Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2020年希望杯二年级竞赛模拟第30题", "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$13$$ " } ] ]

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

[ "$13\\div5=2R3$ There are $3$ odd numbers in each cycle. $3\\times2+2=8$ " ]

[]

[ [ { "aoVal": "A", "content": "$$9180$$ " } ], [ { "aoVal": "B", "content": "$$9090$$ " } ], [ { "aoVal": "C", "content": "$$9000$$ " } ], [ { "aoVal": "D", "content": "$$8910$$ " } ], [ { "aoVal": "E", "content": "$$8190$$ " } ] ]

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

[ "For a number to be a multiple of $$45$$ it must be a multiple of $$5$$ and also of $$9$$. In order to be a multiple of $$5$$, a number\\textquotesingle s units digit must be $$0$$ or $$5$$. However, the units digit of a palindromic number cannot be $$0$$, so it may be deduced that any palindromic number which is a multiple of $$45$$ both starts and ends in the digit $$5$$. In order to make the desired number as large as possible, its second digit should be $$9$$ and for it to be as small as possible its second digit should be $$0$$. So, if possible, the numbers required are of the form \\textquotesingle$$59x95$$\\textquotesingle{} and \\textquotesingle$$50y05$$\\textquotesingle{} . In addition, both numbers are to be multiples of $$9$$ which means the sum of the digits of both must be a multiple of $$9$$. For this to be the case, $$x=8$$ and $$y=8$$ , giving digit sums of $$36$$ and $$18$$ respectively. So the two required palindromic numbers are $$59895$$ and $$50805$$. Their difference is $$9090$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Combinatorics->Math Game->Sudoku" ]

[ "$$(70-1)\\div (1+4)=17\\cdots 1$$ $$A$$ takes one bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0.615$$ " } ], [ { "aoVal": "B", "content": "$$6.0015$$ " } ], [ { "aoVal": "C", "content": "$$6.015$$ " } ], [ { "aoVal": "D", "content": "$$6.105$$ " } ], [ { "aoVal": "E", "content": "$$6.15$$ " } ] ]

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

[ "$$6$$ ones $=6$ $$15$$ thousandths $=$ $1$ hundredth and $5$ thousandths $=0.015$ $\\textasciitilde$ Total $=6.015$ " ]

[]

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

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

[ "The number of students behind Mike plus the position of Mike equals the total number of students in the class. So there are $$13$$ students behind Mike: $$20-7=13$$. However Mike is not included in these $$13$$ students, which means Mike is the $$14$$\\textsuperscript{th} student counting from back to front. " ]

[]

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

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

[ "$\\dfrac{5}{15}=\\dfrac{1}{3}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

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

[ "Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$2\\frac{1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac{17}{36}$ " } ], [ { "aoVal": "C", "content": "$\\frac{73}{36}$ " } ], [ { "aoVal": "D", "content": "$\\frac{14}{9}$ " } ] ]

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

[ "Sum $=\\frac{5}{4}+\\frac{7}{9}=\\frac{73}{36}$ Difference $=\\frac{5}{4}-\\frac{7}{9}=\\frac{17}{36}$ Difference (Sum and difference) $=\\frac{73}{36}-\\frac{17}{36}=\\frac{56}{36}=\\frac{14}{9}=1\\frac{5}{9}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{36}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{18}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{6}$ " } ], [ { "aoVal": "D", "content": "$\\frac{11}{36}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{3}$ " } ] ]

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

[ "This is equivalent to asking for the probability that at least one of the numbers is a multiple of 5 , since if one of the numbers is a multiple of 5 , then the product with it and another integer is also a multiple of 5 , and if a number is a multiple of 5 , then since 5 is prime, one of the factors must also have a factor of 5 , and 5 is the only multiple of 5 on a die, so one of the numbers rolled must be a 5 . To find the probability of rolling at least one 5 , we can find the probability of not rolling a 5 and subtract that from 1, since you either roll a 5 or not roll a 5 . The probability of not rolling a 5 on either dice is $\\left(\\frac{5}{6}\\right)\\left(\\frac{5}{6}\\right)=\\frac{25}{36}$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of 5 , is $1-\\frac{25}{36}=\\frac{11}{36}, \\text{D}$ " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{10}$$. " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{5}$$. " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{6}$$. " } ], [ { "aoVal": "D", "content": "$$\\frac{13}{30}$$. " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{3}$$ " } ] ]

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

[ "Prime numbers from $$1$$ to $$30$$: $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$29$$ Thus, there are $$10$$ prime numbers from $$1$$ to $$30$$. Therefore, the probability that the number she chooses is a prime number is $$\\frac{10}{30}=\\frac{1}{3}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ] ]

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

[ "The product of two whole numbers is $$30$$. If the numbers are $$5$$ and $$6$$, their sum is $$5+6=11$$. " ]

[]

[ [ { "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": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

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

[ "$$6, 16, 26, 36, 46$$ $$50$$ numbers minus $$5$$ numbers -\\/-\\textgreater{} $$45$$ numbers " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "There are $3$ numbers that can be formed between $1000$ and $1999$, and $3\\times2\\times1=6$ numbers that can be formed with $2$ in the first place. So the answer is $3+6=9$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

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

[ "There are only 2 people who can go in the driver\\textquotesingle s seat-Bonnie and Carlo. Any of the 3 remaining people can go in the front passenger seat. There are 2 people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2 \\cdot 3 \\cdot 2$ or 12 ways. The answer is then (D) 12 . " ]

[]

[ [ { "aoVal": "A", "content": "$$21\\text{st}$$ " } ], [ { "aoVal": "B", "content": "$$60\\text{th}$$ " } ], [ { "aoVal": "C", "content": "$$70\\text{th}$$ " } ], [ { "aoVal": "D", "content": "$$80\\text{th}$$ " } ] ]

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

[ "The sum of the digits of $$1993$$ is $$1+9+9+3$$, or $$22$$. At some time in the future, the sum of the digits of a year will be $$33$$. Since $$9+9+9 = 27$$, this will \\emph{first} occur in $$6999$$, the $$70\\text{th}$$ century. " ]

[]

[ [ { "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": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "The possible groups of three integers with product $$36$$ are $$(1,1,36)$$, $$(1,2,18)$$, $$(1,3,12)$$, $$(1,4,9)$$, $$(1,6,6)$$, $$(2,2,9)$$, $$(2,3,6)$$ and $$(3,3,4)$$ with sums $$38$$, $$21$$, $$16$$, $$14$$, $$13$$, $$13$$, $$11$$ and $$10$$ respectively. The only value for the sum that occurs twice is $$13$$. Hence, since Topaz does not know what the three integers chosen are, the sum of Harriet\\textquotesingle s three integers is $$13$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

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

[ "16,25,34,43,52,61,70 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "$3\\times6=18$, $3+6=9$ $2\\times9=18$, $9+2=11$ Thus, the two digits of the number is $9$ and $2$. The difference is $9-2=7$. " ]

[]

[ [ { "aoVal": "A", "content": "$$11.30\\text{pm}$$ " } ], [ { "aoVal": "B", "content": "$$12.30\\text{pm}$$ " } ], [ { "aoVal": "C", "content": "$$4.30\\text{pm}$$ " } ], [ { "aoVal": "D", "content": "$$5.30\\text{pm}$$ " } ], [ { "aoVal": "E", "content": "$$11.30\\text{pm}$$ " } ] ]

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

[ "Each hour, the hour hand turns $$30{}^{}\\circ$$, so it will take $$5\\frac{1}{2}$$ hours to turn $$165{}^{}\\circ$$; the time will therefore be $$4.30\\text{pm}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{61}{216}$ " } ], [ { "aoVal": "B", "content": "$\\frac{367}{1296}$ " } ], [ { "aoVal": "C", "content": "$\\frac{41}{144}$ " } ], [ { "aoVal": "D", "content": "$\\frac{185}{648}$ " } ], [ { "aoVal": "E", "content": "$\\frac{11}{36}$ " } ] ]

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

[ "Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a 5 is just the probability that the highest roll in the first round is 5 . Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P($ No one rolls a 6$)-P($ No one rolls a 5 or 6$)=P($ The highest roll is a 5$)$, so $$ \\begin{gathered} P(\\text { No one rolls a } 6)=\\left(\\frac{5}{6}\\right)^{4}, \\textbackslash\\textbackslash{} P(\\text { No one rolls a } 5 \\text { or } 6)=\\left(\\frac{2}{3}\\right)^{4}, \\textbackslash\\textbackslash{} P(\\text { The highest roll is a } 5)=\\left(\\frac{5}{6}\\right)^{4}-\\left(\\frac{4}{6}\\right)^{4}=\\frac{5^{4}-4^{4}}{6^{4}}=\\frac{369}{1296}=(\\text { C }) \\frac{41}{144} \\text {. } \\end{gathered} $$ " ]

[]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

[ "Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding a Whole Given a Part and the Percentage", "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$\\frac{1}{7}$ of her keychains $=15-8=7$ $\\textasciitilde$ Given to friends $=\\frac{6}{7}$ of her keychains $=7\\times6$ $=42$ " ]

[]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

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

[ "$4\\times4\\times3=48$ " ]

[]

[ [ { "aoVal": "A", "content": "$$10\\times 16$$ " } ], [ { "aoVal": "B", "content": "$$8\\times 10$$ " } ], [ { "aoVal": "C", "content": "$$8\\times 24$$ " } ], [ { "aoVal": "D", "content": "$$16\\times 24$$ " } ] ]

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

[ "Ann\\textquotesingle s awake $$16$$ hours each day. In $$10$$ days, that\\textquotesingle s $$(10\\times 16)$$ hours. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac29$ " } ], [ { "aoVal": "B", "content": "$\\dfrac49$ " } ], [ { "aoVal": "C", "content": "$\\dfrac59$ " } ], [ { "aoVal": "D", "content": "$\\dfrac69$ " } ], [ { "aoVal": "E", "content": "$\\dfrac79$ " } ] ]

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

[ "$1-\\dfrac49=\\dfrac59$ " ]

[ "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 remaining $4$ letters\\textquotesingle{} positions. So the answer is $\\_4P\\_4\\times \\_5C\\_3 =240$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{61}{216}$ " } ], [ { "aoVal": "B", "content": "$\\frac{367}{1296}$ " } ], [ { "aoVal": "C", "content": "$\\frac{41}{144}$ " } ], [ { "aoVal": "D", "content": "$\\frac{185}{648}$ " } ], [ { "aoVal": "E", "content": "$\\frac{11}{36}$ " } ] ]

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

[ "Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a $5$ is just the probability that the highest roll in the first round is $5$. Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P$ (No one rolls a $6$)-$P$ (No one rolls a $5$ or $6$)$=P$ (The highest roll is a $5$), so $$ \\begin{gathered} P(\\text {No one rolls a } 6)=\\left(\\frac{5}{6}\\right)^{4}, \\textbackslash\\textbackslash{} P(\\text {No one rolls a } 5 \\text { or } 6)=\\left(\\frac{2}{3}\\right)^{4}, \\textbackslash\\textbackslash{} P(\\text {The highest roll is a } 5)=\\left(\\frac{5}{6}\\right)^{4}-\\left(\\frac{4}{6}\\right)^{4}=\\frac{5^{4}-4^{4}}{6^{4}}=\\frac{369}{1296}=(\\text {C}) \\frac{41}{144} \\text {. } \\end{gathered} $$ " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "$$45$$ " } ] ]

