Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$0.725$$ " } ], [ { "aoVal": "B", "content": "$$0.785$$ " } ], [ { "aoVal": "C", "content": "$$0.825$$ " } ], [ { "aoVal": "D", "content": "$$0.875$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules", "Overseas In-curriculum->Knowledge Point->Knowing Numbers->Decimals->Converting Between Fractions and Decimals->Converting Fractions to Decimals" ]

[ "Perform long division. $7\\div8=0.875$ " ]

[]

[ [ { "aoVal": "A", "content": "$$90 $$ " } ], [ { "aoVal": "B", "content": "$$ 100 $$ " } ], [ { "aoVal": "C", "content": "$$ 110 $$ " } ], [ { "aoVal": "D", "content": "$$120 $$ " } ], [ { "aoVal": "E", "content": "$$130$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "It is $$75$$ minutes from $$22:45$$ to midnight and then another $$35$$ minutes from midnight until $$00:35$$. So the required number of minutes is $$75 + 35 = 110$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "Impossible events are definite events. $$\\text{Statement 2}$$ is wrong. $$\\text{Statement 1}$$, $$\\text{Statement 3}$$, and $$\\text{Statement 4}$$ are right. Thus, the answer is $$\\text{D}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "$$26+18-38=6$$, $$26-6=20$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$109$$ " } ], [ { "aoVal": "C", "content": "$$999$$ " } ], [ { "aoVal": "D", "content": "$$1009$$ " } ], [ { "aoVal": "E", "content": "$$9999$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$9+1000=1009$ " ]

[]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "Only $$18$$ is one of the multiples of $$5+1$$, and the second player can ensure victory. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "Impossible events are definite events. $$\\text{Statement 4}$$ is wrong. $$\\text{Statement 1}$$, $$\\text{Statement 2}$$, and $$\\text{Statement 3}$$ are right. Thus, the answer is $$\\text{D}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$123$$ " } ], [ { "aoVal": "B", "content": "$$247$$ " } ], [ { "aoVal": "C", "content": "$$427$$ " } ], [ { "aoVal": "D", "content": "$$472$$ " } ], [ { "aoVal": "E", "content": "$$742$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value in Enumeration Problems" ]

[ "The first digits of the two numbers will need to be as close as possible to each other. Since they cannot be equal, they will have to differ by $$1$$; say they are $$n$$ and $$n + 1$$. The difference between the two numbers will then be minimised by making the four digits after $$n + 1$$ as small as possible and the four digits after $$n$$ as large as possible. The smallest four-digit number available is $$123$$ and the largest is $$876$$. So we need to make $$n= 4$$ and then the required diference is $$5123 - 4876 = 247$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums" ]

[ "$$16=1+2+3+4+6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$180$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$90$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "There are $$10$$ players, so there are $$10\\times9\\div2=45$$ games in total. The sum of the points of two players in each game must be $2$. Thus, the sum of all points earned by $10$ players in $45$ games is $2\\times45=90$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "$6\\times9=54$ " ]

[]

[ [ { "aoVal": "A", "content": "Abel " } ], [ { "aoVal": "B", "content": "Ben " } ], [ { "aoVal": "C", "content": "Charles " } ], [ { "aoVal": "D", "content": "more information needed " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "Abby " } ], [ { "aoVal": "B", "content": "Bret " } ], [ { "aoVal": "C", "content": "Carl " } ], [ { "aoVal": "D", "content": "Dana " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 $$2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration" ]

[ "49, 58, 67, 76, 85 " ]

[ "其它" ]

[ [ { "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": "$$\\frac{7}{10}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability" ]

[ "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 $(\\mathbf{B}) \\frac{2}{5}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "Ali, Barb, and Cal \\emph{were} all born on April $$1$$, in \\emph{different} years. This coming Apr $$1$$, if I add all their ages, I\\textquotesingle ll get $$9$$. Since the youngest possible ages are $$1$$ and $$2$$, the oldest possible age is $$6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$86$$ " } ], [ { "aoVal": "B", "content": "$$88$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$94$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "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 $$688\\div8=86$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "girls = $10\\times2=20$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{6}{13}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{8}{13}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8}{15}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{11}{15}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$$\\frac{1}{3}\\times \\frac{1}{5}+\\frac{1}{3}\\times \\frac{2}{5}+\\frac{1}{3}=\\frac{8}{15}$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$31$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$33$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ], [ { "aoVal": "E", "content": "$$35$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "L - F +1 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6+6+6+6+6$$ " } ], [ { "aoVal": "B", "content": "$$5+6$$ " } ], [ { "aoVal": "C", "content": "$$5\\times 6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "By the meaning of multiplication, $$5\\times 6$$ means the sum of five $$6$$\\textquotesingle s. " ]

[]

[ [ { "aoVal": "A", "content": "$$31536$$ " } ], [ { "aoVal": "B", "content": "$$525600$$ " } ], [ { "aoVal": "C", "content": "$$1892160$$ " } ], [ { "aoVal": "D", "content": "$$3536000$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "$$365$$ days have $$365\\times24\\times60\\times60\\div1000=31536$$ thousand seconds． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac17$ " } ], [ { "aoVal": "B", "content": "$\\frac37$ " } ], [ { "aoVal": "C", "content": "$\\frac57$ " } ], [ { "aoVal": "D", "content": "$\\frac47$ " } ], [ { "aoVal": "E", "content": "$\\frac67$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$2$, $4$, and $6$ are even numbers. Thus, the probability is $\\frac37$. " ]

[]

[ [ { "aoVal": "A", "content": "all $4$ are boys " } ], [ { "aoVal": "B", "content": "all $4$ are girls " } ], [ { "aoVal": "C", "content": "$2$ are girls and $2$ are boys " } ], [ { "aoVal": "D", "content": "$3$ are of one gender and $1$ is of the other gender " } ], [ { "aoVal": "E", "content": "all of these outcomes are equally likely " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability" ]

[ "Nil " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$3$ is a factor of $98765$. " } ], [ { "aoVal": "B", "content": "The sum of the interior angles of a triangle is $180$°. " } ], [ { "aoVal": "C", "content": "The age difference between Michael and Candy will increase in $10$ years. " } ], [ { "aoVal": "D", "content": "Choose a $3$-digit number and it can be divided by $9$ without any remainder. " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$A$ is an impossible event. $C$ is an impossible event. $D$ is a random event. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums" ]

[ "$$15=1+2+3+4+5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{15}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{2}{15}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{1}{5}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{4}{15}$ " } ], [ { "aoVal": "E", "content": "$\\dfrac{1}{3}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability" ]

[ "$\\dfrac{A\\_8^{8}\\times A\\_3^{3}}{A\\_{10}^{10}}=\\dfrac{6}{90}=\\dfrac{1}{15}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4450$$ " } ], [ { "aoVal": "B", "content": "$$4540$$ " } ], [ { "aoVal": "C", "content": "$$4500$$ " } ], [ { "aoVal": "D", "content": "$$4505$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition" ]

[ "simple math calculation " ]

[]

[ [ { "aoVal": "A", "content": "$$\\textbackslash$2$$ " } ], [ { "aoVal": "B", "content": "$$\\textbackslash$3$$ " } ], [ { "aoVal": "C", "content": "$$\\textbackslash$5$$ " } ], [ { "aoVal": "D", "content": "$$\\textbackslash$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "The assorted candy weighs $2+3=5$ kg and is priced at $2\\times6+3\\times1=\\textbackslash$15$ in total. Therefore, each kilogram of the assorted candy is priced at $15\\div5=3$ dollars. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle" ]

[ "C " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$17$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$(19\\times4+25\\times2)\\div7=18$ pages. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$8-3=5$ $6+5=11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers" ]

[ "$10=1+2+7=1+3+6=1+4+5=2+3+5$ " ]

[]

[ [ { "aoVal": "A", "content": "Molly　 " } ], [ { "aoVal": "B", "content": "Dolly　 " } ], [ { "aoVal": "C", "content": "Sally　 " } ], [ { "aoVal": "D", "content": "Kelly　 " } ], [ { "aoVal": "E", "content": "Elly　 " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "The probability of taking out a black ball is the smallest. " } ], [ { "aoVal": "B", "content": "It\\textquotesingle s impossible to take out a white ball. " } ], [ { "aoVal": "C", "content": "The probability of taking out a red ball is larger. " } ], [ { "aoVal": "D", "content": "A red ball is surely to be taken out. " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "The number of red balls in the bag is the most compared with other colored balls. If we take out one ball at random, the probability of taking out a red ball is larger. So $$\\text{C}$$ is the answer. " ]

[]

[ [ { "aoVal": "A", "content": "$11:15$ " } ], [ { "aoVal": "B", "content": "$10:45$ " } ], [ { "aoVal": "C", "content": "$10:55$ " } ], [ { "aoVal": "D", "content": "$11:45$ " } ], [ { "aoVal": "E", "content": "$10:35$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation" ]

[ "$7:55+ 3$ hours $$20$$ minutes=$11:15$ $11:15$+$30$ minutes=$11:45$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "3 + 5 = 8 " ]

[]

[ [ { "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": "$\\dfrac{2}{3}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability" ]

[ "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 $\\dfrac{1}{4}+\\dfrac{1}{8}=\\boxed {\\left (\\text{C}\\right )\\dfrac{3}{8}}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2070$$ " } ], [ { "aoVal": "B", "content": "$$1980$$ " } ], [ { "aoVal": "C", "content": "$$2130$$ " } ], [ { "aoVal": "D", "content": "$$2240$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication" ]

[ "$$230$$x(10-1) =230x10-230x1 =2300-230 =2070 " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$69$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "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)" ]

[ "The only such numbers are $$5432$$, $$5431$$, $$5430$$, $$5421$$, $$5420$$, $$5410$$, $$5321$$, $$5320$$, $$5310$$, and $$5210$$. In all, there are $$10$$ such numbers. " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "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 " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$13:28$$ " } ], [ { "aoVal": "B", "content": "$$13:38$$ " } ], [ { "aoVal": "C", "content": "$$13:56$$ " } ], [ { "aoVal": "D", "content": "$$14:02$$ " } ], [ { "aoVal": "E", "content": "$$14:06$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation" ]

[ "Seventeen minutes after $$13:45$$ is $$14:02$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3546$$ " } ], [ { "aoVal": "B", "content": "$$3645$$ " } ], [ { "aoVal": "C", "content": "$$4356$$ " } ], [ { "aoVal": "D", "content": "$$4536$$ " } ], [ { "aoVal": "E", "content": "$$5346$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration" ]

[ "$3456, 3465, 3546, 3564, 3546, 3564$ $\\underline{4356}, 4365, 4536, 4563, 4635, 4653$ $5346, 5364, 5436, 5463, 5634, 5643$ $6345, 6354, 6435, 6453, 6534, 6543$ " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "It cannot be determined. " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$$6+1=7$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 $$(4,1,1)$$. " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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. " ]

[]

[ [ { "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 $ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock" ]

[ "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 $\\rm D$. " ]

[]

[ [ { "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 $40$. " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "So $$\\text{D}$$ is the answer. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ A.M. " } ], [ { "aoVal": "B", "content": "$$7$$ P.M. " } ], [ { "aoVal": "C", "content": "$$7$$ A.M. " } ], [ { "aoVal": "D", "content": "$$7$$ P.M. " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation" ]

[ "Divide $$12012\\div24$$ to get remainder $$12$$. $$12$$ hours after $$7$$ A.M. is $$7$$ P.M. " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "girls = $18\\times2=36$ " ]

[ "其它" ]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "maximum is 6+6+6 = 18. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ], [ { "aoVal": "E", "content": "more information is needed " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 $$14$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition" ]

[ "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 $$9+3+2 = 14$$. ~ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Counting Solid Figures" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "5 * 2 = 10 " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{25}{64}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{64}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$$D$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$$\\left( 1,1,1,1,1,1 \\right)$$；$$\\left( 3,1,1,1 \\right)$$；$$\\left( 1,3,1,1 \\right)$$；$$\\left( 1,1,3,1 \\right)$$；$$\\left( 1,1,1,3 \\right)$$；$$\\left( 3,3 \\right)$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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. " ]

[ "其它" ]

[ [ { "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": "$\\dfrac{2}{3}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability" ]

[ "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 $\\frac{20}{36}=\\frac{5}{9}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems" ]

[ "There are $3\\times 2\\times 1=6$ different ways for three kids to line up. " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$21$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas" ]

[ "It\\textquotesingle s just like a magic square! The sum of all $$12$$ numbers is $$78$$. Hence, the answer is $$78\\div3 = 26$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$37$$ " } ], [ { "aoVal": "C", "content": "$$73$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition" ]

[ "$55 - 28 = 27$ " ]

[ "其它" ]

[ [ { "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": "$\\frac{63}{64}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "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) $\\frac{59}{64}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "$$40$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$$10+22+4=36$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ and $$4$$ " } ], [ { "aoVal": "B", "content": "$$1$$ and $$8$$ " } ], [ { "aoVal": "C", "content": "$$2$$ and $$14$$ " } ], [ { "aoVal": "D", "content": "$$6$$ and $$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "If the average of two numbers is $$7$$, their sum is $$2\\times7 = 14$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1.6$$ " } ], [ { "aoVal": "B", "content": "$$1.8$$ " } ], [ { "aoVal": "C", "content": "$$2.0$$ " } ], [ { "aoVal": "D", "content": "$$2.2$$ " } ], [ { "aoVal": "E", "content": "$$2.4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "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 $$ \\frac{5+2 \\cdot 2+2 \\cdot 1}{5}=\\frac{11}{5}=2.2$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1.76kg$$ " } ], [ { "aoVal": "B", "content": "$$3.44kg$$ " } ], [ { "aoVal": "C", "content": "$$1.68kg$$ " } ], [ { "aoVal": "D", "content": "$$1.72kg$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$$1.72\\times2=3.44$$ $$3.44 -1.68 = 1.76$$ The mass of the other bag of flour is $$1.76\\text{kg}$$. " ]

[ "其它" ]

[ [ { "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": "$\\frac{3}{5}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "$\\frac25\\times\\frac34+\\frac35\\times\\frac24=\\frac{3}{5}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Counting Solid Figures" ]

[ "14 grey cubes and 13 white cubes. " ]

[]

[ [ { "aoVal": "A", "content": "＄$$37$$ " } ], [ { "aoVal": "B", "content": "＄$$65$$ " } ], [ { "aoVal": "C", "content": "＄$$70$$ " } ], [ { "aoVal": "D", "content": "＄$$75$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "I donate a total of ＄$$100+2\\times $$＄$$50+3\\times $$＄$$20+4\\times $$＄$$10+5\\times $$＄$$5=$$＄$$325$$. Each person receives ＄$$325 \\div5 =$$＄$$65$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "The numbers are $16,25,34,43,52,61,70$ which gives us a total of $(\\text{B}) 7$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$48000$$ " } ], [ { "aoVal": "B", "content": "$$49999.5$$ " } ], [ { "aoVal": "C", "content": "$$53332.8$$ " } ], [ { "aoVal": "D", "content": "$$55555$$ " } ], [ { "aoVal": "E", "content": "$$56432.8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "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 $$4.8 \\times 11111 = 53332.8$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$288$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ], [ { "aoVal": "E", "content": "$$252$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations" ]

