math-eval/TAL-SCQ5K huggingface dataset:
Problem: Joshua took a $$3$$ hours $$20$$ minutes train ride from Town $$X$$ to Town $$Y$$. The train departed at $$7:55$$ am, but stopped for $30$ minutes due to heavy rain during his trip. When did Joshua arrive in Town $$Y$$? (2009 Math kangaroo Problems, Level 3-4 , Question \#8) 
Options:
aoVal: A
content: $11:15$ 
aoVal: B
content: $10:45$ 
aoVal: C
content: $10:55$ 
aoVal: D
content: $11:45$ 
aoVal: E
content: $10:35$ 
Answer: D
Analysis: $7:55+ 3$ hours $$20$$ minutes=$11:15$  $11:15$+$30$ minutes=$11:45$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: A whole number has two digits. The product of the digits of this number is 15. The sum of digits of this number is: 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: 3 + 5 = 8 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
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: $\dfrac{1}{4}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{3}{8}$ 
aoVal: D
content: $\dfrac{1}{2}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: The probability of both show a green bean is $\dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{1}{8}$. The probability of both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore, the probability is $\dfrac{1}{4}+\dfrac{1}{8}=\boxed {\left (\text{C}\right )\dfrac{3}{8}}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Calculate:  $$230\times9 =$$． 
Options:
aoVal: A
content: $$2070$$ 
aoVal: B
content: $$1980$$ 
aoVal: C
content: $$2130$$ 
aoVal: D
content: $$2240$$ 
Answer: A
Analysis: $$230$$x(10-1) =230x10-230x1 =2300-230 =2070 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: How many whole numbers between $$5000$$ and $$6000$$ consist of four different digits that decrease from left to right? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$69$$ 
aoVal: D
content: $$120$$ 
Answer: B
Analysis: The only such numbers are $$5432$$, $$5431$$, $$5430$$, $$5421$$, $$5420$$, $$5410$$, $$5321$$, $$5320$$, $$5310$$, and $$5210$$. In all, there are $$10$$ such numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Finding the Shortest Path by Number Notation->Finding the Shortest Path by Number Notation (with specific points or areas)
Problem: $$151$$ traffic ambassadors lined up neatly in a straight line (at equal distances from each other) on a highway to promote traffic laws. After completing their task, at which position of the highway should they assemble in order to minimise their total walking distance from their respective positions to the assembly area? ． 
Options:
aoVal: A
content: $$75^{}\text{th}$$ traffic ambassador\textquotesingle s position 
aoVal: B
content: $$76^{}\text{th}$$ traffic ambassador\textquotesingle s position 
aoVal: C
content: Between the $$75^{}\text{th}$$ and $$76^{}\text{th}$$ traffic ambassadors\textquotesingle~positions 
aoVal: D
content: Any position on the highway 
Answer: B
Analysis: There are an odd number of positions, the assembly area should be in the mid-point, $$\left( 151+1 \right)\div 2=76$$, that is, the $$76^{}\text{th}$$ traffic ambassador\textquotesingle s position 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: One morning, a rabbit, a dog, a cat, and a duck went looking for food outside.  The rabbit says: "If I get food, the dog will also get food."  The dog says: "If I get food, the cat will also get food."  The cat says: "If I get food, the duck will also get food."  In the evening, they find that all of them tell the truth but only two of them get food.~\uline{~~~~~~~~~~}~and~\uline{~~~~~~~~~~}~don\textquotesingle t get food. 
Options:
aoVal: A
content: The rabbit; the dog 
aoVal: B
content: The dog; the cat 
aoVal: C
content: The cat; the duck 
aoVal: D
content: The cat; the rabbit 
Answer: A
Analysis: We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit and the dog don\textquotesingle t get food, the cat and the duck would get food. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Sam\textquotesingle s exam was meant to begin at $$13:45$$ but started $$17$$ minutes late. At what time did it begin?　 
Options:
aoVal: A
content: $$13:28$$ 
aoVal: B
content: $$13:38$$ 
aoVal: C
content: $$13:56$$ 
aoVal: D
content: $$14:02$$ 
aoVal: E
content: $$14:06$$ 
Answer: D
Analysis: Seventeen minutes after $$13:45$$ is $$14:02$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: There are $$24$$ four-digit numbers which is formed using each of the digits $$3$$, $$4$$, $$5$$ and $$6$$ once only. When all of these $$24$$ four-digit numbers are put in order from smallest to largest, which one is in the seventh position? 