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

[ "$$1+2+3+4+5+6+7+8+9=45$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

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

[ "Method 1: $104, 113, 122, 131, 140$; $203, 212, 221, 230$; $302, 311, 320$; $401, 410$; $500$ $5+4+3+2+1=15$ Method 2:$5=0+0+5=0+1+4=0+2+3=1+1+3=1+2+2$ When the three-digit number is formed by $0$, $0$, $5$, it can only be $500$, therefore $1$ number; when the three-digit number is formed by $0$, $1$, $4$, it can be $104$, $140$, $401$ or $410$, therefore $4$ numbers; when the three-digit number is formed by $0$, $2$, $3$, it can be $203$, $230$, $302$, or $320$, therefore $4$ numbers； when the three-digit number is formed by $1$, $1$, $3$, it can be $113$, $131$, or $311$, therefore $3$ numbers; when the three-digit number is formed by $1$, $2$, $2$, it can be $122$, $212$, or $221$, therefore $3$ numbers; in total there are $1+4+4+3+3=15$ numbers that meet the requirement of the question. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$500$$ " } ], [ { "aoVal": "B", "content": "$$505$$ " } ], [ { "aoVal": "C", "content": "$$507$$ " } ], [ { "aoVal": "D", "content": "$$509$$ " } ] ]

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

[ "$97+101+102+104+105=100\\times5-3+1+2+4+5=509$ " ]

[]

[ [ { "aoVal": "A", "content": "$72$ , $24$ " } ], [ { "aoVal": "B", "content": "$96$ , $24$ " } ], [ { "aoVal": "C", "content": "$72$ , $48$ " } ], [ { "aoVal": "D", "content": "$96$ , $48$ " } ] ]

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

[ "①$$3\\times 4\\times 3\\times 2\\times 1=72$$ , ②$$2\\times 4\\times 3\\times 2\\times 1=48$$ . " ]

[]

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

[]

[ [ { "aoVal": "A", "content": "The probability of a certain event to happen is $$1$$. " } ], [ { "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": "Indefinite events include impossible events. " } ] ]

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

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

[]

[ [ { "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 \\#$$2$$. " ]

[]

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

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

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

[ "$$3\\times2\\times 3=18$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

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

[ "$$2\\times2\\times2=8$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$65$$ " } ], [ { "aoVal": "C", "content": "$$95$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ], [ { "aoVal": "E", "content": "$$105$$ " } ] ]

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

[ "Each floor has $$35$$ rooms. On every floor except floor $$2$$, the digit $$2$$ will be used for rooms \\textquotesingle$$n02$$\\textquotesingle, \\textquotesingle$$n12$$\\textquotesingle, \\textquotesingle$$n20$$\\textquotesingle{} to \\textquotesingle$$n29$$\\textquotesingle{} (including \\textquotesingle$$n22$$\\textquotesingle) and \\textquotesingle$$n32$$\\textquotesingle. Hence the digit $$2$$ will be used $$14$$ times on each floor except floor $$2$$. On floor $$2$$, the digit $$2$$ will be used an extra $$35$$ times as the first digit of the room number. Therefore the total number of times the digit $$2$$ will be used is $$5\\times14+35 = 105$$. " ]

[]

[ [ { "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": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$46$$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

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

[ "$$6, 16, 26, 36, 46$$ $$51$$ numbers minus $$5$$ numbers -\\/-\\textgreater{} $$45$$ numbers " ]

[]

[ [ { "aoVal": "A", "content": "$$\\dfrac{3}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{2}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{3}{5}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{7}{10}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of 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 red chip. The answer is $$\\dfrac{4}{10}=\\dfrac{2}{5}$$, or $$\\boxed { (\\text{B})\\dfrac{2}{5}}$$. " ]

[ "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": "Random event " } ], [ { "aoVal": "B", "content": "Certain event " } ], [ { "aoVal": "C", "content": "Impossible event " } ], [ { "aoVal": "D", "content": "None of the above " } ] ]

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

[ "∵$$3+4=7$$, ∴ Three line segments measuring $$3$$, $$4$$, and $$7$$ cannot form a triangle, ∴ Using three line segments measuring $$3$$, $$4$$, and $$7$$ to form a triangle is an impossible event. So $$\\text{C}$$ is the answer. " ]

[]

[ [ { "aoVal": "A", "content": "1，25 " } ], [ { "aoVal": "B", "content": "1，30 " } ], [ { "aoVal": "C", "content": "2，25 " } ], [ { "aoVal": "D", "content": "2，30 " } ] ]

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

[ "1h 30min " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

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

[ "If the ones digit is $$2$$, the $$2$$ numbers are $$202$$ and $$112$$. For each ones digit, the number of possible numbers is the same as the ones digit. In all, there are $$1+2+3+\\cdots+8+9 =45$$ numbers. " ]

[]

[ [ { "aoVal": "A", "content": "Edwin " } ], [ { "aoVal": "B", "content": "Fred " } ], [ { "aoVal": "C", "content": "Gary " } ], [ { "aoVal": "D", "content": "Howard " } ] ]

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

[ "Either Fred or Howard must be lying since what they said did not tally. Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase. " ]

[]

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

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

[ "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 $$75 + 35 = 110$$. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers->Complex Forming Numbers (with special requirements)" ]

[ "17, 28, 39 " ]

[]

[ [ { "aoVal": "A", "content": "$$8:45$$ A.M. " } ], [ { "aoVal": "B", "content": "$$9:45$$ A.M. " } ], [ { "aoVal": "C", "content": "$$10:45$$ A.M. " } ], [ { "aoVal": "D", "content": "$$4:45$$ P.M. " } ] ]

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

[ "One hour ago the time was $$12:15$$ P.M.; two hours ago the time was $$11:15$$ A.M.; three hours ago the time was $$10:15$$ A.M. Three and one-half hours ago the time was $$9:45$$ A.M. " ]

[]

[ [ { "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{9}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{6}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{1}{4}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "E", "content": "$\\dfrac{1}{2}$ " } ] ]

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

[ "There are $3 \\times 2 \\times 1 = 6$ ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is (B) $\\frac{1}{6}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "Since the balls and baskets are all identical, the order does not matter. $$6=1+1+4$$ $$6=1+2+3$$ $$6=2+2+2$$ So, there are $3$ ways to place them. " ]

[]

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

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

[ "Half the numbers from $1$ to $40$ are greater than $20$, Thus, the probability of choosing numbers greater than $20$ is $\\dfrac{1}{2}$. " ]

[]

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

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

[ "Consider the case where two opposite faces are coloured red. Whichever of the four remaining faces is also coloured red, the resulting arrangement is equivalent under rotation to a cube with top, bottom and front faces coloured red. Hence, there is only one distinct colouring of a cube consisting of three red and three blue faces with two opposite faces coloured red. Now consider the case where no two opposite faces are coloured red. This is only possible when the three red faces share a common vertex and, however these faces are arranged, the resulting arrangement is equivalent under rotation to a cube with top, front and right-hand faces coloured red. Hence there is also only one distinct colouring of a cube consisting of three red and three blue faces in which no two opposite faces are coloured red. Therefore there are exactly two different colourings of the cube as described in the question. " ]

[]

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

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

[ "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. " ]

[ "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 that 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 $C$. " ]

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

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

[ "$$6$$, $$3$$, $$9$$, $$1$$, $$4$$, $$8$$, $$2$$, $$10$$, $$5$$. Only $$7$$, which is a prime number and doesn\\textquotesingle t have multiples except itself among $$1$$\\textasciitilde$$10$$, doesn\\textquotesingle t appear. " ]

[]

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

[]

[ [ { "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": "$$50$$ " } ], [ { "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": "$$69$$ " } ], [ { "aoVal": "B", "content": "$$78$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ], [ { "aoVal": "E", "content": "$$49$$ " } ] ]

