math-eval/TAL-SCQ5K huggingface dataset:
Problem: A three-digit integer contains one of each of the digits $1,3$ , and $5$ . What is the probability that the integer is divisible by $5$ ? (2009 AMC 8 Problem, Question \#13) 
Options:
aoVal: A
content: $\frac{1}{6}$ 
aoVal: B
content: $\frac{1}{3}$ 
aoVal: C
content: $\frac{1}{2}$ 
aoVal: D
content: $\frac{2}{3}$ 
aoVal: E
content: $\frac{5}{6}$ 
Answer: B
Analysis: The three digit numbers are $135,153,351,315,513,531$. The numbers that end in 5 are divisible are 5 , and the probability of choosing those numbers is (B) $\frac{1}{3}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Timi has $8$ paintings: $3$ of them are drawing landscape, and $5$ of them are drawing figure. Among the $5$ figure paintings, there are $3$ drawing the whole family of Timi, and the other $2$ are drawing himself. Now, Timi wants to put those painting in a line. The $3$ landscape paintings cannot be adjacent. How many ways can he do this? 
Options:
aoVal: A
content: $$288$$ 
aoVal: B
content: $$72$$ 
aoVal: C
content: $$144$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$252$$ 
Answer: C
Analysis: $\_2P\_2\times \_3P\_3 \times \_2P\_2 \times \_3P\_3=144$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Pip and Bud are playing a game. Each time Pip has to flip a £$$1$$ coin. If Pip gets a head, Bud will pay him £$$1$$. If Pip gets a tail, Bud will take one of her coins. After they played $$50$$ rounds, how much would you expect Pip to gain or lose? 
Options:
aoVal: A
content: Gaining £$$50$$ 
aoVal: B
content: Losing £$$50$$ 
aoVal: C
content: Gaining £$$25$$ 
aoVal: D
content: Losing £$$25$$ 
aoVal: E
content: Break Even 
Answer: E
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A cup costs ￡$$8$$. Which of the following payment is not correct?． 
Options:
aoVal: A
content: One £5 note and three~£1 coins 
aoVal: B
content: Eight £1 coins 
aoVal: C
content: One £5 note and four 50p coins 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: Molly, Dolly, Sally, Elly and Kelly are sitting on a park bench.  Molly is not sitting on the far right and Dolly is not sitting on the far left.  Sally is not sitting at either end.  Kelly is not sitting next to Sally and Sally is not sitting next to Dolly.  Elly is sitting to the right of Dolly but not necessarily next to her.  Who is sitting at the far right end? 
Options:
aoVal: A
content: Molly　 
aoVal: B
content: Dolly　 
aoVal: C
content: Sally　 
aoVal: D
content: Kelly　 
aoVal: E
content: Elly　 
Answer: E
Analysis: The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions $$2$$, $$3$$ and $$4$$ from the left-hand end. Were Sally to sit in place $$2$$, neither Dolly nor Kelly could sit in places $$1$$ or $$3$$ as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places $$4$$ and $$5$$ which is impossible. Similarly, were Sally to sit in place $$3$$, Dolly could not sit in place $$2$$ or $$4$$ and the question also tells us she cannot sit in place $$1$$ so Dolly would have to sit in place $$5$$ making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place $$4$$, Dolly could sit in place $$2$$, Kelly in place $$1$$, Molly (who cannot sit in place $$5$$) in place $$3$$ leaving Elly to sit in place $$5$$ at the right-hand end. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: Only $$1$$ of the $$3$$ boys Abel, Ben and Cain can swim. Abel says, "I can swim." Ben says, "I cannot swim." Cain says, "Abel cannot swim." Only $$1$$ boy is telling the truth. Who can swim? ． 
Options:
aoVal: A
content: Abel 
aoVal: B
content: Ben 
aoVal: C
content: Cain 
Answer: B
Analysis: Only $$1$$ of the $$3$$ boys can swim. Only $$1$$ of the $$3$$ boys is telling the truth!  Since Abel and Cain contradict each other, there must be one who is telling the truth!  If Abel is true, Ben is lying and Ben can swim. Then we have $$2$$ boys (Abel and Ben)  who can swim. Contradiction.  Hence, Cain is true. Abel is lying and so is Ben. Then, Ben can swim. Abel cannot swim and we are not sure if Cain can swim. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many $3$-digit positive integers have digits whose product equals $24$ ? (2009 AMC 8 Problem, Question \#16) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$24$$ 
Answer: D
Analysis: With the digits listed from least to greatest, the $3$-digit integers are $138,146,226,234.$ $226$ can be arranged in $3$ ways, and the other three can be arranged in $6$ ways. There are $3+6(3)=(\mathbf{D}) 21$ 3-digit positive integers. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Summer has $100$ number cards each with a different number from $0$ to $99$. What is the probability that when she chooses two cards randomly, the sum of those two is an even number? 
Options:
aoVal: A
content: $\frac12$ 
aoVal: B
content: $\frac13$ 
aoVal: C
content: $\frac14$ 
aoVal: D
content: $\frac16$ 
Answer: A
Analysis: There are $50$ even number cards and $50$ odd number cards here. If she chooses odd+odd or even+even, she will get an even sum. But if she chooses odd+even or even+odd, she will get an odd sum. So each of the two probabilities is equal. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Karl, Lim, Mary and Navin are each competing in a different sporting event at the Olympic Games - basketball, swimming, taekwondo and women\textquotesingle s rhythmic gymnastics. Karl and Navin do not know how to swim. Mary is the only female. Karl\textquotesingle s event does not require a ball. Which event is Navin competing in? 
Options:
aoVal: A
content: Basketball　 
aoVal: B
content: Swimming　 
aoVal: C
content: Taekwondo　 
aoVal: D
content: Rhythmic gymnastics　 
Answer: A
Analysis: Since \textbf{Mary} is the only female, she must have participated in the women\textquotesingle s\textbf{~Rhythmic gymnastics} as the rest are males.  Since Karl and Navin do not know how to swim, \textbf{Lim} is the only male left and thus he participated in the S\textbf{wimming} event.  And since \textbf{Karl}\textquotesingle s event does not require a ball, he must have participated in \textbf{Taekwondo}.  We can thus conclude that \textbf{Navin}~participated in \textbf{Basketball}. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The average of the first ten odd whole numbers is． 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
Answer: C
Analysis: We have $$\left( 1+3+5+7+9+11+13+15+17+19 \right)\div 10 = 10$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas
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 two 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: Eight students from Think Academy School take a Mathematics test. Unfortunately, none of the students wrote his/her name on the test paper. As a result, the tests are handed back to the students at random. In how many ways can exactly $5$ of the $8$ students get the correct test back? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$56$$ 
aoVal: D
content: $$112$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test.  No. of ways to choose $5$ pupils to get the correct test is $$\frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1}-56.$$  To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$.  Hence, the total no, of ways $=56 \times2=112$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In how many ways can the letters in $BEEKEEPER$ be rearranged so that two or more $E$s do not appear together? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$120$$ 
Answer: D
Analysis: There are $5$ $E$s in total, which have $4$ intervals leaving for the other $4$ letters. Thus, the answer is $\_4P\_4=24$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Two tiles numbered $5$ and $6$ 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 smaller than $6$? 
Options:
aoVal: A
content: $\frac {1}{2}$ 
aoVal: B
content: $\frac {1}{3}$ 
aoVal: C
content: $\frac {1}{4}$ 
aoVal: D
content: $\frac {1}{6}$ 
aoVal: E
content: $\frac {1}{12}$ 
Answer: E
Analysis: There are $2\times6=12$ different combinations. The product of two numbers is smaller than $6$ will be $5\times1$. Thus, the probability is $\frac 1{12}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: The mean, median, and unique mode of the positive integers $3,4,5,6,6,7$, and $x$ are all equal. What is the value of $x$ ? (2012 AMC8, Question \#11) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: unique mode: $6$  median: $6$  $(3+4+5+6+6+7+x)\div7=6$  $x=11$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Basic Concepts of Statistics
Source: 其它
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 $\frac{1}{2}$.  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.  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: Calculate the value of 7 + 16 + 34 + 45 + 50 - 6 - 15 - 4 - 7 
Options:
aoVal: A
content: $$70$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$127$$ 
aoVal: D
content: $$124$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: 7-7 + 16-6 + 34-4 + 45-15 +50  = 0+10+30+30+50  = 120 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: How many whole numbers between $$1$$ and $$500$$ are divisible by $$6$$ but are not divisible by $$8$$? 
Options:
aoVal: A
content: $$83$$ 
aoVal: B
content: $$73$$ 
aoVal: C
content: $$63$$ 
aoVal: D
content: $$53$$ 
Answer: C
Analysis: Every $$6$$th number $$\left( 83 \right.$$ of them$$\left. {} \right)$$ is divisible by $$6$$. Every $$24$$th number $$\left( 20 \right.$$ of them$$\left. {} \right)$$ is divisible by $$6$$ and $$8$$, and $$83-20 = 63$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: There is a ball in a box and three kids are guessing what colour it is.  Val says: "The ball is white."  John says: "The ball is blue."  Elvis says: "I agree with Val."  They open the box and find only one of them guessed right.  What colour is the ball? 
Options:
aoVal: A
content: white 
aoVal: B
content: blue 
aoVal: C
content: Uncertain 
Answer: B
Analysis: We can spot that Val\textquotesingle s guess and Elvis\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, John is right. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: A train was supposed to arrive at a $$MRT$$ station at $$5:30 \rm pm.$$ But the train arrived half an hour early because of a traffic accident and changed another line. What time did the train reach the $$MRT$$ station? (adapted from 2008 Math kangaroo Problems, Level 3-4 , Question \#6） 
Options:
aoVal: A
content: $5:00$ pm 
aoVal: B
content: $4:30$ pm 
aoVal: C
content: $4:00$ pm 
aoVal: D
content: $5:30$ pm 
aoVal: E
content: $4:40$ pm 
Answer: A
Analysis: $5:30$-$30$ minutes=$5:00$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
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: Rose is $$150\rm cm$$ tall. Rose\textquotesingle s older brother Quentin is $$10 \rm cm$$ taller than Rose is. Rose\textquotesingle s younger brother Sam is $$4 \rm cm$$ shorter than Rose is. What is the average of the heights of Rose, Sam, and Quentin? 
Options:
aoVal: A
content: $$148\rm cm$$ 
aoVal: B
content: $$150\rm cm$$ 
aoVal: C
content: $$152\rm cm$$ 
aoVal: D
content: $$154\rm cm$$ 
Answer: C
Analysis: Rose is $$150\rm cm$$ tall. Quentin is $$10 \rm cm$$ taller than Rose, so Quentin is $$160 \rm cm$$ tall. Sam is $$4 \rm cm$$ shorter than Rose, so Sam is $$146 \rm cm$$ tall. Their average height is $$ (150+160+146)\div3 =152 \rm cm$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are $$24$$ four-digit numbers which use each of the digits $$3$$, $$5$$, $$6$$ and $$9$$ once only.  When all of these $$24$$ four-digit numbers are put in order from smallest to largest, which one is in \emph{eighth} position? 
Options:
aoVal: A
content: $$3569$$ 
aoVal: B
content: $$5369$$ 
aoVal: C
content: $$5396$$ 
aoVal: D
content: $$5639$$ 
aoVal: E
content: $$5936$$ 
Answer: C
Analysis: When put in order, the numbers are: $$3569$$, $$3596$$, $$3659$$, $$3695$$, $$3956$$, $$3965$$, $$5369$$, $$5396$$, $$5639$$, $$5693$$, $$\ldots $$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: A dataset of $9$ numbers has an average of $72$. After removing one of the numbers, the average of the remaining numbers becomes $78$. The number that gets removed is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$60$$ 
Answer: B
Analysis: $$9\times72=648$$  $$78\times8=624$$  $$648-624=24$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Basic Concepts of Statistics
Problem: Joann has a robot toy that can walk. The robot can only walk in straight line, and each step can only walk forward for $$1$$ cm or $$3$$ cm. If the robot walks $$4$$ steps forward, it can travel~\uline{~~~~~~~~~~}~different distances. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: $1+1+1+1=4$  $3+3+3+3=12$  $1+1+1+3=6$  $1+1+3+3=8$  $1+3+3+3=10$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: The numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, $$9$$, $$10$$, $$11$$, and $$12$$ are arranged in $3$ columns of $4$ numbers eadh 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->Counting Modules->Statistics and Probability
Problem: Math Town is hosting a triathlon competition (a combination of swimming, cycling, and running). The competition starts at $$9:30$$ A.M.. It takes the winner $$25$$ minutes to swim, $$1$$ hour $$10$$ minutes to cycle, and $$35$$ minutes to run. At what time does the winner complete the competition? 
Options:
aoVal: A
content: $$11:30$$ A.M. 
aoVal: B
content: $$10:40$$ A.M. 
aoVal: C
content: $$11:40$$ A.M. 
aoVal: D
content: $$11:50$$ A.M. 
aoVal: E
content: $$11:40$$ P.M. 
Answer: C
Analysis: It takes $25+35=60$ min, which is equal to $1$ hour to swim and run.  It takes $1$ h + $1$ h $10$ min = $2$ h $10$ min in total.  Thus, the end time is $9:30+$$2$ h $10$ min$=11:40$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Joann and Sana are practicing writing stories. Joann writes $6$ stories each day, and she writes $2$ more than that of Sana each day. How many stories do they write in total in one week? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$28$$ 
aoVal: C
content: $$42$$ 
aoVal: D
content: $$70$$ 
aoVal: E
content: $$98$$ 
Answer: D
Analysis: $$6-2=4$$  $$6+4=10$$  $$7\times 10=70$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: A win, a loss and a draw are three outcomes of a football game: a win scores $$2$$ points, a draw scores $$1$$ point for each of two teams and a loss scores $$0$$ points. Now, $15$ teams run a single round-robin tournament. How many scores in total will all the $15$ teams get? 
Options:
aoVal: A
content: $$150$$ 
aoVal: B
content: $$420$$ 
aoVal: C
content: $$105$$ 
aoVal: D
content: $$225$$ 
aoVal: E
content: $$210$$ 
Answer: E
Analysis: $15\times14\div2\times2=210$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: On a fine morning, a rabbit, a dog, a cat, and a duck went out to look for food.  The rabbit said: "If I get food, the dog will also get food."  The dog said: "If I get food, the cat will also get food."  The cat said: "If I get food, the duck will also get food."  That evening, they found that all of them were telling the truth but only two of them did get food.~\uline{~~~~~~~~~~}~and~\uline{~~~~~~~~~~}~didn\textquotesingle t get any food. 
Options:
aoVal: A
content: The rabbit; the dog 
aoVal: B
content: The dog; the cat 
aoVal: C
content: The cat; the duck 
aoVal: D
content: The cat; the rabbit 
Answer: A
Analysis: We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit, and the dog don\textquotesingle t get food, the cat and the duck would get food. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Seattle is $3$ hours behind New York. For example, when Seattle is $6$ AM, New York is $9$ AM. One day, Rosie took a flight from Seattle to New York. The flight took $$5$$ hours to reach New York, and the time was $9:40$ PM in New York when Rosie arrived. What was the time in Seattle when Rosie departed? 
Options:
aoVal: A
content: $3:40$ PM 
aoVal: B
content: $3:20$ PM 
aoVal: C
content: $2:40$ PM 
aoVal: D
content: $1:40$ PM 
aoVal: E
content: $1:20$ PM 
Answer: D
Analysis: It\textquotesingle s $9:40$ PM in New York, which means it\textquotesingle s $6:40$ PM in Seattle.  The plane departed at $6:40$-$5$ hours=$1:40$ PM. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Fiona spent $$1$$ h $45$ min at the library. At the library, she spent $$30$$ min looking for reading materials. She spent the rest of the time reading them. How many minutes did Fiona spend reading? (adapted from 2011 Math kangaroo Problems, Level 3-4 , Question \#4) 
Options:
aoVal: A
content: $$45$$ min 
aoVal: B
content: $$35$$ min 
aoVal: C
content: $$75$$ min 
aoVal: D
content: $$55$$ min 
aoVal: E
content: $$65$$ min 
Answer: C
Analysis: Time spent reading  $$\rm =1h45min-30min$$  $$\rm =1h15min$$  $$\rm =60min+ 15min$$  $$\rm =75min$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations