math-eval/TAL-SCQ5K huggingface dataset:
Problem: Jack built a cube using 27 small cubes which are colored either black or white (see figure). No two of the small cubes which are colored in the same color have a common face. How many white cubes did Jack use? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$15$$ 
Answer: C
Analysis: 14 grey cubes and 13 white cubes. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Counting Solid Figures
Source: 其它
Problem: I donate a ＄$$100$$ bill, $$2$$＄$$50$$ bills, $$3$$＄$$20$$ bills, $$4$$＄$$10$$ bills, and $$5$$＄$$5$$ bills. If $$5$$ people divide my money equally, each person receives． 
Options:
aoVal: A
content: ＄$$37$$ 
aoVal: B
content: ＄$$65$$ 
aoVal: C
content: ＄$$70$$ 
aoVal: D
content: ＄$$75$$ 
Answer: B
Analysis: I donate a total of ＄$$100+2\times $$＄$$50+3\times $$＄$$20+4\times $$＄$$10+5\times $$＄$$5=$$＄$$325$$. Each person receives ＄$$325 \div5 =$$＄$$65$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Find the number of two-digit positive integers whose digits total $7$. (2004 AMC 8 Problem, Question \#8) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: The numbers are $16,25,34,43,52,61,70$ which gives us a total of $(\text{B}) 7$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Rabbit Borya likes cabbages and carrots very much. In a day he eats either 9 carrots, or 2 cabbages, or 1 cabbage and 4 carrots. During one week Borya has eaten 30 carrots. How mnay cabbages has he eaten during this week? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: Borya ate 30 carrots in a week, which is a number divisible by 3.  On the days that he ate 9 carrots, the total number of carrots is also divisible by 3.  Hence, the total number of carrots he ate on days that he ate 4 carrots must also be divisible by 3, meaning the number of days he ate 4 carrots is divisible by 3.  If there were no days he ate 4 carrots, he ate 9 carrots in 30/9 days, which is impossible. If there were 6 days he ate 4 carrots, he was left with only 30-6 x 4 = 6 carrots to eat on other days, which is also impossible.  therefore, he ate 4 carrots on 3 days, meaning there are \_\_\_ 2 days he ate 9 carrots. There were 3 days Borya ate 1 cabbage, and 7 \_ days he ate 2 cabbages.  He ate a total of 3 + 2~ x 2 = 7 cabbages in a week. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Source: 其它
Problem: What is the average (mean) of all $$5-$$digit numbers that can be formed by using each of the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$8$$ exactly once? ($$2005$$ AMC $$10B$$ Problem, Question \#$$20$$) 
Options:
aoVal: A
content: $$48000$$ 
aoVal: B
content: $$49999.5$$ 
aoVal: C
content: $$53332.8$$ 
aoVal: D
content: $$55555$$ 
aoVal: E
content: $$56432.8$$ 
Answer: C
Analysis: Method $$1$$: We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $$4!=24$$ ways to arrange the other numbers, so each number appears in each spot $$24$$ times. Therefore, the sum of all such numbers is $$24\times (1+3+5+7+8)\times (11111)=24\times24\times11111=6399936$$.  Since there are $$5!=120$$ such numbers, we divide $$6399936\div120$$~ to get $$53332.8$$.  Method $$2$$: We can first solve for the mean for the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$9$$ since each is $$2$$ away from each other. The mean of the numbers than can be solved using these digits is $$55555$$. The total amount of numbers that can be formed using these digits is $$5!=120$$. The sum of these numbers is $$55555(120)=6666600$$. Now we can find out the total value that was gained by replacing the $$8$$ with a $$9$$. We can start by calculating the gain when the $$8$$ was in the ones digit. Since there are $$4!=24$$ numbers with the $$8$$ in the ones digit and $$1$$ was gain from each of them, $$24$$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $$24+240+2400+24000+240000=266664$$ as the total amount that was gained. Subtract this amount from the sum of the digits using the~ $$9$$ instead of the $$8$$ to get $$6666600-266664=6399936$$. Finally, we divide this by $$120$$ to get the average $$\frac{6399936}{120}=53332.8$$.  Method $$3$$: The average value of the digits is $$\frac{(1 + 3 + 5 + 7 + 8)}{5} = 4.8$$. Values occur in every place so $$4.8 \times 11111 = 53332.8$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: How many digits are there from $8$ to $78$? 
Options:
aoVal: A
content: $$70$$ 
aoVal: B
content: $$71$$ 
aoVal: C
content: $$138$$ 
aoVal: D
content: $$140$$ 
Answer: D
Analysis: $8$ to $9 = 2$ digits  $10$ to $78: 78 - 10 + 1 = 69$ numbers, $69 \times 2 = 138$ digits  $138 + 2 = 140$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: There are $$124$$ teams participating in a volleyball match held in the Boston. Using the single elimination method, how many games will be played in total? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$62$$ 
aoVal: C
content: $$124$$ 
aoVal: D
content: $$248$$ 
aoVal: E
content: $$123$$ 
Answer: E
Analysis: Single elimination tournament: $$124-1=123$$ games 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Sports Competition
Problem: SASMO 2016 P2 Q6  Uncle John has a farm. His wife and his two sons are staying with him in the farm. They raise 10 cows and 20 chickens. How many total legs are there in the farm? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$80$$ 
aoVal: C
content: $$86$$ 
aoVal: D
content: $$88$$ 
aoVal: E
content: $$120$$ 
Answer: D
Analysis: 10 x 4 + 20 x 2 + 2 + 2 + 4 = 88. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Abby, Bret, Carl, and Dana are seated 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 seat $$3$$. Who is sitting in seat $$2$$? 
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->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132$ . (2016 AMC 8 Problem, Question \#11) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: We can write the two digit number in the form of $10 a+b$; reverse of $10 a+b$ is $10 b+a$. The sum of those numbers is: $$ \begin{gathered} (10 a+b)+(10 b+a)=132 \textbackslash\textbackslash{} 11 a+11 b=132 \textbackslash\textbackslash{} a+b=12 \end{gathered} $$ We can use brute force to find order pairs $(a, b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than 10 . Thus our ordered pairs are $(3,9) ;(4,8) ;(5,7) ;(6,6) ;(7,5) ;(8,4) ;(9,3)$ or $(\text{B}) 7$ ordered pairs. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Ms. Osborne asks each student in her class to draw a rectangle with integral side lengths and a perimeter of $$50$$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles? 
Options:
aoVal: A
content: $$76$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$128$$ 
aoVal: D
content: $$132$$ 
aoVal: E
content: $$136$$ 
Answer: D
Analysis: As we know, the sum of the length and width is $$25$$. The largest area is $$13\times12=156$$ and the smallest area is $$24\times1=24$$, so the difference is $$156-24=132$$. 
Knowledge points: Overseas Competition->Knowledge Point->Geometry Modules->Objects with Straight Sides->Knowing Graphs
Problem: A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won $$4$$ games, Ines won $$3$$ games, Janet won $$2$$ games, Kendra won $$2$$ games and Lara won $2$ games, how many games did Monica (the sixth player) win? ($$2006 \text{ AMC } 8$$ Problem, Question \#$$20$$) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: C
Analysis: There are $$6$$ players, so there are $$6\times5\div2=15$$ games in total. By now, $$4+3+2+2+2=13$$ games have been finished (there is one winner in each game), so Monica needs to win $$15-13=2$$ games. Therefore, the answer is $$\rm C$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Each of $$300$$ students belongs to exactly $$2$$ of the $$5$$ school clubs. What is the average number of students in each club? 
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$120$$ 
aoVal: D
content: $$150$$ 
Answer: C
Analysis: Since $$300$$ students belong to $$2$$ clubs each, there are $$600$$ club memberships. The average membership of the $$5$$ clubs is $$600\div 5 = 120$$ students. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: How many terms are there in the arithmetic sequence~$2,\textbackslash{} 5,\textbackslash{} 8,\textbackslash{} 11,\textbackslash{} 14\cdots \textbackslash{} 95,\textbackslash{} 98$$?$terms. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$33$$ 
aoVal: E
content: $$34$$ 
Answer: D
Analysis: $(98-2)\div3+1=33$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Three friends, Ms Raja, Ms Omar and Ms Beatty all live in the same street. They are a doctor, an engineer and a musician in some order. The youngest one, the doctor, does not have a brother. Ms Beatty is older than the engineer and is married to Ms Omar\textquotesingle s brother. What are the names, in order, of the doctor and the engineer? 
