math-eval/TAL-SCQ5K huggingface dataset:
Problem: A fair $6$-sided die is rolled once. What is the probability that the number on the top is an odd number? (adapted from 2011 AMC 8 Problem, Question \#18) 
Options:
aoVal: A
content: $\frac16$ 
aoVal: B
content: $\frac13$ 
aoVal: C
content: $\frac12$ 
aoVal: D
content: $\frac23$ 
aoVal: E
content: $\frac56$ 
Answer: C
Analysis: There are $3$ odd numbers out of $6$, so the probability is $\frac36=\frac12$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: How many two-digit numbers are there where the ones digit is greater than the tens digit? (2008 Math Kangaroo Problem, Level 3-4, Question \#21) 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: $$8+7+6+5+4+3+2+1=36$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Amy, Bill and Celine are friends with different ages.  $$\rm I$$. Bill is the oldest.  $$\rm II$$. Amy is not the oldest.  $$\rm III$$. Celine is not the youngest.  Rank the friends from youngest to oldest. 
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: $$30$$ children going to Adventure Park took part in at least one of two events. $$15$$ of them took part in the "moving bridge" contest, and $$20$$ of them went down the zip-wire. How many children from Adventure Park took part in both events? ($$2013$$ Math Kangaroo Problems, Level $$3-4$$, Question \#$$14$$) 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$5$$ 
Answer: E
Analysis: $15+20-30=5$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
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 die with 2 evens and 4 odds on both dies. For the sum to be even, the 2 rolls 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: How many different four-digit odd numbers can be made by using digits $$1$$, $$2$$, $$3$$, $$4$$, and $$5$$ (without digits be repeated)? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$72$$ 
Answer: D
Analysis: Rule of product: $$3\times 4\times 3\times 2=72$$; therefore we can choose D. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers
Problem: How many ways are there of making a total of 10 using three different positive numbers? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$1$$ 
aoVal: E
content: $$0$$ 
Answer: A
Analysis: $$1+2+7=10$$，  $$1+3+6=10$$，  $$1+4+5=10$$，  $$2+3+5=10$$．  So the answer is $$\text{A}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Source: 其它
Problem: A conductor wanted to make a trio consisting of a fiddler, a pianist, and a drummer. He had to choose one of two fiddlers, one of two pianists, and one of two drummers. He decided to try each of the possible trios. How many attempts did he have to make? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$25$$ 
Answer: C
Analysis: $$2\times2\times2=8$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: Eve brings $$12$$ pieces of candy, Alice brings $$9$$ pieces of candy and Irene doesn\textquotesingle t bring any candy. They put all the pieces of candy together on a table and divide them equally among themselves. How many pieces of candy does each of the girls get? (2012 Math Kangaroo Problem, Levels 1-2, Question \#15） 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: They have $12+9=21$ pieces of candy in total. $21\div3=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Given that only one of the following statement is corect, which one is correct?  ($$1$$) All of the statements below are correct.  ($$2$$) None of the statement below is corect.  ($$3$$) One of the statements above is corect.  ($$4$$) All the statements above are correct.  ($$5$$) None of the statement above is corect. 
