Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$$$Jack " } ], [ { "aoVal": "B", "content": "Sarah " } ], [ { "aoVal": "C", "content": "Jimmy " } ] ]

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

[ "We can spot that Jack\\textquotesingle s statement and Jimmy\\textquotesingle s statement contradict each other, so one of them is telling the truth. Therefore, Sarah tells a lie. " ]

[]

[ [ { "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)" ]

[ "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": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ], [ { "aoVal": "E", "content": "$$288$$ " } ] ]

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

[ "$$\\_{3}C\\_1\\times \\_{2}C\\_1\\times \\_{3}A\\_3\\times \\_{2}C\\_1=72$$, $$\\_{3}C\\_1\\times \\_{2}C\\_1\\times \\_{3}A\\_3\\times \\_{2}C\\_1\\times \\_{3}C\\_1=216$$, $$72+216=288$$. " ]

[]

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

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

[ "We can spot that Val\\textquotesingle s guess and Elvis\\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, John guessed it right. " ]

[]

[ [ { "aoVal": "A", "content": "$$30000$$ " } ], [ { "aoVal": "B", "content": "$$32800$$ " } ], [ { "aoVal": "C", "content": "$$34800$$ " } ], [ { "aoVal": "D", "content": "$$38800$$ " } ], [ { "aoVal": "E", "content": "$$40000$$ " } ] ]

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

[ "$$(6\\times7+2\\times6)\\times6!=54\\times720=38800$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1440$$ " } ], [ { "aoVal": "B", "content": "$$2880$$ " } ], [ { "aoVal": "C", "content": "$$5760$$ " } ], [ { "aoVal": "D", "content": "$$17280$$ " } ], [ { "aoVal": "E", "content": "$$34560$$ " } ] ]

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

[ "Since the three 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 $\\_3P\\_3\\times \\_4P\\_4$~because there are $\\_3P\\_3$~ways to arrange just three Arabic books, and $\\_4P\\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $D$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$26$$ " } ] ]

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

[ "$4+6+8+8=26$ " ]

[ "其它" ]

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

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

[ "There are $4$ ways to choose three from four. Two $B$s with one $O$ can form Bob\\textquotesingle s name. Thus, $2$ of the $4$ ways can get Bob\\textquotesingle s name. " ]

[]

[ [ { "aoVal": "A", "content": "$$260$$ " } ], [ { "aoVal": "B", "content": "$$280$$ " } ], [ { "aoVal": "C", "content": "$$300$$ " } ], [ { "aoVal": "D", "content": "$$320$$ " } ] ]

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

[ "There are $200$ multiples of $3$ and there are $120$ multiples of of $5$. There are also $40$ multiples of $15$. By the Inclusion Exclusion Principle, there are $200+120-40=280$ numbers which are multiples of $3$ or $5$. " ]

[]

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

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

[ "$$2\\times 3=6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$17$$ " } ] ]

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

[ "Among these numbers, there are $$8$$ pairs of numbers can get the sum of $$18$$($$1+17=2+16=3+15=4+14=5+13=6+12=7+11=$$$$8+10$$), and $$9$$ is useless. So in the worst case, after we choose $$9$$, we need $$8+1=9$$ more numbers to make sure a pair appears. Thus, the answer is $$1+8+1=10$$. Copyrighted material used with permission from Math Kangaroo in USA, NFP Inc. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\textgreater$$ " } ], [ { "aoVal": "B", "content": "$$\\textless$$ " } ], [ { "aoVal": "C", "content": "$$=$$ " } ] ]

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

[ "Isolate the decimals and whole number parts " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$5:00$ PM " } ], [ { "aoVal": "B", "content": "$4:30$ PM " } ], [ { "aoVal": "C", "content": "$4:00$ PM " } ], [ { "aoVal": "D", "content": "$5:30$ PM " } ], [ { "aoVal": "E", "content": "$4:45$ PM " } ] ]

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

[ "$5:30$-$45$ minutes=$4:45$ " ]

[]

[ [ { "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->Permutations and Combinations->Combinations" ]

[ "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": "$$96$$ " } ], [ { "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": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