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

[ "To divide $15+1+1=17$ into three groups, there are $\\_{16}C\\_{2}=120$ ways in total. But, since all digits cannot be larger than $9$, except the situations that don\\textquotesingle t meet the requirements, there are $120-\\_7C\\_2-\\_6C\\_2\\times 2=69$ numbers that match the condition. " ]

[]

[ [ { "aoVal": "A", "content": "$$753$$ " } ], [ { "aoVal": "B", "content": "$$861$$ " } ], [ { "aoVal": "C", "content": "$$951$$ " } ], [ { "aoVal": "D", "content": "$$1086$$ " } ], [ { "aoVal": "E", "content": "$$1110$$ " } ] ]

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

[ "If we want to get the sum as large as possible, the hundred digits for both numbers should be as large as possible. Therefore, they should be $$5$$ and $$4$$. The sum of the hundreds digits is $$9$$. For the same reason, the sum of the tens digits should be $$3+2=5$$ and the sum of the ones digits should be $$1 + 0 = 1$$. Therefore, $$\\rm C$$ is the correct answer. " ]

[]

[ [ { "aoVal": "A", "content": "$\\frac 1{10}$ " } ], [ { "aoVal": "B", "content": "$\\frac 3{10}$ " } ], [ { "aoVal": "C", "content": "$\\frac 2{5}$ " } ], [ { "aoVal": "D", "content": "$\\frac 1{2}$ " } ], [ { "aoVal": "E", "content": "$\\frac 3{5}$ " } ] ]

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

[ "$3, 6, 15,$ and $18$ are multiples of $3$. Thus, the probability is $\\frac 2{5}$. " ]

[]

[ [ { "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" ]

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

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

[]

[ [ { "aoVal": "A", "content": "$$100$$ cm " } ], [ { "aoVal": "B", "content": "$$105$$ cm " } ], [ { "aoVal": "C", "content": "$$110$$ cm " } ], [ { "aoVal": "D", "content": "$$111$$ cm " } ], [ { "aoVal": "E", "content": "$$125$$ cm " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Extreme Value in Inclusion-Exclusion for Multi-sets" ]

[ "$$131-21=110$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$68$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ], [ { "aoVal": "E", "content": "$$32$$ " } ] ]

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

[ "$24+24\\div2-4=32$ " ]

[]

[ [ { "aoVal": "A", "content": "$72$ , $24$ " } ], [ { "aoVal": "B", "content": "$96$ , $24$ " } ], [ { "aoVal": "C", "content": "$72$ , $48$ " } ], [ { "aoVal": "D", "content": "$96$ , $48$ " } ] ]

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

[ "①$$3\\times 4\\times 3\\times 2\\times 1=72$$ , ②$$2\\times 4\\times 3\\times 2\\times 1=48$$ . " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$130$$ " } ], [ { "aoVal": "B", "content": "$$1300$$ " } ], [ { "aoVal": "C", "content": "$$260$$ " } ], [ { "aoVal": "D", "content": "$$2600$$ " } ] ]

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

[ "$$=13\\times(21+46+33)=1300$$ " ]

[]

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

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

[ "$$(70-1)\\div (1+4)=13\\cdots 4$$ $$A$$ takes four bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{10}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{12}$$ " } ] ]

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

[ "∵ There are $$10$$ possible situations. However, there\\textquotesingle s only one time for him to get the first digit right at the first try. ∴ The probability that he can get the first digit right at the first try is $$\\frac{1}{10}$$. So $$\\text{D}$$ is the answer. " ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$190$$ " } ], [ { "aoVal": "B", "content": "$$290$$ " } ], [ { "aoVal": "C", "content": "$$390$$ " } ], [ { "aoVal": "D", "content": "$$490$$ " } ] ]

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

[ "$56+127+34+73=(56+34)+(127+73)=90+200=290$ " ]

[]