Options:
aoVal: A
content: (1) 
aoVal: B
content: (2) 
aoVal: C
content: (3) 
aoVal: D
content: (4) 
aoVal: E
content: (5) 
Answer: E
Analysis: Suppose ($$1$$) is correct, then ($$2$$) must be wrong which contradicts that only one statement is correct.  Suppose ($$2$$) is correct, then ($$5$$) is correct which contradicts that only one statement is correct.  Suppose ($$3$$) is correct, it also contradicts that only one statement is correct.  Suppose ($$4$$) is correct, it also contradicts that only one statement is correct.  Hence ($$5$$) is the correct statement. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: The farmer has a fish, a cat, and a dog. He wants to take them cross the river by boat, and he can only take one animal each time. When the farmer is away, the cat cannot be put with the fish or the dog, or they will fight. The boat makes one trip from one side of the river to the other side, so it takes~\uline{~~~~~~~~~~}~trips in total to carry all the animals and farmers cross the river. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: E
Analysis: We can use $A$ to represent the side of the river where the farmer is at the beginning, and use $B$ to represent the other side.  The first time, take the cat from $A$ to $B$.  The second time, the farmer go back to $A$.  The third time, take the fish from $A$ to $B$.  The fourth time, take the cat from $B$ to $A$.  The fifth time, take the dog from $A$ to $B$.  The sixth time, the farmer go back to $A$.  The seventh time, take the cat from $A$ to $B$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: Two dice are thrown. What is the probability that the product of the two numbers is a multiple of $$5$$? ($$2001$$ AMC $$8$$ Problem, Question \#$$ 18$$) 
Options:
aoVal: A
content: $$\frac{1}{36}$$ 
aoVal: B
content: $$\frac{1}{18}$$ 
aoVal: C
content: $$\frac{1}{6}$$ 
aoVal: D
content: $$\frac{11}{36}$$ 
aoVal: E
content: $$\frac{1}{3}$$ 
Answer: D
Analysis: $$5$$ is the only multiple of $$5$$ on a die, so one of the numbers rolled must be a $$5$$. To find the probability of rolling at least one $$5$$, we can find the probability of not rolling a $$5$$ and subtract that from $$1$$, since you either roll a $$5$$ or not roll a $$5$$. The probability of not rolling a $$5$$ on either dice is $$\left( \frac{5}{6} \right)\times\left( \frac{5}{6} \right)=\frac{25}{36}$$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $$5$$, is $$1-\frac{25}{36}=\frac{11}{36}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Mom has $190$ coins in total and wants to give them to Sana as the pocket money. She gives Sana $30$ coins for the first month, and $40$ coins for each of the following months. How many months in total can Sana get the pocket money? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: C
Analysis: $190-30=160$  $40+40+40+40=160$  $4+1=5$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
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? (2016 AMC 8 Problems, Question \#21) 
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: My train leaves Southampton and arrives in Birmingham at $$08:48$$ that morning, the duration of the whole journey is $$52$$ minutes, what time does the train leave Southampton? 
Options:
aoVal: A
content: $$09:40$$ 
aoVal: B
content: $$09:30$$ 
aoVal: C
content: $$07:56$$ 
aoVal: D
content: $$07:04$$ 
aoVal: E
content: $$07:40$$ 
Answer: C
Analysis: The departure time is $$52$$ minutes forward from $$8:48$$. You can push it forward $$48$$ minutes, which is $$8:00$$, and then push it forward $$4$$ minutes, which is $$7:56$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: In the following events, the certain event is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: Toss two identical coins and both land on heads. 
aoVal: B
content: Throw a fair die and number of the dot shown is $$3$$. 
aoVal: C
content: Sun sets in the west. 
aoVal: D
content: It must rain in cloudy days. 
Answer: C
Analysis: $$\text{A}$$. Throw a coin randomly. It has a $$50\textbackslash\%$$ chance of landing head up and a $$50\textbackslash\%$$ chance of landing tail up.  $$\text{B}$$. Throw a die, and number of the dot shown may be any number from $$1$$ to $$6$$.  $$\text{C}$$. Sun sets in the west.  $$\text{D}$$. It may not rain in cloudy days.  So $$\text{C}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: The correct understanding of the statement "there is a $$75\textbackslash\%$$ chance that it will rain tomorrow in New York City" is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: It will rain $$75\textbackslash\% $$ of the time tomorrow in New York City. 
aoVal: B
content: It will rain in $$75\textbackslash\%$$ of the regions in New York City tomorrow. 
aoVal: C
content: It will definitely rain tomorrow in New York City. 
aoVal: D
content: The probability of raining in New York City tomorrow is high. 