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

[ "$$9=2+7=3+6=4+5$$ $$9=2+3+4$$ So there are $$3+1=4$$ ways " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{6}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{5}{12}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{1}{2}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{7}{12}$ " } ], [ { "aoVal": "E", "content": "$\\dfrac{5}{6}$ " } ] ]

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

[ "There are $6\\cdot 6=36$ ways to roll the two dice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $\\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\\dfrac{15+6}{36}=\\dfrac{21}{36}=\\boxed {\\left (\\text{D}\\right )\\dfrac{7}{12}}$. " ]

[]

[ [ { "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": "$$26$$ " } ], [ { "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

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

[ "She picks two $13$s, one $14$, three $15$s, five $19$s, and four $20$s. The mode is $19.$ The median is $19.$ Thus, their difference should be $0.$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$11:05$ " } ], [ { "aoVal": "B", "content": "$12:05$ " } ], [ { "aoVal": "C", "content": "$11:45$ " } ], [ { "aoVal": "D", "content": "$11:55$ " } ], [ { "aoVal": "E", "content": "$12:00$ " } ] ]

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

[ "$9:35+1:30+1=12:05$ " ]

[]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

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

[ "Try to consider the problem of the youngest Alice in such way: make each grandchild has a similar age as possible, and the age of each grandchild should be different, that is, $$1 + 2 + 3 +\\cdots 9 + 10 = 55$$; then $$180 - 55 = 125$$, $$125 \\div 10 = 12R5$$, and $$5$$ is left. If every child\\textquotesingle s age is added by $$12$$, then $$5$$ is left. Give the extra year to each of the $$5$$ oldest children. In doing so, the minimum age of Alice is $$10+12+1=23$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$8866$$ " } ], [ { "aoVal": "B", "content": "$$8976$$ " } ], [ { "aoVal": "C", "content": "$$8736$$ " } ], [ { "aoVal": "D", "content": "$$8636$$ " } ] ]

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

[ "624x14=624x(10+4)=624x10+624x4=6240+2496=8736 " ]

[]

[ [ { "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. " ]

[]

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

[ "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": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ], [ { "aoVal": "E", "content": "$$38$$ " } ] ]

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

[ "Since the product of the two positive integers is $$240$$, the possible pairs of integers are $$\\left( 1,240 \\right)$$, $$\\left( 2,120 \\right)$$, $$\\left( 3,80 \\right)$$, $$\\left( 4,60 \\right)$$, $$\\left( 5,48 \\right)$$, $$\\left( 6,40 \\right)$$, $$\\left( 8,30 \\right)$$, $$\\left( 10,24 \\right)$$, $$\\left( 12,20 \\right)$$ and $$\\left( 15,16 \\right)$$. The respective sums of these pairs are $$241$$, $$122$$, $$83$$, $$64$$, $$53$$, $$46$$, $$38$$, $$34$$, $$32$$ and $$31$$. Of these, the smallest value is $$31$$. " ]

[ "其它" ]

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

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

[ "The three digit numbers are $135,153,351,315,513,531$. The numbers that end in 5 are divisible are 5 , and the probability of choosing those numbers is (B) $\\frac{1}{3}$. " ]

[ "其它" ]

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

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "Gaining £$$50$$ " } ], [ { "aoVal": "B", "content": "Losing £$$50$$ " } ], [ { "aoVal": "C", "content": "Gaining £$$25$$ " } ], [ { "aoVal": "D", "content": "Losing £$$25$$ " } ], [ { "aoVal": "E", "content": "Break Even " } ] ]

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

[ "Nil " ]

[]

[ [ { "aoVal": "A", "content": "$$8866$$ " } ], [ { "aoVal": "B", "content": "$$8976$$ " } ], [ { "aoVal": "C", "content": "$$8736$$ " } ], [ { "aoVal": "D", "content": "$$8636$$ " } ] ]

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

[ "624x14=624x(10+4)=624x10+624x4=6240+2496=8736 " ]

[]