Options:
aoVal: A
content: $$3546$$ 
aoVal: B
content: $$3645$$ 
aoVal: C
content: $$4356$$ 
aoVal: D
content: $$4536$$ 
aoVal: E
content: $$5346$$ 
Answer: C
Analysis: $3456, 3465, 3546, 3564, 3546, 3564$  $\underline{4356}, 4365, 4536, 4563, 4635, 4653$  $5346, 5364, 5436, 5463, 5634, 5643$  $6345, 6354, 6435, 6453, 6534, 6543$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: Jack enters the classroom and sees all the $6$ classmates in the classroom. How many students are there in the classroom now? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: It cannot be determined. 
Answer: B
Analysis: $$6+1=7$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: In how many different ways can six identical coins be distributed among Al, Bo, and Carl so that each gets at least $$1$$ oin? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: The only possible distributions (Al, Bo, Carl) are these ten: $$(1,1,4)$$, $$(1,2,3)$$, $$(1,3,2)$$, $$(1,4,1)$$, $$(2,1,3)$$, $$(2,2,2)$$, $$(2,3,1)$$, $$(3,1,2)$$, $$(3,2,1)$$, and $$(4,1,1)$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: If $$a⊕b= \frac{1}{ \dfrac{1}{a}+ \dfrac{1}{b}}$$, then what is the value of $$ (1\times2)⊕(2\times3)⊕(3 \times4)⊕\ldots⊕(2013\times2014)$$? 
Options:
aoVal: A
content: $$\frac {2013}{2014}$$ 
aoVal: B
content: $$\frac{2014}{2013}$$ 
aoVal: C
content: $$\frac{2014}{2015}$$ 
aoVal: D
content: $$\frac{2015}{2014}$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: Notice that $$x\_{1}⊕x\_{2}⊕\ldots⊕x\_{n}={{(\sum\nolimits\_{i=1}^{n}{x\_{i}^{-1}})}^{-1}}$$,  $$\sum\nolimits\_{i=1}^{n}{\frac{1}{i(i+1)}}=\sum\nolimits\_{i=1}^{n}{(\frac{1}{1}}-\frac{1}{i+1})=1-\frac{1}{n+1}=\frac{1}{n+1}$$ and so  $$(1\times 2)\oplus (2\times 3)\oplus (3\times 4)\oplus \cdots \oplus (2013\times 2014)={{(\frac{2013}{2014})}^{-1}}=\frac{2014}{2013}$$,  This is the reason behind the pattern. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock?~ ~ ．(Only consider angles less than 180^{}\circ) 
Options:
aoVal: A
content: $50^{}\circ $ 
aoVal: B
content: $120^{}\circ $ 
aoVal: C
content: $135^{}\circ $ 
aoVal: D
content: $150^{}\circ $ 
aoVal: E
content: $165^{}\circ $ 
Answer: D
Analysis: The smaller angle is $\frac 5{12}$ of a full circle.  A full circle has $360$ degrees, so the angle is $\frac 5{12}\times 360^{}\circ =150^{}\circ $.  So, the answer is $\rm D$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock
Problem: If we throw two $6-$sided dice of the same shape and size, which statements among the following options is an impossible event? 
Options:
aoVal: A
content: The sum of dots is $$12$$. 
aoVal: B
content: The product is a prime number. 
aoVal: C
content: The sum of dots is larger than $$1$$ but smaller than $$3$$. 
aoVal: D
content: The product of is $40$. 
Answer: D
Analysis: So $$\text{D}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: The time twelve thousand and twelve hours after $$7$$ A.M. is． 
Options:
aoVal: A
content: $$1$$ A.M. 
aoVal: B
content: $$7$$ P.M. 
aoVal: C
content: $$7$$ A.M. 
aoVal: D
content: $$7$$ P.M. 
Answer: D
Analysis: Divide $$12012\div24$$ to get remainder $$12$$. $$12$$ hours after $$7$$ A.M. is $$7$$ P.M. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: There are $$18$$ 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: $$9$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$54$$ 
Answer: C
Analysis: girls = $18\times2=36$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: Kevin has 3 regular dice. Each dice has numbers from 1 to 6. Which of the following could not be the sum of the numbers on top of the 3 dice? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: All the above numbers are possible sum 
Answer: D
Analysis: maximum is 6+6+6 = 18. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Veronica rolls three six-sided dice. The product of the three numbers is $$90$$.  What is the sum of the three numbers? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: more information is needed 
Answer: B
Analysis: Given that the numbers on a die are $$1$$, $$2$$, $$3$$, $$4$$, $$5$$ and $$6$$, to achieve a product of $$90$$, one of the numbers must be $$5$$ and the other two must have a product of $$18$$. Thus the numbers can only be $$3$$, $$5$$ and $$6$$ , whose sum is $$14$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Mike and Sara are looking for a place to eat lunch. They know there are nine Chinese restaurants, three Mexican restaurants, and two fast food restaurants nearby. How many different choices do they have in total to eat one meal? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$54$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$14$$ 
Answer: E
Analysis: They can only choose one place, so it can only be Chinese, Mexican, or fast food. Therefore, we can add each one up to get $$9+3+2 = 14$$.  ~ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Problem: Mary had equal number of green, yellow and red tokens. She used some of those tokens to make a pile. You can see all used tokens in the figure. She still has five tokens which are not on the pile. How many yellow tokens did she have have at the beginning?  insert pic 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$18$$ 
Answer: B
Analysis: The used tokens are 5 red tokens at the bottom, 4 green tokens and 4 yellow tokens. Now we distribute the unused 5 tokens to the 3 colors such that each color has the same number of tokens.  Firstly, we need 1 more green token and 1 more yellow token to make all colors equal in number of tokens.  We are left with 3 tokens, and we will equally distribute these tokens to the colors, i.e. each color will have 1 more token.  Hence, the unused tokens are 1 red, 2 green and 2 yellow. There are 4+2 = 6 yellow tokens in total. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Counting Solid Figures
Source: 其它
Problem: Josip has 4 toys - a car, a doll, a ball and a ship. He wants to put them on a line on a shelf. The ship has to be next to the car and the doll has to be next to the car. In how many ways can he arrange them so all the conditions would be fulfilled? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: B
Analysis: The car is in the middle of the ship and the doll. There are 2 ways to arrange the three toys, i.e. door-call-ship and ship-car-doll.  The last toy, the ball, can be placed on either of the 2 ends.  Therefore, there are 2 x 2 = 4 ways to arrange the toys. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
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 * 2 = 10 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: A box contains $5$ red balls and $3$ white balls that are identical in all aspects except color. One ball is drawn at random from the box and then replaced. The box is then thoroughly shaken so that the balls are arranged at random again and a second ball is drawn randomly from the box. What is the probability of drawing white ball on both draws? 
Options:
aoVal: A
content: $$\frac{5}{8}$$ 
aoVal: B
content: $$\frac{3}{8}$$ 
aoVal: C
content: $$\frac{25}{64}$$ 
aoVal: D
content: $$\frac{9}{64}$$ 
Answer: D
Analysis: $$D$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $6$ stairs, and you can walk for one or three staris each time. There are~\uline{~~~~~~~~~~}~different ways to walk the $6$ stairs. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: E
Analysis: $$\left( 1,1,1,1,1,1 \right)$$；$$\left( 3,1,1,1 \right)$$；$$\left( 1,3,1,1 \right)$$；$$\left( 1,1,3,1 \right)$$；$$\left( 1,1,1,3 \right)$$；$$\left( 3,3 \right)$$． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
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: $$11$$ 
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. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
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 (Case 1 ): 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 (Case 2 ): Similarly, we have $2 * 2=4$ ways to obtain 2 evens. 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: Three kids line up to play games. In 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 $3\times 2\times 1=6$ different ways for three kids to line up. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: The numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, $$9$$, $$10$$, $$11$$, and $$12$$ are arranged in $3$ columns of $4$ numbers each so that the sum of the numbers in each column is the same. The sum of the numbers in each column is． 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$32$$ 
Answer: C
Analysis: It\textquotesingle s just like a magic square! The sum of all $$12$$ numbers is $$78$$. Hence, the answer is $$78\div3 = 26$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas
Problem: Mother has $$12$$ identical peaches. She wants to place them into $$3$$ identical baskets. How many ways can she place the peaches if the number of peaches in each basket cannot be zero and must be different from one other? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$12$$ 
Answer: C
Analysis: Order does not matter as peaches and baskets are identical. But we need to have different number of peaches in each basket and none of the baskets can be empty.  $$12=1+2+9$$  $$12=1+3+8$$  $$12=1+4+7$$  $$12=1+5+6$$  $$12=2+3+7$$  $$12=2+4+6$$  $$12=3+4+5$$  There are $$7$$ ways she can place the peaches. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: Fill in the blank:~\uline{~~~~~~~~~~}~is $2$ tens $8$ ones less than $5$ tens $5$ ones. 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$37$$ 
aoVal: C
content: $$73$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: $55 - 28 = 27$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: Una rolls 6 standard 6 -sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4 ? 