[ [ { "aoVal": "A", "content": "$$8.10\\text{p}.\\text{m}.$$ " } ], [ { "aoVal": "B", "content": "$$8.50\\text{p}.\\text{m}.$$ " } ], [ { "aoVal": "C", "content": "$$9.10\\text{p}.\\text{m}.$$ " } ], [ { "aoVal": "D", "content": "$$9.50\\text{p}.\\text{m}.$$ " } ], [ { "aoVal": "E", "content": "$$1.46\\text{p}.\\text{m}.$$ " } ] ]

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

[ "$$8.50\\text{p}.\\text{m}.$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$33$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "$\\_4C\\_2\\times3+\\_6C\\_2=33$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$72$$ " } ], [ { "aoVal": "B", "content": "$$108$$ " } ], [ { "aoVal": "C", "content": "$$140$$ " } ], [ { "aoVal": "D", "content": "$$144$$ " } ] ]

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

[ "Even number between 30 and 40: 32,34,36,38 $$32+34+36+38=140$$ " ]

[]

[ [ { "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": "$$5$$ " } ], [ { "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" ]

[ "Two matching pairs of socks could be obtained by choosing four socks, but this is not certain. For instance, two of one colour, one of a second colour and one of a third colour could be drawn. Combinations of five chosen socks would give two matching pairs unless three socks of one colour, one of a second colour and one of a third colour were chosen. When this is the case, drawing a sixth sock would guarantee that there would be two matching pairs as there would now be either four socks of one colour plus two other socks or three socks of one colour, two of a second colour plus one of a third colour. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$68$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$65$$ " } ], [ { "aoVal": "D", "content": "$$63$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Figures Formed by Grid Points" ]

[ "Each small square is 36/9 = 4 dm on a side. there are 60/4 = 15 squares along the length. Both dimensions, 15 and 9, are odd. Also the corners are suns, so there is one more sun than there are moons. Thus, there are a total of (15 x 9 - 1)/2 = 67 moons in the carpet when fully rolled. " ]

[]

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

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

[ "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 also sees $$3$$ 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": "$$\\frac{3}{10}$$ " } ], [ { "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": "$$90 $$ " } ], [ { "aoVal": "B", "content": "$$ 100 $$ " } ], [ { "aoVal": "C", "content": "$$ 110 $$ " } ], [ { "aoVal": "D", "content": "$$120 $$ " } ], [ { "aoVal": "E", "content": "$$130$$ " } ] ]

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

[ "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": "Box A " } ], [ { "aoVal": "B", "content": "Box B " } ], [ { "aoVal": "C", "content": "Box C " } ] ]

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

[ "If Box $$\\rm A$$ has the wrong label, it contains blue marbles. This indicates that Box $$\\rm B$$ has the wrong label too, because only one box has blue marbles. It also means that Box $$\\rm C$$ has the wrong label. We will then have a scenario of three wrong labels, so Box $$\\rm A$$ cannot be the one having the wrong label. If Box $$\\rm A$$ has the right label, Box $$\\rm B$$ and Box $$\\rm C$$ will have the wrong labels. Box $$\\rm A\\rightarrow$$ white marbles Box $$\\rm B\\rightarrow$$ white marbles Box $$\\rm C\\rightarrow$$ blue marbles Box $$\\rm C$$ contains blue marbles. " ]

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

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

[ "In my hometown, city streets are numbered with odd numbers in increasing order from south to north. I must travel from north to south to go directly from $$241\\text{st}$$ Street to $$225\\text{th}$$ Street.~ When I travel from north to south, I travel south. " ]

[]

[ [ { "aoVal": "A", "content": "$3$ hours $$45$$ min " } ], [ { "aoVal": "B", "content": "$3$ hours $$35$$ min " } ], [ { "aoVal": "C", "content": "$2$ hours $$45$$ min " } ], [ { "aoVal": "D", "content": "$3$ hours $$25$$ min " } ], [ { "aoVal": "E", "content": "$3$ hours $$45$$ min " } ] ]

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

[ "$$12:00-8:00=4$$ h $4$h- $15$min =$3$ h $45$min " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{7}{36}$. " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{6}$. " } ], [ { "aoVal": "C", "content": "$\\dfrac{5}{36}$. " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{9}$. " } ], [ { "aoVal": "E", "content": "$\\dfrac{5}{12}$. " } ] ]