Answer: D
Analysis: The statement that "there is a $$75\textbackslash\% $$ chance that it will rain tomorrow in New York City"~shows that it is more likely to rain tomorrow in New York city. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: The digits $1$, $2$, and $3$ can make~\uline{~~~~~~~~~~}~three-digit numbers without repeating digits. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$30$$ 
Answer: A
Analysis: $$3\times 2\times 1=6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers
Problem: If a turkey facing north turns $$225^{}\circ $$ clockwise, it will then face． 
Options:
aoVal: A
content: southwest　 
aoVal: B
content: southeast　 
aoVal: C
content: northwest　 
aoVal: D
content: northeast　 
Answer: A
Analysis: Since $$225^{}\circ =180^{}\circ +45^{}\circ $$, the bird turns $$45^{}\circ $$ past south. That\textquotesingle s southwest. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Directions and Coordinates->Directions
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. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: $71+82+93+104+195=$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$500$$ 
aoVal: B
content: $$540$$ 
aoVal: C
content: $$545$$ 
aoVal: D
content: $$550$$ 
Answer: C
Analysis: $$71+82+93+104+195=70+80+90+100+200+5=545$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: The number of days in July plus the number in August is twice the number of days in． 
Options:
aoVal: A
content: March　 
aoVal: B
content: April　 
aoVal: C
content: June　 
aoVal: D
content: November　 
Answer: A
Analysis: Together, July \& August have $$62$$ days. That\textquotesingle s twice the $$31$$ days in March. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Let $f$ be a linear function for which $f(3)-f(2)=5$. What is $f(8)-f(2)$? (  Adapted From 2003 AMC 12B Problems, Question \#9) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$40$$ 
Answer: D
Analysis: Let $f$ be a linear function with slope $m$.  $$ \begin{gathered} m=\frac{f(3)-f(2)}{\Delta x}=\frac{5}{3-2}=5\textbackslash\textbackslash{} f(8)-f(2)=m \Delta x=5(8-2)=30 \Rightarrow(D) \end{gathered}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the $$100\rm th$$ number in the arithmetic sequence $$1$$, $$5$$, $$9$$, $$13$$, $$17$$, $$21$$, $$25$$, $$\cdots$$? 
Options:
aoVal: A
content: $$397$$ 
aoVal: B
content: $$399$$ 
aoVal: C
content: $$401$$ 
aoVal: D
content: $$403$$ 
aoVal: E
content: $$405$$ 
Answer: A
Analysis: $$1+(5-1)\times 99=397$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: What is the correct ordering of the three numbers $$\dfrac{5}{19}$$, $$\dfrac{7}{21}$$, and $$\dfrac{9}{23}$$, in increasing order? ($$2012$$ AMC $$8$$ Problem, Question \#$$4$$) 
Options:
aoVal: A
content: $$\dfrac{9}{23}\textless{} \dfrac{7}{21}\textless\dfrac{5}{19}$$ 
aoVal: B
content: $$\dfrac{5}{19}\textless{} \dfrac{7}{21}\textless{} \dfrac{9}{23}$$ 
aoVal: C
content: $$\dfrac{9}{23}\textless{} \dfrac{5}{19}\textless{} \dfrac{7}{21}$$ 
aoVal: D
content: $$\dfrac{5}{19}\textless{} \dfrac{9}{23}\textless{} \dfrac{7}{21}$$ 
aoVal: E
content: $$\dfrac{7}{21}\textless{} \dfrac{5}{19}\textless{} \dfrac{9}{23}$$ 
Answer: B
Analysis: $$\rm Method$$ $$1$$:  The value of $$\dfrac{7}{21}$$ is $$\dfrac{1}{3}$$. Now we give all the fractions a common denominator.  $$\dfrac{5}{19} \Rightarrow \dfrac{345}{1311}$$,  $$\dfrac{1}{3} \Rightarrow \dfrac{437}{1311}$$,  $$\dfrac{9}{23} \Rightarrow \dfrac{513}{1311}$$.  Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, $$\frac{5}{19}\textless{} \frac{7}{21}\textless{} \frac{9}{23}$$.  $$\rm Method$$ $$2$$:  Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus,  $$1- \dfrac{5}{19}= \dfrac{14}{19}$$,  $$1- \dfrac{7}{21}= \dfrac{14}{21}$$,  $$1- \dfrac{9}{23}= \dfrac{14}{23}$$.  All three fraction have common numerator $$14$$. Now the order of the fractions is obvious . $$\dfrac{14}{19}\textgreater\dfrac{14}{21}\textgreater\dfrac{14}{23}\Rightarrow\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.  $$\rm Method$$ $$3$$:  Change $$\frac7{21}$$ into $$\frac13$$, $$\dfrac{1}{3} \times \dfrac{5}{5}=\dfrac{5}{15}$$, $$\dfrac{5}{15}\textgreater\dfrac{5}{19}$$, $$\dfrac{7}{21}\textgreater\dfrac{5}{19}$$, and $$\dfrac{1}{3} \times \dfrac{9}{9}=\dfrac{9}{27}$$, $$\dfrac{9}{27}\textless\dfrac{9}{23}$$, $$\dfrac{7}{21} \textless{} \dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.  $$\rm Method$$ $$4$$:  When $$\dfrac{x}{y}\textless1$$ and $$z\textgreater0$$, $$\dfrac{x+z}{y+z}\textgreater\dfrac{x}{y}$$. Hence, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: Which of the following statements is true? 