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

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

[ "The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions $$2$$, $$3$$ and $$4$$ from the left-hand end. Were Sally to sit in place $$2$$, neither Dolly nor Kelly could sit in places $$1$$ or $$3$$ as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places $$4$$ and $$5$$ which is impossible. Similarly, were Sally to sit in place $$3$$, Dolly could not sit in place $$2$$ or $$4$$ and the question also tells us she cannot sit in place $$1$$ so Dolly would have to sit in place $$5$$ making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place $$4$$, Dolly could sit in place $$2$$, Kelly in place $$1$$, Molly (who cannot sit in place $$5$$) in place $$3$$ leaving Elly to sit in place $$5$$ at the right-hand end. " ]

[]

[ [ { "aoVal": "A", "content": "Abel " } ], [ { "aoVal": "B", "content": "Ben " } ], [ { "aoVal": "C", "content": "Cain " } ] ]

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

[ "Only $$1$$ of the $$3$$ boys can swim. Only $$1$$ of the $$3$$ boys is telling the truth! Since Abel and Cain contradict each other, there must be one who is telling the truth! If Abel is true, Ben is lying and Ben can swim. Then we have $$2$$ boys (Abel and Ben) who can swim. Contradiction. Hence, Cain is true. Abel is lying and so is Ben. Then, Ben can swim. Abel cannot swim and we are not sure if Cain can swim. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

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

[ "With the digits listed from least to greatest, the $3$-digit integers are $138,146,226,234.$ $226$ can be arranged in $3$ ways, and the other three can be arranged in $6$ ways. There are $3+6(3)=(\\mathbf{D}) 21$ 3-digit positive integers. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac12$ " } ], [ { "aoVal": "B", "content": "$\\frac13$ " } ], [ { "aoVal": "C", "content": "$\\frac14$ " } ], [ { "aoVal": "D", "content": "$\\frac16$ " } ] ]

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

[ "There are $50$ even number cards and $50$ odd number cards here. If she chooses odd+odd or even+even, she will get an even sum. But if she chooses odd+even or even+odd, she will get an odd sum. So each of the two probabilities is equal. " ]

[]

[ [ { "aoVal": "A", "content": "Basketball　 " } ], [ { "aoVal": "B", "content": "Swimming　 " } ], [ { "aoVal": "C", "content": "Taekwondo　 " } ], [ { "aoVal": "D", "content": "Rhythmic gymnastics　 " } ] ]

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

[ "Since \\textbf{Mary} is the only female, she must have participated in the women\\textquotesingle s\\textbf{~Rhythmic gymnastics} as the rest are males. Since Karl and Navin do not know how to swim, \\textbf{Lim} is the only male left and thus he participated in the S\\textbf{wimming} event. And since \\textbf{Karl}\\textquotesingle s event does not require a ball, he must have participated in \\textbf{Taekwondo}. We can thus conclude that \\textbf{Navin}~participated in \\textbf{Basketball}. " ]

[]

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

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

[ "We have $$\\left( 1+3+5+7+9+11+13+15+17+19 \\right)\\div 10 = 10$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{3}{4}$ " } ], [ { "aoVal": "B", "content": "$\\frac{57}{64}$ " } ], [ { "aoVal": "C", "content": "$\\frac{59}{64}$ " } ], [ { "aoVal": "D", "content": "$\\frac{187}{192}$ " } ], [ { "aoVal": "E", "content": "$\\frac{63}{64}$ " } ] ]

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

[ "We will use complementary counting to find the probability that the product is not divisible by $4$ . Then, we can find the probability that we want by subtracting this from $1$. We split this into two cases. Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\\frac{1}{2}$, our probability is $\\left(\\frac{1}{2}\\right)^{6}=\\frac{1}{64}$. Case 2: The product is divisible by $2$, but not by $4$. We need 5 numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\\frac{1}{2}$ chance that an odd number is rolled, a $\\frac{1}{3}$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\\left(\\frac{1}{2}\\right)^{5}\\left(\\frac{1}{3}\\right) \\cdot 6=\\frac{1}{16}$. Therefore, the probability the product is not divisible by 4 is $\\frac{1}{64}+\\frac{1}{16}=\\frac{5}{64}$. Our answer is $1-\\frac{5}{64}=$ (C) $\\frac{59}{64}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$112$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test. No. of ways to choose $5$ pupils to get the correct test is $$\\frac{8 \\times 7 \\times 6 \\times 5 \\times 4}{5 \\times 4 \\times 3 \\times 2 \\times 1}-56.$$ To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$. Hence, the total no, of ways $=56 \\times2=112$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

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

[ "There are $5$ $E$s in total, which have $4$ intervals leaving for the other $4$ letters. Thus, the answer is $\\_4P\\_4=24$. " ]

[ "其它" ]

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

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

[ "There are $2\\times6=12$ different combinations. The product of two numbers is smaller than $6$ will be $5\\times1$. Thus, the probability is $\\frac 1{12}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "unique mode: $6$ median: $6$ $(3+4+5+6+6+7+x)\\div7=6$ $x=11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{9}{16}$ " } ], [ { "aoVal": "B", "content": "$\\frac{5}{8}$ " } ], [ { "aoVal": "C", "content": "$\\frac{3}{4}$ " } ], [ { "aoVal": "D", "content": "$\\frac{25}{32}$ " } ], [ { "aoVal": "E", "content": "$\\frac{13}{16}$ " } ] ]

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

[ "We will use complementary counting. First, the frog can go left with probability $\\frac{1}{4}$. We observe symmetry, so our final answer will be multiplied by $4$ for the $4$ directions, and since $4 \\cdot \\frac{1}{4}=1$, we will ignore the leading probability. From the left, she either goes left to another edge $\\left(\\frac{1}{4}\\right)$ or back to the center $\\left(\\frac{1}{4}\\right)$. Time for some casework. Case 1: She goes back to the center. Now, she can go in any $4$ directions, and then has $2$ options from that edge. This gives $\\frac{1}{2}$. Case 2: She goes to another edge (rightmost). Subcase 1: She goes back to the left edge. She now has $2$ places to go, giving $\\frac{1}{2}$ Subcase 2: She goes to the center. Now any move works. $\\frac{1}{4} \\cdot \\frac{1}{2}+\\frac{1}{4} \\cdot 1=\\frac{1}{8}+\\frac{1}{4}=\\frac{3}{8}$ for this case. She goes back to the center in Case 1 with probability $\\frac{1}{4}$, and to the right edge with probability $\\frac{1}{4}$ So, our answer is $\\frac{1}{4} \\cdot \\frac{1}{2}+\\frac{1}{4} \\cdot \\frac{3}{8}=\\frac{1}{4}\\left(\\frac{1}{2}+\\frac{3}{8}\\right)=\\frac{1}{4} \\cdot \\frac{7}{8}=\\frac{7}{32}$ But, don\\textquotesingle t forget complementary counting. So, we get $1-\\frac{7}{32}=\\frac{25}{32} \\Longrightarrow D$. " ]

[ "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": "$$83$$ " } ], [ { "aoVal": "B", "content": "$$73$$ " } ], [ { "aoVal": "C", "content": "$$63$$ " } ], [ { "aoVal": "D", "content": "$$53$$ " } ] ]

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

[ "Every $$6$$th number $$\\left( 83 \\right.$$ of them$$\\left. {} \\right)$$ is divisible by $$6$$. Every $$24$$th number $$\\left( 20 \\right.$$ of them$$\\left. {} \\right)$$ is divisible by $$6$$ and $$8$$, and $$83-20 = 63$$. " ]

[]

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

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

[ "We can spot that Val\\textquotesingle s guess and Elvis\\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, John is right. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$5:00$ pm " } ], [ { "aoVal": "B", "content": "$4:30$ pm " } ], [ { "aoVal": "C", "content": "$4:00$ pm " } ], [ { "aoVal": "D", "content": "$5:30$ pm " } ], [ { "aoVal": "E", "content": "$4:40$ pm " } ] ]

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

[ "$5:30$-$30$ minutes=$5:00$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

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

[ "18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 " ]

[]

[ [ { "aoVal": "A", "content": "$$148\\rm cm$$ " } ], [ { "aoVal": "B", "content": "$$150\\rm cm$$ " } ], [ { "aoVal": "C", "content": "$$152\\rm cm$$ " } ], [ { "aoVal": "D", "content": "$$154\\rm cm$$ " } ] ]

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

[ "Rose is $$150\\rm cm$$ tall. Quentin is $$10 \\rm cm$$ taller than Rose, so Quentin is $$160 \\rm cm$$ tall. Sam is $$4 \\rm cm$$ shorter than Rose, so Sam is $$146 \\rm cm$$ tall. Their average height is $$ (150+160+146)\\div3 =152 \\rm cm$$. " ]

[]

[ [ { "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": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "$$9\\times72=648$$ $$78\\times8=624$$ $$648-624=24$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "$1+1+1+1=4$ $3+3+3+3=12$ $1+1+1+3=6$ $1+1+3+3=8$ $1+3+3+3=10$ " ]

[]

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

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$11:30$$ A.M. " } ], [ { "aoVal": "B", "content": "$$10:40$$ A.M. " } ], [ { "aoVal": "C", "content": "$$11:40$$ A.M. " } ], [ { "aoVal": "D", "content": "$$11:50$$ A.M. " } ], [ { "aoVal": "E", "content": "$$11:40$$ P.M. " } ] ]

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

[ "It takes $25+35=60$ min, which is equal to $1$ hour to swim and run. It takes $1$ h + $1$ h $10$ min = $2$ h $10$ min in total. Thus, the end time is $9:30+$$2$ h $10$ min$=11:40$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ], [ { "aoVal": "E", "content": "$$98$$ " } ] ]

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

[ "$$6-2=4$$ $$6+4=10$$ $$7\\times 10=70$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$420$$ " } ], [ { "aoVal": "C", "content": "$$105$$ " } ], [ { "aoVal": "D", "content": "$$225$$ " } ], [ { "aoVal": "E", "content": "$$210$$ " } ] ]

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

[ "$15\\times14\\div2\\times2=210$. " ]

[]

[ [ { "aoVal": "A", "content": "The rabbit; the dog " } ], [ { "aoVal": "B", "content": "The dog; the cat " } ], [ { "aoVal": "C", "content": "The cat; the duck " } ], [ { "aoVal": "D", "content": "The cat; the rabbit " } ] ]

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

[ "We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit, and the dog don\\textquotesingle t get food, the cat and the duck would get food. " ]

[]

[ [ { "aoVal": "A", "content": "$3:40$ PM " } ], [ { "aoVal": "B", "content": "$3:20$ PM " } ], [ { "aoVal": "C", "content": "$2:40$ PM " } ], [ { "aoVal": "D", "content": "$1:40$ PM " } ], [ { "aoVal": "E", "content": "$1:20$ PM " } ] ]

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

[ "It\\textquotesingle s $9:40$ PM in New York, which means it\\textquotesingle s $6:40$ PM in Seattle. The plane departed at $6:40$-$5$ hours=$1:40$ PM. " ]

[]

[ [ { "aoVal": "A", "content": "$$45$$ min " } ], [ { "aoVal": "B", "content": "$$35$$ min " } ], [ { "aoVal": "C", "content": "$$75$$ min " } ], [ { "aoVal": "D", "content": "$$55$$ min " } ], [ { "aoVal": "E", "content": "$$65$$ min " } ] ]

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

[ "Time spent reading $$\\rm =1h45min-30min$$ $$\\rm =1h15min$$ $$\\rm =60min+ 15min$$ $$\\rm =75min$$. " ]

[]

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

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

[ "Either Fred or Howard must be lying since what they said is conflicting. Since only one person was lying, Gary was telling the truth, which means Fred broke the vase. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

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

[ "To guarantee he get a red pen, the worse cases scenario is he picked all the green and black pens. Which is $$12+10=22$$. Then, the next pen must be the red. Thus, $$22+1=23$$. " ]

[]

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

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

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$11:03$$ A.M. " } ], [ { "aoVal": "B", "content": "$$11:30$$ A.M. " } ], [ { "aoVal": "C", "content": "$$1:00$$ P.M. " } ], [ { "aoVal": "D", "content": "$$2:00$$ P.M. " } ] ]

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

[ "It takes $$3$$ hours for the \\emph{minute} hand to go around $$3$$ times, so the time will be $$2$$ P.M. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$480$$ " } ], [ { "aoVal": "E", "content": "$$720$$ " } ] ]

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

[ "$3\\times2\\times1=6$ $4\\times3\\times2=24$ $24\\times6=144$ " ]