Options:
aoVal: A
content: $\frac{3}{4}$ 
aoVal: B
content: $\frac{57}{64}$ 
aoVal: C
content: $\frac{59}{64}$ 
aoVal: D
content: $\frac{187}{192}$ 
aoVal: E
content: $\frac{63}{64}$ 
Answer: C
Analysis: We will use complementary counting to find the probability that the product is not divisible by 4 . Then, we can find the probability that we want by subtracting this from 1 . We split this into 2 cases. Case 1: The product is not divisible by 2. We need every number to be odd, and since the chance we roll an odd number is $\frac{1}{2}$, our probability is $\left(\frac{1}{2}\right)^{6}=\frac{1}{64}$ Case 2: The product is divisible by 2 , but not by 4 . We need 5 numbers to be odd, and one to be divisible by 2 , but not by 4 . There is a $\frac{1}{2}$ chance that an odd number is rolled, a $\frac{1}{3}$ chance that we roll a number satisfying the second condition (only 2 and 6 work), and 6 ways to choose the order in which the even number appears. Our probability is $\left(\frac{1}{2}\right)^{5}\left(\frac{1}{3}\right) \cdot 6=\frac{1}{16}$. Therefore, the probability the product is not divisible by 4 is $\frac{1}{64}+\frac{1}{16}=\frac{5}{64}$. Our answer is $1-\frac{5}{64}=$ (C) $\frac{59}{64}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A data set consists of~~(not distinct) positive integers: 1,7,5,2,5 and x. The average (arithmetic mean) of the~~numbers equals a value in the data set. What is the sum of all positive values of x? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$36$$ 
aoVal: E
content: $$40$$ 
Answer: D
Analysis: $$10+22+4=36$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: If the average of two numbers is $$7$$, the numbers could be． 
Options:
aoVal: A
content: $$3$$ and $$4$$ 
aoVal: B
content: $$1$$ and $$8$$ 
aoVal: C
content: $$2$$ and $$14$$ 
aoVal: D
content: $$6$$ and $$8$$ 
Answer: D
Analysis: If the average of two numbers is $$7$$, their sum is $$2\times7 = 14$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris\textquotesingle s. What is the expected number of balls that occupy their original positions after these two successive transpositions? 
Options:
aoVal: A
content: $$1.6$$ 
aoVal: B
content: $$1.8$$ 
aoVal: C
content: $$2.0$$ 
aoVal: D
content: $$2.2$$ 
aoVal: E
content: $$2.4$$ 
Answer: D
Analysis: After the first swap, we do casework on the next swap.  Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves $5$ balls occupying their original position.  Case 2: Silva swaps one ball that has just been swapped with one that hasn\textquotesingle t swapped There are two ways for Silva to do this, and it leaves $2$ balls occupying their original positions.  Case 3 : Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves $1$ ball occupying their original positions. Our answer is the average of all $5$ possible swaps, so we get $$ \frac{5+2 \cdot 2+2 \cdot 1}{5}=\frac{11}{5}=2.2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: The average mass of $$2$$ bags of flour is $$1.72 \text{kg}$$. The mass of one of the bag of flour is $$1.68 \text{kg}$$. What is the mass of the other bag of flour? 
Options:
aoVal: A
content: $$1.76kg$$ 
aoVal: B
content: $$3.44kg$$ 
aoVal: C
content: $$1.68kg$$ 
aoVal: D
content: $$1.72kg$$ 
Answer: A
Analysis: $$1.72\times2=3.44$$  $$3.44 -1.68 = 1.76$$  The mass of the other bag of flour is $$1.76\text{kg}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
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?~ 
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