a total of $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Four students asked their teacher, Mr Carter, to line up with them to take a picture.  ① If Mr Carter does not want to stand on either end, how many different ways can they line up for the picture?  ② If Mr Carter insists on standing on one of the $2$ ends, how many different ways can they line up for the picture? 
Options:
aoVal: A
content: $72$ , $24$ 
aoVal: B
content: $96$ , $24$ 
aoVal: C
content: $72$ , $48$ 
aoVal: D
content: $96$ , $48$ 
Answer: C
Analysis: ①$$3\times 4\times 3\times 2\times 1=72$$ ,  ②$$2\times 4\times 3\times 2\times 1=48$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: When Pinocchio lies, his nose gets 6 cm longer. When he tells the truth, his nose gets 2 cm shorter. When his nose was 9cm long, he told three lies and made twotrue statments. How long was Pinnochio\textquotesingle s nose afterwards? 
Options:
aoVal: A
content: $$14$$cm 
aoVal: B
content: $$15$$cm 
aoVal: C
content: $$19$$cm 
aoVal: D
content: $$23$$cm 
aoVal: E
content: $$31$$cm 
Answer: D
Analysis: 9 + (3 x 6) - (2 x 2) = 23cm. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Four boys each prepared $1$ gift for a party. How many ways can these $4$ gifts be \textsubscript{} distributed among the $4$ kids so that no one receives his own gift? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$1$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: Let the boys be A, B,~C and D and their gifts be 1, 2, 3 and 4. List down all possible arrangements:  (A2, B1, C4, D3), (A2, B3, C4, D1), (A2, B4, C1, D3), (A3, B1, C4, D2), (A3, B4, C2, D1), (A3, B4, C1, D2), (A4, B1, C2, D3), (A4, B3, C2, D1), (A4, B3, C1, D2)  There are \textbf{9} possible arrangements. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration
Source: 其它
Problem: What is the sum of all the digits in the number $2023$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: D
Analysis: $2+0+2+3=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: The correct information James can get from the statement ``there is a $11\textbackslash\%$ chance that it will rain tomorrow in Los Angeles'' is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: It will rain $11\textbackslash\%$ of the time tomorrow in Los Angeles. 
aoVal: B
content: It will rain in $11\textbackslash\%$ of the regions in Los Angeles tomorrow. 
aoVal: C
content: It will definitely rain tomorrow in Los Angeles. 
aoVal: D
content: The probability of raining in Los Angeles is high. 
aoVal: E
content: The probability of raining in Los Angeles is low. 
Answer: E
Analysis: It is less likely to rain tomorrow in Los Angeles. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Ron\textquotesingle s work day is half over at $$1:15$$ P.M. if he starts work at $$9:30$$ A.M., his work day ends at． 
Options:
aoVal: A
content: $$4:30$$ P.M. 
aoVal: B
content: $$5:00$$ P.M. 
aoVal: C
content: $$5:30$$ P.M. 
aoVal: D
content: $$9:30$$ P.M. 
Answer: B
Analysis: From $$9:30$$ A.M. until $$1:15$$ P.M. is $$3$$ hours and $$45$$ minutes. Add $$3$$ hours and $$45$$ minutes to $$1:15$$ P.M. to get $$5:00$$ P.M.. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: What time is $$11999$$ hours after $$3$$ P.M.? 
Options:
aoVal: A
content: $$2$$ A.M. 
aoVal: B
content: $$2$$ P.M. 
aoVal: C
content: $$4$$ A.M. 
aoVal: D
content: $$4$$ P.M. 
Answer: B
Analysis: $$12000$$ hours after $$3$$ P.M. is $$3$$ P.M. $$11999$$ hours after $$3$$ P.M is $$2$$ P.M. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Moon and Archie played chess competitively. Both of them are of the same level in terms of skill. They agreed to play in a best-of-seven games, where the one who wins four games first would be the ultimate winner. They have already played three games, with Moon winning two games and Archie winning just one game. What is the probability that Moon will be the ultimate winner? 
Options:
aoVal: A
content: $$\frac{3}{8}$$ 
aoVal: B
content: $$\frac{11}{16}$$ 
aoVal: C
content: $$\frac{3}{16}$$ 
aoVal: D
content: $$\frac{7}{16}$$ 
Answer: B
Analysis: $$\rm B$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability