Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

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

[ "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 $$4\\times3\\times2\\times1=24$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{3}{10}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{7}{20}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{2}{5}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{9}{20}$ " } ] ]

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

[ "There are $8$ prime numbers in total. So, answer $= \\frac{8}{20} = \\frac{2}{5}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$39$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ], [ { "aoVal": "E", "content": "$$51$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Regular Figures->Counting Triangles" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

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

[ "All natural numbers divisible by $3$ in the range of $1-75$ are：$75\\div3=25$． All natural numbers divisible by $5$ in the range of $1-75$ are：$75\\div5=15$． The natural numbers divisible by 3 and 5 that are divisible by $15$：$75\\div15=5$． The natural numbers divisible by $3$ or $5$ are：$25+15-5=35$． " ]

[]

[ [ { "aoVal": "A", "content": "$$1001$$ " } ], [ { "aoVal": "B", "content": "$$1003$$ " } ], [ { "aoVal": "C", "content": "$$1005$$ " } ], [ { "aoVal": "D", "content": "$$1007$$ " } ], [ { "aoVal": "E", "content": "$$1009$$ " } ] ]

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

[ "The prime factors of $$209$$ are $$11$$ and $$19$$, so these must have been the reversed numbers that Miruna multiplied. The correct multiplication was $$11 \\times 91=1001$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$78$$ " } ], [ { "aoVal": "C", "content": "$$81$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ], [ { "aoVal": "E", "content": "$$87$$ " } ] ]

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

[ "Given that the product, $$15600$$, is a multiple of $$25$$, the three numbers between them must contribute two factors of $$5$$; since they are consecutive and there are only three of them, this can happen only if one number is itself a multiple of $$25$$. It is now worth observing that the product of three consecutive numbers is roughly the same as the cube of the middle of the three. Since $$20^{3}\\textless{} 15 600 \\textless{} 30^{3}$$, the middle number must lie between $$20$$ and $$30$$, hence one of the numbers must be $$25$$. The numbers can therefore be $$\\left\\textbackslash{ {23, 24, 25} \\right\\textbackslash}$$,~ $$\\left\\textbackslash{ {24, 25, 26} \\right\\textbackslash}$$ or $$\\left\\textbackslash{ {25, 26, 27} \\right\\textbackslash}$$. The product $$15600 =13 \\times1200$$, so it has factors both $$4$$ and $$13$$. The triple $$\\left\\textbackslash{ {25, 26, 27} \\right\\textbackslash}$$ has no factor of $$4$$, and $$\\left\\textbackslash{ {23, 24, 25} \\right\\textbackslash}$$ no factor of $$13$$. So, by elimination, the numbers are $$24$$, $$25$$ and $$26$$, and their total is $$75$$. " ]

[]

[ [ { "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" ]

[ "There are $6\\times6=36$ ways to roll a die twice, 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 time is greater than the second should be the same as the number of rolls where the second time 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 $\\dfrac{15+6}{36}=\\dfrac{21}{36}=\\frac{7}{12}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

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

[ "Ones: $$13, 23, 33, 43, 53, 63, 73$$ Tens: $$30, 31, 32, 33, 34, 35, 36, 37, 38, 39$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "$$12+8+8+4=32$$ $$32\\div4=8$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$-18.5$$ " } ], [ { "aoVal": "B", "content": "$$-13.5$$ " } ], [ { "aoVal": "C", "content": "$$0$$ " } ], [ { "aoVal": "D", "content": "$$13.5$$ " } ], [ { "aoVal": "E", "content": "$$18.5$$ " } ] ]

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

[ "The formula for expected values is $$ \\text { Expected Value }=\\sum(\\text { Outcome } \\cdot \\text { Probability }) . $$ We have $$ \\begin{aligned} t \\& =50 \\cdot \\frac{1}{5}+20 \\cdot \\frac{1}{5}+20 \\cdot \\frac{1}{5}+5 \\cdot \\frac{1}{5}+5 \\cdot \\frac{1}{5} \\textbackslash\\textbackslash{} \\& =(50+20+20+5+5) \\cdot \\frac{1}{5} \\textbackslash\\textbackslash{} \\& =100 \\cdot \\frac{1}{5} \\textbackslash\\textbackslash{} \\& =20, \\textbackslash\\textbackslash{} s \\& =50 \\cdot \\frac{50}{100}+20 \\cdot \\frac{20}{100}+20 \\cdot \\frac{20}{100}+5 \\cdot \\frac{5}{100}+5 \\cdot \\frac{5}{100} \\textbackslash\\textbackslash{} \\& =25+4+4+0.25+0.25 \\textbackslash\\textbackslash{} \\& =33.5 . \\end{aligned} $$ Therefore, the answer is $t-s=(\\mathbf{B})-13.5$. " ]