[ "$\\_2P\\_2\\times \\_3P\\_3 \\times \\_2P\\_2 \\times \\_3P\\_3=144$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$70$$ " } ], [ { "aoVal": "B", "content": "$$71$$ " } ], [ { "aoVal": "C", "content": "$$138$$ " } ], [ { "aoVal": "D", "content": "$$140$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$8$ to $9 = 2$ digits $10$ to $78: 78 - 10 + 1 = 69$ numbers, $69 \\times 2 = 138$ digits $138 + 2 = 140$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$62$$ " } ], [ { "aoVal": "C", "content": "$$124$$ " } ], [ { "aoVal": "D", "content": "$$248$$ " } ], [ { "aoVal": "E", "content": "$$123$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Sports Competition" ]

[ "Single elimination tournament: $$124-1=123$$ games " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$80$$ " } ], [ { "aoVal": "C", "content": "$$86$$ " } ], [ { "aoVal": "D", "content": "$$88$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "10 x 4 + 20 x 2 + 2 + 2 + 4 = 88. " ]

[]

[ [ { "aoVal": "A", "content": "Abby " } ], [ { "aoVal": "B", "content": "Bret " } ], [ { "aoVal": "C", "content": "Carl " } ], [ { "aoVal": "D", "content": "Dana " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing" ]

[ "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 $$2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$76$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$128$$ " } ], [ { "aoVal": "D", "content": "$$132$$ " } ], [ { "aoVal": "E", "content": "$$136$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Geometry Modules->Objects with Straight Sides->Knowing Graphs" ]

[ "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 $$156-24=132$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 $$\\rm C$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ], [ { "aoVal": "E", "content": "$$34$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$(98-2)\\div3+1=33$ " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$27-20=7$ $7+7=14$ $27-14=13$ " ]

[ "其它" ]

[ [ { "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": "$\\dfrac{5}{6}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability" ]

[ "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 $\\dfrac{15+6}{36}=\\dfrac{21}{36}=\\boxed {\\left (\\text{D}\\right )\\dfrac{7}{12}}$. " ]

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[ [ { "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 $ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 $\\rm D$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$69$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "The only such numbers are $$5432$$, $$5431$$, $$5430$$, $$5421$$, $$5420$$, $$5410$$, $$5321$$, $$5320$$, $$5310$$, and $$5210$$. In all, there are $$10$$ such numbers. " ]

[ "其它" ]

[ [ { "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": "$\\frac{13}{16}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "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 $1-\\frac{7}{32}=\\frac{25}{32} \\Longrightarrow D$. " ]

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[ [ { "aoVal": "A", "content": "$$13:38$$ " } ], [ { "aoVal": "B", "content": "$$13:56$$ " } ], [ { "aoVal": "C", "content": "$$14:02$$ " } ], [ { "aoVal": "D", "content": "$$14:06$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation" ]

[ "Seventeen minutes after $$13:45$$ is $$14:02$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$11$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication" ]

[ "7 = 3 + 3 + 1 + (1) + (1) + (1) = 10 " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "\"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. " ]

[]

[ [ { "aoVal": "A", "content": "$6$ " } ], [ { "aoVal": "B", "content": "$8$ " } ], [ { "aoVal": "C", "content": "$10$ " } ], [ { "aoVal": "D", "content": "$12$ " } ], [ { "aoVal": "E", "content": "$14$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "$$2\\times3\\times2\\times1=12$$ ways. " ]

[]

[ [ { "aoVal": "A", "content": "$$$$north " } ], [ { "aoVal": "B", "content": "$$$$east " } ], [ { "aoVal": "C", "content": "$$$$south " } ], [ { "aoVal": "D", "content": "$$$$west " } ], [ { "aoVal": "E", "content": "$$$$west-south-west " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Directions and Coordinates->Directions" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "RB, RG, RY, RP, RW BG, BY, BP, BW GY, GP, GW, YP, YW, PW " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$22$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Coloring Problems" ]

[ "Remove the two cubes in both ends, and then we can get: $[(3-2)+(4-2)+(5-2)]\\times4=24$ " ]

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[ [ { "aoVal": "A", "content": "$$86$$ " } ], [ { "aoVal": "B", "content": "$$88$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$94$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)" ]

[ "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 $$688\\div8=86$$. " ]