Options:
aoVal: A
content: Raja and Omar 
aoVal: B
content: Omar and Beatty 
aoVal: C
content: Beatty and Omar 
aoVal: D
content: Raja and Beatty 
aoVal: E
content: Omar and Raja 
Answer: A
Analysis: The doctor is the youngest and does not have a brother. Since Ms Omar has a brother and Ms Beatty is older than the engineer, the doctor is Ms Raja. Also, since Ms Beatty is older than the engineer she cannot be the engineer. Hence the engineer is Ms Omar. Therefore the doctor and the engineer in order are Raja and Omar. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: There were $20$ ducks, pigs, and sheep in total in Sam\textquotesingle s farm. After Sam bought some new sheep, the number of sheep has doubled. There are $27$ ducks, pigs, and sheep in total. Originally, how many ducks and pigs were there? 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: A
Analysis: $27-20=7$  $7+7=14$  $27-14=13$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? 
Options:
aoVal: A
content: $\dfrac{1}{6}$ 
aoVal: B
content: $\dfrac{5}{12}$ 
aoVal: C
content: $\dfrac{1}{2}$ 
aoVal: D
content: $\dfrac{7}{12}$ 
aoVal: E
content: $\dfrac{5}{6}$ 
Answer: D
Analysis: There are $6\cdot 6=36$ ways to roll the two dice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\boxed {\left (\text{D}\right )\dfrac{7}{12}}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Source: 其它
Problem: What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock?． 
Options:
aoVal: A
content: $50^{}\circ $ 
aoVal: B
content: $120^{}\circ $ 
aoVal: C
content: $135^{}\circ $ 
aoVal: D
content: $150^{}\circ $ 
aoVal: E
content: $165^{}\circ $ 
Answer: D
Analysis: The smaller angle is $\frac 5{12}$ of a full circle.  A full circle has $360$ degrees, so the angle is $\frac 5{12}\times 360^{}\circ =150^{}\circ $.  So, the answer is $\rm D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many whole numbers between $$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->Permutations and Combinations->Combinations
Problem: Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops? 
Options:
aoVal: A
content: $\frac{9}{16}$ 
aoVal: B
content: $\frac{5}{8}$ 
aoVal: C
content: $\frac{3}{4}$ 
aoVal: D
content: $\frac{25}{32}$ 
aoVal: E
content: $\frac{13}{16}$ 
Answer: D
Analysis: We will use complementary counting. First, the frog can go left with probability $\frac{1}{4}$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4 \cdot \frac{1}{4}=1$, we will ignore the leading probability. From the left, she either goes left to another edge $\left(\frac{1}{4}\right)$ or back to the center $\left(\frac{1}{4}\right)$. Time for some casework. Case 1: She goes back to the center. Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac{1}{2}$. -End case 1 Case 2: She goes to another edge (rightmost). Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac{1}{2}$ Subcase 2: She goes to the center. Now any move works. $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot 1=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ for this case. -End case 2 She goes back to the center in Case 1 with probability $\frac{1}{4}$, and to the right edge with probability $\frac{1}{4}$ So, our answer is $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot \frac{3}{8}=\frac{1}{4}\left(\frac{1}{2}+\frac{3}{8}\right)=\frac{1}{4} \cdot \frac{7}{8}=\frac{7}{32}$ But, don\textquotesingle t forget complementary counting. So, we get $1-\frac{7}{32}=\frac{25}{32} \Longrightarrow D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Sam\textquotesingle s exam was meant to begin at $$13:45$$ but started $$17$$ minutes late. At what time did it begin?　 
Options:
aoVal: A
content: $$13:38$$ 
aoVal: B
content: $$13:56$$ 
aoVal: C
content: $$14:02$$ 
aoVal: D
content: $$14:06$$ 
Answer: C
Analysis: Seventeen minutes after $$13:45$$ is $$14:02$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: SASMO 2014 P2 Q6  A shop sells sweets where every 3 sweet wrappers can be excanged for one more sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of sweets that he can get from the shop? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$11$$ 
Answer: D
Analysis: 7 = 3 + 3 + 1 + (1) + (1) + (1) = 10 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: One day, Pip asks his parents: "What day is it today?"  His mother says: "Today is Monday."  His father says: "Today is Tuesday."  Which of the following is true? 