[]

[ [ { "aoVal": "A", "content": "Julia " } ], [ { "aoVal": "B", "content": "Kasia " } ], [ { "aoVal": "C", "content": "Susanna " } ], [ { "aoVal": "D", "content": "Helena " } ] ]

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

[ "Susanna: March 20\\textsuperscript{th} Kasia: March 1\\textsuperscript{st} Julia: July 20\\textsuperscript{th} Helena: May 17\\textsuperscript{th} " ]

[]

[ [ { "aoVal": "A", "content": "$$1001$$ " } ], [ { "aoVal": "B", "content": "$$1111$$ " } ], [ { "aoVal": "C", "content": "$$1221$$ " } ], [ { "aoVal": "D", "content": "$$1101$$ " } ] ]

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

[ "7x(10+1)x13 =(70+7)x13 =77x(10+3) =77x10+77x3 =770+231 =1001 " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{3}{4}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{6}$ " } ] ]

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

[ "The probability that both show a green bean is $\\dfrac{1}{2}\\cdot \\dfrac{1}{3}=\\dfrac{1}{6}$. " ]

[ "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": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "$$243$$ " } ] ]

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

[ "$3^{}x \\cdot 9^{}y=3^{x+2y}=3^{3}=27$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$66$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$74$$ " } ], [ { "aoVal": "D", "content": "$$77$$ " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

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

[ "The sum of $3$ odd numbers is an odd number. The sum of $4$ even numbers is an even number. The sum of an odd number and an even number is an odd number, so the total score of the $7$ students is an odd number. The total score of $8$ students is an even number, and the sum of $2$ odd numbers is an even number. Therefore, the score of the last one should be an odd number. " ]

[]

[ [ { "aoVal": "A", "content": "$$I$$ " } ], [ { "aoVal": "B", "content": "$$II$$ " } ], [ { "aoVal": "C", "content": "$$III$$ " } ] ]

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

[ "I. Both cards are in the same colour: $$ \\dfrac{5}{9}$$; II. Both cards are red: $$\\dfrac{4}{9}$$; III. Two cards are in different colours: $$\\dfrac{4}{9}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{2}{3}$ " } ], [ { "aoVal": "B", "content": "$\\frac{105}{128}$ " } ], [ { "aoVal": "C", "content": "$\\frac{125}{128}$ " } ], [ { "aoVal": "D", "content": "$\\frac{253}{256}$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

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

[ "Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals. Case 1: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. Case $2: 3$ are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either. Case $3: 2$ are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are $\\left(\\begin{array}{l}4 \\textbackslash\\textbackslash{} 2\\end{array}\\right)$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6 \\cdot 2=12$ ways to color the pentagon so that no such triangle has the same color for all of its sides. These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\\frac{12}{1024}=1-\\frac{3}{256}=\\frac{253}{256}=D$ " ]

[]

[ [ { "aoVal": "A", "content": "Amos, James " } ], [ { "aoVal": "B", "content": "James, Eugene " } ], [ { "aoVal": "C", "content": "James, Leo " } ], [ { "aoVal": "D", "content": "Amos, Eugene " } ] ]

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

[ "From clue $$2$$, James is taller than Leo and Leo is taller than Eugene. From clue $$3$$, Amos is taller than James. Rank from tallest to shortest: \\textbf{Amos}, James, Leo, \\textbf{Eugene}. " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$13$$ " } ] ]

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

[ "$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10+11+12 = 78$$, $$80 -- 78 = 2$$. Since all the numbers should be different, the remaining \\textquotesingle$$2$$\\textquotesingle{} cannot make up a new positive integer. It can only be added to the number(s) before. Therefore, there are at most $$12$$ different positive integers that can add up to $$80$$. " ]

[]

[ [ { "aoVal": "A", "content": "Indefinite events include impossible events. " } ], [ { "aoVal": "B", "content": "The probability of an impossible event to happen is $$0$$. " } ], [ { "aoVal": "C", "content": "The probability of an indefinite event to happen is between $$0$$ and $$1$$. " } ], [ { "aoVal": "D", "content": "The probability of a certain event to happen is $$1$$. " } ] ]

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

[ "Impossible events are definite events. $$\\text{B}$$, $$\\text{C}$$, and $$\\text{D}$$ are right. Thus, the answer is $$\\text{A}$$. " ]

[]

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

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

[ "The combinations of digits that give multiples of $$3$$ are $$(1$$, $$2$$, $$3)$$ and $$(2$$, $$3$$, $$4)$$. For each of them, there are $3\\times2\\times1=6$ possibilities. Thus, there are $6+6=12$ possibilities in total. The number of ways to choose three digits out of four is $$4\\times3\\times2=24$$. Therefore, the probability is $$\\frac{12}{24}=\\frac12$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "$$8+7+6+5+4+3+2+1=36$$. " ]