[]

[ [ { "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$576$$ " } ], [ { "aoVal": "D", "content": "$$2880$$ " } ], [ { "aoVal": "E", "content": "$$14400$$ " } ] ]

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

[ "$8-4+1=5$ $5\\times4\\times3\\times2\\times1=120$ $4\\times3\\times2\\times1=24$ $120\\times24=2880$ " ]

[ "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": "$72$ , $24$ " } ], [ { "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$$2\\times 4\\times 3\\times 2\\times 1=48$$ . " ]

[]

[ [ { "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. " ]

[]

[ [ { "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": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$112$$ " } ] ]

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

[ "If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test. No. of ways to choose $5$ pupils to get the correct test is $$\\frac{8 \\times 7 \\times 6 \\times 5 \\times 4}{5 \\times 4 \\times 3 \\times 2 \\times 1}-56.$$ To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$. Hence, the total no, of ways $=56 \\times2=112$. " ]

[]

[ [ { "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": "$\\dfrac{1}{27}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{7}{9}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{7}{36}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{9}$ " } ], [ { "aoVal": "E", "content": "$\\dfrac{5}{18}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "There are $7$ out of $36$ outcomes are perfect squares: $1+3$, $2+2$, $3+1$, $3+6$, $4+5$, $5+4$, $6+3$. " ]

[]

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

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$49200$$ " } ], [ { "aoVal": "B", "content": "$$94080$$ " } ], [ { "aoVal": "C", "content": "$$564480$$ " } ], [ { "aoVal": "D", "content": "$$1800$$ " } ], [ { "aoVal": "E", "content": "$$98400$$ " } ] ]

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

[ "There are $3$ $E$s in total now with other $8$ letters remaining. But pay attention to $B$ and $P$: there are $3$ $B$s and $3$ $P$ here. There are $\\_8P\\_5 \\div \\_3P\\_3$ ways for us to arrange the $8$ letters\\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $9$ intervals. So the answer is $\\_8P\\_5 \\div \\_3P\\_3 \\times \\_9C\\_3=94080$. " ]

[]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

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

[ "$$85+77-150=12$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$11:05$ am. " } ], [ { "aoVal": "B", "content": "$12:05$ am. " } ], [ { "aoVal": "C", "content": "$11:45$ am. " } ], [ { "aoVal": "D", "content": "$11:55$ am. " } ], [ { "aoVal": "E", "content": "$12:00$ am. " } ] ]

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

[ "$9:35+1:30+1=12:05$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$111109$$ " } ], [ { "aoVal": "B", "content": "$$111119$$ " } ], [ { "aoVal": "C", "content": "$$111100$$ " } ], [ { "aoVal": "D", "content": "$$111105$$ " } ] ]

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

[ "$9+99+999+9999+99999=10+100+1000+10000+100000-5=111105$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

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

[ "The average is the middle number, $$5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$109$$ " } ], [ { "aoVal": "C", "content": "$$191$$ " } ], [ { "aoVal": "D", "content": "$$200$$ " } ] ]

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

[ "Pat and Lee counted leaves on two plants. Pat\\textquotesingle s got a $$1$$-digit number. Lee got a $$3$$-digit number. If the dīfference of the numbers was $$91$$, the numbers were $$100$$ and $$9$$, and the sum is $$109$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$512$$ " } ], [ { "aoVal": "B", "content": "$$648$$ " } ], [ { "aoVal": "C", "content": "$$720$$ " } ], [ { "aoVal": "D", "content": "$$728$$ " } ], [ { "aoVal": "E", "content": "$$800$$ " } ] ]

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

[ "One-digit number: $9-1=8$ Two-digit number: $8\\times9=72$ Three-digit number: $8\\times9\\times9=648$ So, there are $8+72+648=728$ numbers that meet the requirements. " ]

[]

[ [ { "aoVal": "A", "content": "$$7190$$ " } ], [ { "aoVal": "B", "content": "$$7290$$ " } ], [ { "aoVal": "C", "content": "$$7390$$ " } ], [ { "aoVal": "D", "content": "$$7490$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles->Number Puzzles (horizontal forms)" ]