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

[ "There are $7$ out of $36$ outcomes which are prime numbers under $6$: $1+1$, $1+2$, $1+4$, $2+1$, $2+3$, $3+2$, $4+1$. " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{3}{4}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{6}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{9}$ " } ] ]

[ "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": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

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

[ "There are 3 different people here, so order does matter. Each of them must shoot at least once: $$7=1+1+5$$ $$7=1+2+4$$ $$7=1+3+3$$ $$7=1+4+2$$ $$7=1+5+1$$ $\\textasciitilde$ $$7=2+1+4$$ $$7=2+2+3$$ $$7=2+3+2$$ $$7=2+4+1$$ $\\cdots$ And so on. If we keep listing, total will be: $$5+4+3+2+1=15$$ ways " ]

[]

[ [ { "aoVal": "A", "content": "Among the $3$ balls taken out, at least $1$ is white. " } ], [ { "aoVal": "B", "content": "Among the $3$ balls taken out, at least $1$ is black. " } ], [ { "aoVal": "C", "content": "Among the $3$ balls taken out, at least $2$ are black. " } ], [ { "aoVal": "D", "content": "Among the $3$ balls taken out, at least $2$ are white. " } ] ]

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

[ "$$\\text{A}$$ is a certain event. $$\\text{B}$$ is a random event, so it\\textquotesingle s wrong. $$\\text{C}$$ is a random event, so it\\textquotesingle s wrong. $$\\text{D}$$ is a random event, so it\\textquotesingle s wrong. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "4.16 million dollars " } ], [ { "aoVal": "B", "content": "4.17 million dollars " } ], [ { "aoVal": "C", "content": "4.18 million dollars " } ], [ { "aoVal": "D", "content": "unable to determine " } ] ]

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

[ "The sample is biased becuase Black Friday is a promotion which allowed the store to earn way much more than they usually earn. " ]

[]

[ [ { "aoVal": "A", "content": "Edwin " } ], [ { "aoVal": "B", "content": "Fred " } ], [ { "aoVal": "C", "content": "Gary " } ], [ { "aoVal": "D", "content": "Howard " } ] ]

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

[ "Either Fred or Howard must be lying since what they said did not tally. Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$210$$ " } ], [ { "aoVal": "B", "content": "$$215$$ " } ], [ { "aoVal": "C", "content": "$$220$$ " } ], [ { "aoVal": "D", "content": "$$225$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "7 x 30 + 15 = 210 + 15 = 225 " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ $$\\text{kg}$$ " } ], [ { "aoVal": "B", "content": "$$15$$ $$\\text{kg}$$ " } ], [ { "aoVal": "C", "content": "$$24$$ $$\\text{kg}$$ " } ], [ { "aoVal": "D", "content": "$$30$$ $$\\text{kg}$$ " } ] ]

[ "Overseas In-curriculum->Knowledge Point->Algebra-> Numbers, Letters and Equations->Equivalent Substitution->Direct Substitution", "Overseas Competition->Knowledge Point->Counting Modules" ]

[ "$$240\\div16=15\\text{kg}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "To obtain an even number when adding two integers, both integers must be even or both integers must be odd. Therefore the four integers remaining once Avani has removed her three integers must all be odd or all be even or there would be a possibility that the sum of Niamh\\textquotesingle s two integers could be odd. Since there were four odd integers and three even integers on the cards in the box initially, the integers on the cards remaining once Avani has removed her cards are all odd. Therefore the cards Avani removed had the three even integers $$2$$, $$4$$ and $$6$$ written on them which have sum $$12$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

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

[ "Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels. " ]

[]

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

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

[ "The girls must be in the odd-numbered seats; otherwise, two girls would be seated next to each other. This leaves only the even-numbered seats for the boys, so a boy is seated in seat $$4$$. " ]