[]

[ [ { "aoVal": "A", "content": "white " } ], [ { "aoVal": "B", "content": "Blue " } ], [ { "aoVal": "C", "content": "Uncertain " } ] ]

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

[ "Val\\textquotesingle s point and Elvis\\textquotesingle{} points are identical, so both of them guessed incorrectly. Therefore, John guessed correctly. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "The number is a multiple of $3$, $6$ and $9$. " } ], [ { "aoVal": "B", "content": "The number is a multiple of $3$ and $6$ but not a multiple of $9$. " } ], [ { "aoVal": "C", "content": "The number is a multiple of $3$ and $9$ but not a multiple of $6$. " } ], [ { "aoVal": "D", "content": "The number is a multiple of $3$ but not a multiple of $6$. " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "The sum of the digits of 202220222022 is 2 x 9 = 18. Hence this number is divisible by 3 and 9. As its last digit is 2, it is also divisible by 6. The answer is \\textbf{Option A.} " ]

[]

[ [ { "aoVal": "A", "content": "$$\\rm~~154 cm$$ " } ], [ { "aoVal": "B", "content": "$$\\rm~~157 cm$$ " } ], [ { "aoVal": "C", "content": "$$\\rm~~162 cm$$ " } ], [ { "aoVal": "D", "content": "$$\\rm~~164 cm$$ " } ] ]

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

[ "The total height is $160\\times 3 = 480\\text{cm}$ and hence the height of both June and Linda is $480-166=314\\text{cm}$. Therefore the height of Linda is $314 \\div 2 = 157\\text{cm}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "7A = 24 + 32 7A = 56 A = 8 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac12$ " } ], [ { "aoVal": "B", "content": "$\\frac5{11}$ " } ], [ { "aoVal": "C", "content": "$\\frac3{10}$ " } ], [ { "aoVal": "D", "content": "$\\frac25$ " } ], [ { "aoVal": "E", "content": "$\\frac{6}{11}$ " } ] ]

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

[ "There are $11$ numbers in total. Among them, $4,6, 8, 9,$ and $10$ are composite numbers. " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

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

[ "Let the number of pupils who like neither subject be $$x$$. Hence the number who like both subjects is $$2x$$. Therefore the number of pupils who like only Maths is $$20−2x$$ and the number who like only English is $$18−2x$$. Since there are $$30$$ pupils in my class, we have $$\\left( 20-2x \\right)+2x+\\left( 18-2x \\right)+x=30$$ and hence $$38−x = 30$$. This has solution $$x = 8$$ and hence the number of pupils who like only Maths is $$20-2\\times 8=4$$. " ]

[]

[ [ { "aoVal": "A", "content": "Amos, James " } ], [ { "aoVal": "B", "content": "James, Eugene " } ], [ { "aoVal": "C", "content": "James, Leo " } ], [ { "aoVal": "D", "content": "Amos, Eugene " } ] ]

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

[ "From tallest to shortest: \\textbf{Amos}, James, Leo, \\textbf{Eugene}. " ]

[ "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: 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: Similarly, we have $2 * 2=4$ ways to obtain $2$ evens. The probability is $\\frac{20}{36}=\\frac{5}{9}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$114$$ " } ], [ { "aoVal": "B", "content": "$$204$$ " } ], [ { "aoVal": "C", "content": "$$239$$ " } ], [ { "aoVal": "D", "content": "$$242$$ " } ] ]

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

[ "$5 - 9: 9 - 5 + 1 = 5$ $10 - 99: 99 - 10 + 1 = 90$ $100 - 118: 118 - 100 + 1 = 19$ $5\\times1 + 90\\times2 + 19\\times3 = 5 + 180 + 57 = 242$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$70$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$127$$ " } ], [ { "aoVal": "D", "content": "$$124$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$7-7 + 16-6 + 34-4 + 45-15 +50$$ $$= 0+10+30+30+50$$ $$= 120$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ am " } ], [ { "aoVal": "B", "content": "$$4$$ am " } ], [ { "aoVal": "C", "content": "$$5$$ am " } ], [ { "aoVal": "D", "content": "$$11$$ am " } ], [ { "aoVal": "E", "content": "$$7$$ pm " } ] ]

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

[ "Eight hours behind $$11$$ am is $$3$$ am. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$3\\times3=9$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{6}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{5}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{4}$ " } ], [ { "aoVal": "D", "content": "$\\frac{1}{3}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{2}$ " } ] ]

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

[ "The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are 15 ways of choosing $2$ numbers randomly. Thus $\\frac{5}{15} = \\frac{1}{3}$. The answer is $D$. " ]

[]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

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

[ "$\\frac{6!}{3!\\times2!}=60$. " ]

[]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$96$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

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

[ "The first student has six choices of books; the second has five; and the third has four. By the Rule of product, there is a total of $$6\\times5\\times4=120$$ways. " ]

[]

[ [ { "aoVal": "A", "content": "$$960$$ " } ], [ { "aoVal": "B", "content": "$$950$$ " } ], [ { "aoVal": "C", "content": "$$940$$ " } ], [ { "aoVal": "D", "content": "$$930$$ " } ] ]

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

[ "25x38 =25x(30+8) =25x30+25x8 =750+200 =950 " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{11}{16}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{16}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{7}{16}$$ " } ] ]

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

[ "$$\\rm B$$ " ]

[]

[ [ { "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->Combinatorics->Logical Reasoning->Reasoning by Comparing" ]

[ "\"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": "$\\dfrac{1}{4}$ " } ], [ { "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 that both show a green bean is $\\dfrac{1}{2}\\times\\dfrac{1}{4}=\\dfrac{1}{8}$. The probability that both show a red bean is $\\dfrac{1}{2}\\times \\dfrac{2}{4}=\\dfrac{1}{4}$. Therefore the probability is $\\frac{1}{4}+\\frac{1}{8}=\\frac{3}{8}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac {1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac {1}{6}$ " } ], [ { "aoVal": "C", "content": "$\\frac {1}{3}$ " } ], [ { "aoVal": "D", "content": "$\\frac {1}{4}$ " } ], [ { "aoVal": "E", "content": "$\\frac {1}{12}$ " } ] ]

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

[ "There are $12$ different combinations. The product of two numbers is greater than $12$ will be $2\\times6$. Thus, the probability is $\\frac 1{12}$ . " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "1494 " } ], [ { "aoVal": "B", "content": "1490 " } ], [ { "aoVal": "C", "content": "1485 " } ], [ { "aoVal": "D", "content": "1484 " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$299 + 297 + 295 + 296 + 298$$ = $300 - 1 + 300 - 3 + 300 - 5 + 300 - 4 + 500 - 2$~ = $1485$~ " ]

[]

[ [ { "aoVal": "A", "content": "$$$$None " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "$(12+18)-(29-3)=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

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

[ "$(19\\times6+26)\\div7=140\\div7=20$ pages. " ]

[]

[ [ { "aoVal": "A", "content": "Julia " } ], [ { "aoVal": "B", "content": "Kasia " } ], [ { "aoVal": "C", "content": "Zuzanna " } ], [ { "aoVal": "D", "content": "Helena " } ] ]

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

[ "Given that Kasia and Zuzanna were born in the same month, their birth month must be March. Given that Julia and Zuzanna were born on the same day of a month, they must be born on the $$20^{\\rm th}$$. Hence, Zuzanna was born on March $$20$$; Kasia was born on March $$1$$; and Julia was born on July $$20$$. Helena is therefore the one born on May $$17$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

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

[ "$$ 1 + 1+1 + 21 = 24$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$29$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

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

[ "Since the product of the two positive integers is $$100$$, the possible pairs of integers are $$\\left( 1,100 \\right)$$, $$\\left( 2,50 \\right)$$, $$\\left( 4,25 \\right)$$, $$\\left( 5,20 \\right)$$, $$\\left( 10,10 \\right)$$, the smaller the difference, the smaller the sum, so the smallest sum is $$10+10=20$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ], [ { "aoVal": "E", "content": "$$28$$ " } ] ]

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

[ "Count layer by layer. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$64$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

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

[ "There are $3$ $C$s in total now with other $4$ letters remaining. There are $\\_4P\\_4$ ways for us to arrange the $4$ letters\\textquotesingle{} positions. So the answer is $\\_4P\\_4=24$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems" ]

[ "Consider the worst case scenario: first 3 are drawn, all of different colors, when the fourth is drawn, no matter which color is drawn, a pair of the same color will definitely be matched. Continue to draw the fifth one. The worst case is that the last color socks are drawn again, and at this point there are three socks of the same color and one sock of each of the other two colors. If you continue to draw the sixth one, no matter which color you draw, you will be able to form two pairs of socks of the same color. So the answer is to draw 6 times. " ]

[]

[ [ { "aoVal": "A", "content": "$\\frac12$ " } ], [ { "aoVal": "B", "content": "$\\frac13$ " } ], [ { "aoVal": "C", "content": "$\\frac1{12}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{3}{20}$ " } ] ]

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

[ "$\\dfrac{\\_4C\\_2 \\times~~\\_3C\\_1}{\\_{10}C\\_3}=\\dfrac{6\\times3}{120}=\\dfrac{3}{20}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Regular Figures->Counting Triangles" ]

[ "[insert pic] " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac3{16}$ " } ], [ { "aoVal": "B", "content": "$\\frac14$ " } ], [ { "aoVal": "C", "content": "$\\frac38$ " } ], [ { "aoVal": "D", "content": "$\\frac12$ " } ], [ { "aoVal": "E", "content": "$\\frac34$ " } ] ]

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

[ "In order for $a \\cdot d-b \\cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\\times2+2\\times4=12$ ways to pick numbers to obtain an even product. There are $2 \\cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \\cdot d-b \\cdot c$ odd is $2 \\cdot(12 \\cdot 4)=96.$ Thus, the answer is $\\frac{96}{256}=\\frac38$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

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

[ "The prime factorization of $12 !$ is $2^{10} \\cdot 3^{5} \\cdot 5^{2} \\cdot 7 \\cdot 11$. This yields a total of $11 \\cdot 6 \\cdot 3 \\cdot 2 \\cdot 2$ divisors of $12 !$. In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \\cdot 3 \\cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc. The probability that the divisor chosen is a perfect square is $$ \\frac{6 \\cdot 3 \\cdot 2}{11 \\cdot 6 \\cdot 3 \\cdot 2 \\cdot 2}=\\frac{1}{22} \\Longrightarrow \\frac{m}{n}=\\frac{1}{22} \\Longrightarrow m+n=1+22=\\text { (E) } 23 $$. " ]

[]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$95$$ " } ], [ { "aoVal": "C", "content": "$$94$$ " } ], [ { "aoVal": "D", "content": "$$93$$ " } ] ]

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

[ "I scored a total of $$720$$ on all $$8$$ tests. The total of $$435$$ on the first $$5$$ tests leaves a total of $$285$$ for the last $$3$$ tests, so the average is $$285\\div3 = 95$$. " ]

[ "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->Statistics and Probability" ]

[ "Rolling a pair of fair 6 -sided dice, the probability of getting a sum of 7 is $\\frac{1}{6}$ : Regardless what the first die shows, the second die has exactly one outcome to make the sum 7 . We consider the complement: The probability of not getting a sum of 7 is $1-\\frac{1}{6}=\\frac{5}{6}$. Rolling the pair of dice $n$ times, the probability of getting a sum of 7 at least once is $1-\\left(\\frac{5}{6}\\right)^{}n$. Therefore, we have $1-\\left(\\frac{5}{6}\\right)^{}n\\textgreater\\frac{1}{2}$, or $$ \\left(\\frac{5}{6}\\right)^{}n\\textless\\frac{1}{2} $$ Since $\\left(\\frac{5}{6}\\right)^{4}\\textless\\frac{1}{2}\\textless\\left(\\frac{5}{6}\\right)^{3}$, the least integer $n$ satisfying the inequality is (C) 4 . " ]

[]

[ [ { "aoVal": "A", "content": "The ball taken out must be a black ball. " } ], [ { "aoVal": "B", "content": "It is impossible that a white ball will be taken out. " } ], [ { "aoVal": "C", "content": "It is very likely that a red ball will be taken out. " } ], [ { "aoVal": "D", "content": "It is certain that a red ball will 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 coloured 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": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

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

[ "16, 27, 38, 49, 50, 61, 72, 83, 94 " ]

[]

[ [ { "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" ]

[ "There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$. How to solve the problem: The probability of getting an odd number first is $\\dfrac{4}{6}=\\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\\dfrac{2}{6}=\\dfrac{1}{3}$. Now we multiply the two fractions: $\\dfrac{2}{3}\\times\\dfrac{1}{3}=\\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\\dfrac{2}{9}\\times2=\\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\\dfrac{4}{9}=\\left (\\text{C}\\right )\\dfrac{5}{9}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "There are $$6$$ possible cases when taking $$2$$ balls from the box. In the worst case, each of the first $$6$$ students takes different balls to the others. When the $${{7}^{\\text{th}}}$$ student takes the balls, he/she would definitely take the same balls with someone among the first $$6$$ students. $$6+1=7$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$13$$ minutes " } ], [ { "aoVal": "B", "content": "$$23$$ minutes " } ], [ { "aoVal": "C", "content": "$$39$$ minutes " } ], [ { "aoVal": "D", "content": "$$47$$ minutes " } ], [ { "aoVal": "E", "content": "$$73$$ minutes " } ] ]

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

[ "The rider travels for $$13$$ hours over $$60$$ rides, which is anaverage time of $$\\frac{13}{60}$$ of an hour per ride, hence $$13$$ minutes. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac3{16}$ " } ], [ { "aoVal": "B", "content": "$\\frac14$ " } ], [ { "aoVal": "C", "content": "$\\frac38$ " } ], [ { "aoVal": "D", "content": "$\\frac12$ " } ], [ { "aoVal": "E", "content": "$\\frac34$ " } ] ]

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

[ "In order for $a \\cdot d-b \\cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\\times2+2\\times4=12$ ways to pick numbers to obtain an even product. There are $2 \\cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \\cdot d-b \\cdot c$ odd is $2 \\cdot(12 \\cdot 4)=96.$ Thus, the answer is $\\frac{96}{256}=\\frac38$ " ]

[]

[ [ { "aoVal": "A", "content": "$$22$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$37$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ] ]

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

[ "Nil " ]

[ "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 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1440$$ " } ], [ { "aoVal": "B", "content": "$$2880$$ " } ], [ { "aoVal": "C", "content": "$$5760$$ " } ], [ { "aoVal": "D", "content": "$$182440$$ " } ], [ { "aoVal": "E", "content": "$$362880$$ " } ] ]

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

[ "Since the two Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\\textquotesingle re trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $\\_5P\\_5$. Now we multiply this product by $\\_2P\\_2\\times \\_4P\\_4$~because there are $\\_2P\\_2$~ways to arrange just two Arabic books, and $\\_4P\\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $C$. " ]

[]

[ [ { "aoVal": "A", "content": "Bill, Amy, Celine " } ], [ { "aoVal": "B", "content": "Amy, Bill, Celine " } ], [ { "aoVal": "C", "content": "Celine, Amy, Bill " } ], [ { "aoVal": "D", "content": "Celine, Bill, Amy " } ], [ { "aoVal": "E", "content": "Amy, Celine, Bill " } ] ]

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

[ "If Bill is the oldest, then Amy is not the oldest, and both statements $$\\rm I$$ and $$\\rm II$$ are true, so statement $$\\rm I$$ is not the true one. If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements $$\\rm II$$ and $$\\rm III$$ are true, so statement $$\\rm II$$ is not the true one. Therefore, statement $$\\rm III$$ is the true statement, and both $$\\rm I$$ and $$\\rm II$$ are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is $$\\rm E$$: Amy, Celine, Bill. " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

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

[ "Note that the digit \\textquotesingle$$0$$\\textquotesingle{} could not be the highest digit of a four-digit number. Therefore, the smallest number is $$1502$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2009$$ " } ], [ { "aoVal": "B", "content": "$$147$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

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

[ "$$2009=7^{2}\\times 41$$, and the average of $$7$$ and $$41$$ is $$(7+41)\\div 2=24$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

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

[ "If $$m$$ and $$n$$ are positive integers, then $$mn{}\\textgreater m+n$$ unless at least one of $$m$$ or $$n$$ is equal to $$1$$, or $$m=n=2$$. So, to maximise the expression, we need to place multiplication signs between $$2$$ and $$3$$ and between $$3$$ and $$4$$. However, we need to place an addition sign between $$1$$ and $$2$$ because $$1+2\\times3\\times4=25$$, whereas $$1\\times2\\times3\\times4=24$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem" ]

[ "In the first weighing with the balance, put coins $$1$$, $$2$$, $$3$$, and $$4$$ on one end of the balance and coins $$5$$, $$6$$, $$7$$, and $$8$$ on the other end of the balance. The balance has two situations: balanced or not. Analyze the situation of the balance: if balanced, the fake coin is among the remaining $$4$$ coins. In the second weighing with the balance, randomly take $$3$$ coins from coin $$1$$ to coin $$8$$ and put them on the left end of the balance and randomly take $$3$$ coins from coin $$9$$ to coin $$12$$ and put them on the right end of the balance (such as $$9$$, $$10$$, $$11$$). The balance also has two situations: balanced or not. If balanced, coin $$12$$ is the coin of different weight. In the third weighing with the balance, comparing No. $$12$$ coin with any other coin, whether the coin is lighter or heavier can be known. If not, it can be known that the coin of different weight is among the three coins $$9$$, $$10$$, and $$11$$, and that whether it is lighter or heavier than other coins can be known. In the third weighing with the balance, randomly take two of the coins (such as $$9$$ and $$10$$) and put them on the both ends of the balance. If balanced, the remaining coin (coin $$11$$) is the one we are looking for; if not, based on the previous judgement that whether the coin is lighter or heavier, it can be determined that which one of the coins on the balance is what we are looking for. Analyze the first imbalanced situation as follows: There are two situations: the right end weighs more or the left end weighs more. Assume the left end weighs more (which is the same for the situation that the right end weighs more.) In the second weighing with the balance, take off $$3$$ coins randomly from the left end (such as $$1$$, $$2$$, and $$3$$) and move $$3$$ coins from the right end to the left end (such as $$5$$, $$6$$, and $$7$$), then take $$3$$ coins randomly from the $$4$$ coins left in the first weighing (such as $$9$$, $$10$$, and $$11$$) to the right end, and there sees $$3$$ situations for the balance: ① the left end weighs more, ② the two ends strike a balance, ③ the right end weighs more. Analyze the situations one by one as follows: ① If the left end weighs more, the coin we are looking for must be coin $$4$$ or coin $$8$$. In the third time weighing with the balance, take one of the coins (such as coin $$4$$) and put it on the left end of the balance. Randomly take one of the remaining $$10$$ coins and put it on the right end. There are also $$3$$ situations. $$a$$: If balanced, coin $$8$$ is the one we are looking for. Based on the result of using the balance for the second time, it is known that the coin weighs less than other coins. $$b$$: If the left end weighs more, coin $$4$$ is the one we are looking for and it weighs more than other coins. $$c$$: If the right end weighs more, coin $$4$$ is the one we are looking for and it weighs less than other coins. ② If the two ends strike a balance, the coin we are looking for is among the three coins ($$1$$, $$2$$, and $$3$$) taken from the left end. Since the left end weighs more in the first weighing, it is known that the coin weighs more than other coins. The following analysis is the same as previous one and will not be repeated. ③ If the right end weighs more, the coin we are looking for is among the three coins ($$5$$, $$6$$, and $$7$$) moved from the right end to the left end. Based on the results of weighing in the first two times (the left end weighs more in the first weighing and the right end weighs more in the second weighing), it is known that the coin weighs less than other coins. The following analysis is the same. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac {1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac {1}{6}$ " } ], [ { "aoVal": "C", "content": "$\\frac {1}{3}$ " } ], [ { "aoVal": "D", "content": "$\\frac {1}{4}$ " } ], [ { "aoVal": "E", "content": "$\\frac {1}{12}$ " } ] ]

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

[ "There are $12$ different combinations. The product of two numbers is greater than $20$ will be $4\\times5$ and $4\\times6$. Thus, the probability is $\\frac 2{12}$ = $\\frac 16$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

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

[ "$$ 1 + 1+1 + 21 = 24$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$240$$ " } ], [ { "aoVal": "B", "content": "$$180$$ " } ], [ { "aoVal": "C", "content": "$$64$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ], [ { "aoVal": "E", "content": "$$140$$ " } ] ]

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

[ "There are $3$ $C$s in total now with other $4$ letters remaining. There are $\\_4P\\_4$ ways for us to arrange the $4$ letters\\textquotesingle{} positions. So the answer is $\\_4P\\_4\\times \\_5C\\_3 =240$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

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

[ "$$3\\times 3\\times 3=27$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

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

[ "I have $$5$$ sweaters and $$6$$ pairs of pants. For each sweater, there are $$6$$ pairs of pants with which that sweater can be paired. There are $$5$$ sweaters, so there are $$5\\times6=30$$ different possible outfits. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "$27-20=7$ $7+7=14$ $20-14=6$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac {1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac {1}{3}$ " } ], [ { "aoVal": "C", "content": "$\\frac {1}{4}$ " } ], [ { "aoVal": "D", "content": "$\\frac {1}{5}$ " } ], [ { "aoVal": "E", "content": "$\\frac {1}{6}$ " } ] ]

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

[ "There are $2\\times6=12$ different combinations. The product of two numbers is smaller than $8$ will be $5\\times1$ and $6\\times1$. Thus, the probability is $\\frac 2{12}$ = $\\frac 16$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

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

[ "There are $4$ $A$s in total, which have $3$ intervals leaving for the other $3$ letters. Thus, the answer is $\\_3P\\_3=6$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

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

[ "Total number of days $$=31-17 +1= 15$$ days. " ]

[ "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" ]

[ "$$5 \\times 2 = 10$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$150$$ " } ], [ { "aoVal": "C", "content": "$$800$$ " } ], [ { "aoVal": "D", "content": "$$990$$ " } ], [ { "aoVal": "E", "content": "$$1200$$ " } ] ]

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

[ "D " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "$$4 \\times 3 \\times 1 \\times 2 \\times 1 = 24$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3569$$ " } ], [ { "aoVal": "B", "content": "$$5369$$ " } ], [ { "aoVal": "C", "content": "$$5396$$ " } ], [ { "aoVal": "D", "content": "$$5639$$ " } ], [ { "aoVal": "E", "content": "$$5936$$ " } ] ]

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

[ "When put in order, the numbers are: $$3569$$, $$3596$$, $$3659$$, $$3695$$, $$3956$$, $$3965$$, $$5369$$, $$5396$$, $$5639$$, $$5693$$, $$\\ldots $$ " ]

[]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Dotted Line Arrangement" ]

[ "Each pair has $$2$$ intersection points. The $$6$$ pairs have $$12$$ such points. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1800$$ " } ], [ { "aoVal": "B", "content": "$$1200$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$720$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