[ "$486\\times 15 = 7290$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

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

[ "In total, there are $2+2+2=6$ different pairs. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$130$$ " } ], [ { "aoVal": "B", "content": "$$137$$ " } ], [ { "aoVal": "C", "content": "$$138$$ " } ], [ { "aoVal": "D", "content": "$$129$$ " } ] ]

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

[ "$19+37+81=19+81+37=100+37=137$ " ]

[ "其它" ]

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

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

[ "Volume of cube $=2\\times4\\times8=64\\textasciitilde cm^{3}$ Side length of cube $=$$\\sqrt[3]{64}$$=4\\textasciitilde cm$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5800$$ " } ], [ { "aoVal": "B", "content": "$$4500$$ " } ], [ { "aoVal": "C", "content": "$$4300$$ " } ], [ { "aoVal": "D", "content": "$$4750$$ " } ] ]

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

[ "$$Nil$$ " ]

[]

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

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

[ "The information that in any group of four pencils, at least two have the same colour, tells us that there at most three different coloured pencils in Carl\\textquotesingle s pencil case. The information that in any group of five pencils, at most three have the same colour, tells us that there are at most three pencils of any single colour in the pencil case. Hence there are three pencils of each of the three different colours and so Carl\\textquotesingle s pencil case contains three blue pencils. " ]

[]

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

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

[ "$$2\\times 3=6$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ], [ { "aoVal": "E", "content": "$$35$$ " } ] ]

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

[ "Consider the distribution of the three consecutive heads. We have four cases: 1110xx, 01110x, x01110, xx0111. Since case 1 and case4, case 2 and case 3 are the same, we only need to think about case 1 and case 2. The probability of case 1: $(\\dfrac{1}{2})^{4}=\\frac{1}{16}$. The probability of case 2: $(\\dfrac{1}{2})^{5}=\\frac{1}{32}$. The sum of the probabilities of 4 cases is $2\\times(\\frac{1}{16}+\\frac{1}{32})=\\frac{3}{16}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$192$$ " } ], [ { "aoVal": "B", "content": "$$196$$ " } ], [ { "aoVal": "C", "content": "$$200$$ " } ], [ { "aoVal": "D", "content": "$$204$$ " } ], [ { "aoVal": "E", "content": "$$208$$ " } ] ]

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

[ "$(4+20)\\times17\\div2=204$ " ]

[]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ] ]

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

[ "$$20+25-10=35$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$2.5\\textasciitilde m/s$ " } ], [ { "aoVal": "B", "content": "$5\\textasciitilde m/s$ " } ], [ { "aoVal": "C", "content": "$2\\textasciitilde m/s$ " } ], [ { "aoVal": "D", "content": "$0.4\\textasciitilde m/s$ " } ] ]

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

[ "Average speed $=\\frac{Total distance}{Total Time}=\\frac{1000\\textasciitilde m}{400\\textasciitilde s}=2.5\\textasciitilde m/s$ " ]

[ "其它" ]

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

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

[ "$$6=0+6$$ $$6=1+5$$ $$6=2+4$$ $$6=3+3$$ " ]

[]

[ [ { "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": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "The $9$ pearls should be put in each of the two gems. Thus, just consider the combination of gems: if all red gems are together, she has $1$ way; if two red gems are together, she has $3$ ways; if all the red gems are not together, the blue gems can be put $1-2-3$, $1-1-4$ or $2-2-2$, so there are $3$ ways. There are $7$ ways. " ]

[]

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

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

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

[]

[ [ { "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}. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\textbackslash$48$ " } ], [ { "aoVal": "B", "content": "$\\textbackslash$88$ " } ], [ { "aoVal": "C", "content": "$\\textbackslash$35.20$ " } ], [ { "aoVal": "D", "content": "$\\textbackslash$52.80$ " } ] ]

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

[ "Price after discount $=\\textbackslash$80\\times(100\\textbackslash\\%-40\\textbackslash\\%)=\\textbackslash$48$ Price after service charges $=\\textbackslash$48\\times110\\textbackslash\\%=\\textbackslash$52.80$ " ]

[]

[ [ { "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->Counting Modules->Permutations and Combinations->Combinations" ]

[ "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 Bill is the youngest. " ]