Options:
aoVal: A
content: $$4+7=3$$ 
aoVal: B
content: $$3=4-7$$ 
aoVal: C
content: $$3+4=7$$ 
aoVal: D
content: $$4=7+3$$ 
aoVal: E
content: $$3-7=4$$ 
Answer: C
Analysis: $$3+4=7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Alysha and Julia have some biscuits. Altogether they have $28$ biscuits. Alysha has $4$ more biscuits than Julia. How many biscuits does Alysha have? (Adapted from 2021 Math Kangaroo Problem, Level 3-4, Question \#7) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$16$$ 
Answer: E
Analysis: $(28 + 4) \div 2 = 16$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the simplest form of $5$ minutes$: 30 $ seconds? 
Options:
aoVal: A
content: $5:30$ 
aoVal: B
content: $1:6$ 
aoVal: C
content: $6:1$ 
aoVal: D
content: $10:1$ 
Answer: D
Analysis: We need to make units same first. $5$ minutes equal to $300$ seconds. Now we could remove the same unit, second. We get $300:30$ and simplify it to $10:1$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: Let $f$ be a linear function for which $f(6)-f(2)=12$. What is $f(12)-f(2)$? (  2003 AMC 12B Problems, Question \#9) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$36$$ 
Answer: D
Analysis: Let $f$ be a linear function with slope $m$.  $$ \begin{gathered} m=\frac{f(6)-f(2)}{\Delta x}=\frac{12}{6-2}=3 \textbackslash\textbackslash{} f(12)-f(2)=m \Delta x=3(12-2)=30 \Rightarrow(D) \end{gathered}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The sum of five consecutive integers is $$2015$$. What is the smallest of these integers? 
Options:
aoVal: A
content: $$401$$ 
aoVal: B
content: $$403$$ 
aoVal: C
content: $$405$$ 
aoVal: D
content: $$407$$ 
aoVal: E
content: $$409$$ 
Answer: A
Analysis: Let the five consecutive integers be $$n-2$$, $$n-1$$, $$n$$, $$n + 1$$ and $$n + 2$$. These have a sum of $$5n$$. Hence $$5n=2015$$ and therefore $$n=403$$. Therefore, the smallest integer is $$403-2=401$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: \textbf{Megan wonders how the size of her beagle Herbie compares with other beagles. Herbie is 40.6cm tall. Megan learned on the internet that beagle heights are approximately normally distributed with a mean of 38.5 cm and a standard deviation of 1.25 cm. What is the percentile rank of Herbie's height?} 
Options:
aoVal: A
content: $$59$$ 
aoVal: B
content: $$65$$ 
aoVal: C
content: $$74$$ 
aoVal: D
content: $$92$$ 
aoVal: E
content: $$95$$ 
Answer: E
Analysis: \textbf{Z = $$\frac{40.6-38.5}{1.25}$$=1.68 → percentile = 0.9535} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$(2+4+6+8+10)\div (10+8+6+4+2)=$$． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$60$$ 
Answer: B
Analysis: $$(2+4+6+8+10)\div (10+8+6+4+2)=30\div30=1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: Jack has eight chocolates. He gives Jimmy one, Luna two, and Tim three respectively. Then Jack\textquotesingle s mother gives Jack two more. How many chocolates are left for Jack?~(adapted from $$2005$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$6$$) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: The problem can be regarded as an equation $8$ is the number of chocolates Jack had, $1, 2, 3$ is the number of chocolates Jack gave out, and $2$ is the number of chocolates his mother gave Jack. So the remaining chocolate is $8-1-2-3+2=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers->Questions Involving Addition and Subtraction