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

[ "There are $5$ $3$s in total now with other $5$ digits remaining. There are $\\_5P\\_5$ ways for us to arrange the $5$ letters\\textquotesingle{} positions. Then, we can put the $5$ $3$s in the $6$ intervals. So the answer is $\\_5P\\_5 \\times \\_6C\\_5=720$. " ]

[]

[ [ { "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 " } ] ]

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

[ "omitted " ]

[]

[ [ { "aoVal": "A", "content": "$\\frac12$ " } ], [ { "aoVal": "B", "content": "$\\frac13$ " } ], [ { "aoVal": "C", "content": "$\\frac1{12}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{6}$ " } ], [ { "aoVal": "E", "content": "$\\frac14$ " } ] ]

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

[ "$\\dfrac{\\_5C\\_2 \\times~~\\_2C\\_1}{\\_{10}C\\_3}=\\dfrac{10\\times2}{120}=\\dfrac{2}{12}=\\dfrac{1}{6}$. " ]

[]

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

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

[ "$$13$$、$$10$$、$$31$$、$$30$$， notice that $$0$$ cannot be in the first place, so only $$4$$ numbers can be formed． So the answer is $$\\text{B}$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$1680$$ " } ], [ { "aoVal": "B", "content": "$$4096$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$256$$ " } ] ]

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

[ "$$8\\times 7\\times 6\\times 5=1680$$. " ]

[]

[ [ { "aoVal": "A", "content": "The card numbered $$1$$ is not in box $$Q$$ " } ], [ { "aoVal": "B", "content": "Four cards in box $$Q$$ have even numbers on " } ], [ { "aoVal": "C", "content": "The card numbered $$5$$ is in box $$Q$$ " } ], [ { "aoVal": "D", "content": "The card numbered $$2$$ is in box $$Q$$ " } ], [ { "aoVal": "E", "content": "Exactly three cards in box $$Q$$ have odd numbers on " } ] ]

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

[ "Note first that the sum of the numbers on the eight cards is $$36$$. Therefore the sum of the numbers on the cards in each of the boxes is $$18$$. There are only three cards in box $$P$$ and hence the possible combinations for the numbers on the cards in box $$P$$ are $$\\left( 8,7,3 \\right)$$, $$\\left( 8,6,4 \\right)$$ and $$\\left( 7,6,5 \\right)$$ with the corresponding combinations for box $$Q$$ being $$\\left( 6,5,4,2,1 \\right)$$, $$\\left( 7,5,3,2,1 \\right)$$ and $$\\left( 8,4,3,2,1 \\right)$$. The only statement which is true for all three possible combinations for box $$Q$$ is that the card numbered $$2$$ is in box $$Q$$. Hence the only statement which must be true is statement $$\\rm D$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems" ]

[ "$$7+4+2=13$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{4}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{3}$ " } ], [ { "aoVal": "C", "content": "$\\frac{3}{8}$ " } ], [ { "aoVal": "D", "content": "$\\frac{1}{2}$ " } ], [ { "aoVal": "E", "content": "$\\frac{2}{3}$ " } ] ]

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

[ "The probability that both show a green bean is $\\frac{1}{2} \\cdot \\frac{1}{4}=\\frac{1}{8}$. The probability that both show a red bean is $\\frac{1}{2} \\cdot \\frac{2}{4}=\\frac{1}{4}$. Therefore the probability is $\\frac{1}{4}+\\frac{1}{8}=\\left(\\right.$ C) $\\frac{3}{8}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac1{11}$ " } ], [ { "aoVal": "B", "content": "$\\frac4{11}$ " } ], [ { "aoVal": "C", "content": "$\\frac7{11}$ " } ], [ { "aoVal": "D", "content": "$\\frac{10}{11}$ " } ], [ { "aoVal": "E", "content": "$\\frac2{11}$ " } ] ]

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

[ "There are $4$ numbers greater than $7$. Thus, the probability is $\\frac4{11}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$14$ " } ], [ { "aoVal": "B", "content": "$30$ " } ], [ { "aoVal": "C", "content": "$48$ " } ], [ { "aoVal": "D", "content": "$80$ " } ], [ { "aoVal": "E", "content": "$90$ " } ] ]

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

[ "$5\\times3\\times6=90$ " ]

[]

[ [ { "aoVal": "A", "content": "The sum of dots is $$12$$. " } ], [ { "aoVal": "B", "content": "The sum of dots is smaller than $$3$$. " } ], [ { "aoVal": "C", "content": "The sum of dots is larger than $$4$$ but smaller than $$8$$. " } ], [ { "aoVal": "D", "content": "The sum of dots is $$13$$. " } ] ]

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

[ "The maximum sum is $$6+6=12$$, so \"the sum of dots is $$13$$\" is an impossible event; so $$\\text{D}$$ is the answer. " ]