Options:
aoVal: A
content: One of these two sentences is definitely wrong and the other one is correct. 
aoVal: B
content: It is possible that both of Pip\textquotesingle s parents are wrong. 
aoVal: C
content: It is possible that both of Pip\textquotesingle s parents are right. 
aoVal: D
content: If Pip\textquotesingle s mother is wrong, then his father must be right 
Answer: B
Analysis: "Today is Monday" is not the opposite of "Today is Tuesday".i.e. they can both be false.  "Today is Monday" is the direct opposite of "Today is not Monday". One must be true and the other must be false. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Five students are lining up to take a picture. If Jessica is standing in the middle, and Ian is standing on either end, how many ways can the students line up? ． 
Options:
aoVal: A
content: $6$ 
aoVal: B
content: $8$ 
aoVal: C
content: $10$ 
aoVal: D
content: $12$ 
aoVal: E
content: $14$ 
Answer: D
Analysis: $$2\times3\times2\times1=12$$ ways. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A sergeant stands too long in the sun and gets confused. His troops are lined up facing north. Then he gives the order to "Right Turn $90^{}\circ$" $$70$$ times, and his troops do so. In which direction are the troops facing at the end? 
Options:
aoVal: A
content: $$$$north 
aoVal: B
content: $$$$east 
aoVal: C
content: $$$$south 
aoVal: D
content: $$$$west 
aoVal: E
content: $$$$west-south-west 
Answer: C
Analysis: After every $$4$$ "Right turns" the troops will be facing north again; so after $$68$$ turns they are facing north, after $$69$$ turns east, and after $$70$$ turns south. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Directions and Coordinates->Directions
Problem: Hansel wants to buy 2 dice of different colours. If the available colours in a store are red, blue, green, yellow, pink and white, how many different combinations of 2 dice are there in the store? (Example: 1 combination is yellow and white) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: RB, RG, RY, RP, RW  BG, BY, BP, BW  GY, GP, GW,  YP, YW,  PW 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Source: 其它
Problem: There is a cuboid with a dimension of $3\times4\times5$. Now paint all the surfaces red and cut it into many $1\times1\times1$ cubes. How many cubes that have two faces painted red are there? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$22$$ 
Answer: A
Analysis: Remove the two cubes in both ends, and then we can get: $[(3-2)+(4-2)+(5-2)]\times4=24$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Coloring Problems
Source: 其它
Problem: Sophia's average score on six tests is $$82$$. Her average scores on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. What is her average score on all eight tests? ． 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$94$$ 
Answer: A
Analysis: Sophia\textquotesingle s total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: Nina and four of her friends are lining up to take a picture. If Nina has to be in the middle, how many different ways can the five friends line up? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$120$$ 
Answer: C
Analysis: Since Nina has to be in the middle, then there are four friends who can line up in different ways. We can write the equation as $$4\times3\times2\times1=24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: Each of Basil\textquotesingle s friends added the number of the day and the number of the month of their birthdays and obtained $$35$$. Their birthdays all fall on different days. What is the greatest possible number of friends that Basil has? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: Their birthdays could fall on $$5/30$$, $$6/29$$, $$7/28$$, $$8/27$$, $$9/26$$, $$10/25$$, $$11/24$$ and $$12/23$$ to meet the greatest possible number. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Source: 其它
Problem: There are $20$ cards numbered from $1$ to $20$, respectively. What is the probability of taking out a card with a prime number? 
Options:
aoVal: A
content: $\dfrac{3}{10}$ 
aoVal: B
content: $\dfrac{7}{20}$ 
aoVal: C
content: $\dfrac{2}{5}$ 
aoVal: D
content: $\dfrac{9}{20}$ 
Answer: C
Analysis: There are $8$ prime numbers in total.  So, answer $= \frac{8}{20} = \frac{2}{5}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: SASMO 2015 P2 Q7  Study the figures made with matchsticks below. How many matchsticks are needed to make figure 5? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$39$$ 
aoVal: D
content: $$45$$ 
aoVal: E
content: $$51$$ 
Answer: D
Analysis: Figure 1 = 1 triangle = 3 matchsticks  Figure 2 = (1+2) triangle = 3 x 3 = 9 matchsticss.  Figure 3 = (1+2+3) triangle = 6 x 3 = 18 matchsticks.  Figure 4 =  Figure 5 = (1+2+3+4+5) triangle = 15 x 3 = 45 matchsticks. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Regular Figures->Counting Triangles
Source: 其它