Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$399$$ " } ], [ { "aoVal": "B", "content": "$$402$$ " } ], [ { "aoVal": "C", "content": "$$5050$$ " } ], [ { "aoVal": "D", "content": "$$20100$$ " } ], [ { "aoVal": "E", "content": "$$25050$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "The result is equal to the sum of all numbers from $1$ to $200$, which is $(1+200)\\times200\\div2=20100$. " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$98$$ " } ], [ { "aoVal": "C", "content": "$$96$$ " } ], [ { "aoVal": "D", "content": "$$94$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences->Pyramid Sequences" ]

[ "$$1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4$$ $$=1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1-3-2-1$$ $$=10\\times 10-6$$ $$=100-6$$ $$=94$$． " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers" ]

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$98! \\cdot 99!$ " } ], [ { "aoVal": "B", "content": "$98! \\cdot 100!$ " } ], [ { "aoVal": "C", "content": "$99! \\cdot 100!$ " } ], [ { "aoVal": "D", "content": "$99! \\cdot 101!$ " } ], [ { "aoVal": "E", "content": "$100! \\cdot 101!$ " } ] ]

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

[ "C " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$210$$ " } ], [ { "aoVal": "B", "content": "$$420$$ " } ], [ { "aoVal": "C", "content": "$$630$$ " } ], [ { "aoVal": "D", "content": "$$840$$ " } ], [ { "aoVal": "E", "content": "$$1050$$ " } ] ]

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

[ "B " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2\\times7\\times24$$ " } ], [ { "aoVal": "B", "content": "$$(7+7)\\times24\\times60$$ " } ], [ { "aoVal": "C", "content": "$$2\\times7\\times12\\times60$$ " } ], [ { "aoVal": "D", "content": "$$2\\times24\\times60$$ " } ], [ { "aoVal": "E", "content": "$$(7+7)\\times12\\times60$$ " } ] ]

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

[ "One week has $$7$$ days. Two weeks has $$7+7$$ One day has $$24$$ hours. One hour has $$60$$ minutes. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

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

[ "C " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "There are $3+5+1=9$ students in each row, and there are $2+1+1=4$ rows. Thus, there are $9\\times4=36$ students in this class. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers" ]

[ "$$Omitted.$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{8}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{9}{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{8}{15}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{20}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$200$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$2000$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "$$10000\\div 200=50$$; $$50\\times \\underline{200}=10000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\textbackslash$ 2$ " } ], [ { "aoVal": "B", "content": "$\\textbackslash$ 8$ " } ], [ { "aoVal": "C", "content": "$\\textbackslash$ 10$ " } ], [ { "aoVal": "D", "content": "$\\textbackslash$ 12$ " } ] ]

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

[ "$\\textbackslash$2$, as the producer surplus is calculated as the difference between the price the good is sold for and the cost of production, which is $\\textbackslash$10$ - $\\textbackslash$8$ = $\\textbackslash$2$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$-7-4 \\sqrt{3}$ " } ], [ { "aoVal": "B", "content": "$-2-\\sqrt{3}$ " } ], [ { "aoVal": "C", "content": "$$-1$$ " } ], [ { "aoVal": "D", "content": "$-2+\\sqrt{3}$ " } ], [ { "aoVal": "E", "content": "$-7+4 \\sqrt{3}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations" ]

[ "It is given that $a x^{2}+b x+c=0$ has 1 real root, so the discriminant is zero, or $b^{2}=4 a c$. Because $a, b, c$ are in arithmetic progression, $b-a=c-b$, or $b=\\frac{a+c}{2}$. We need to find the unique root, or $-\\frac{b}{2 a}$ (discriminant is 0 ). From $b^{2}=4 a c$, we can get $-\\frac{b}{2 a}=-\\frac{2 c}{b}$ Ignoring the negatives(for now), we have $\\frac{2 c}{b}=\\frac{2 c}{\\frac{a+c}{2}}=\\frac{4 c}{a+c}=\\frac{1}{\\frac{1}{\\frac{4 c}{a+c}}}=\\frac{1}{\\frac{a+c}{4 c}}=\\frac{1}{\\frac{a}{4 c}+\\frac{1}{4}}$. Fortunately, finding $\\frac{a}{c}$ is not very hard. Plug in $b=\\frac{a+c}{2}$ to $b^{2}=4 a c$, we have $a^{2}+2 a c+c^{2}=16 a c$, or $a^{2}-14 a c+c^{2}=0$, and dividing by $c^{2}$ gives $\\left(\\frac{a}{c}\\right)^{2}-14\\left(\\frac{a}{c}\\right)+1=0$, so $\\frac{a}{c}=\\frac{14 \\pm \\sqrt{192}}{2}=7 \\pm 4 \\sqrt{3}$. But $7-4 \\sqrt{3}\\textless1$, violating the assumption that $a \\geq c$. Therefore, $\\frac{a}{c}=7+4 \\sqrt{3}$. Plugging this in, we have $\\frac{1}{\\frac{a}{4 c}+\\frac{1}{4}}=\\frac{1}{2+\\sqrt{3}}=2-\\sqrt{3}$. But we need the negative of this, so the answer is (D). " ]

[]

[ [ { "aoVal": "A", "content": "$$397$$ " } ], [ { "aoVal": "B", "content": "$$399$$ " } ], [ { "aoVal": "C", "content": "$$401$$ " } ], [ { "aoVal": "D", "content": "$$403$$ " } ], [ { "aoVal": "E", "content": "$$405$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "$$1+(5-1)\\times 99=397$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$401$$ " } ], [ { "aoVal": "B", "content": "$$403$$ " } ], [ { "aoVal": "C", "content": "$$404$$ " } ], [ { "aoVal": "D", "content": "$$405$$ " } ], [ { "aoVal": "E", "content": "$$2001$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "Based on the middle term rule, we can find the middle number is $$401$$, and the greatest number is $$401+1+1=403$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$10^{2}$$ " } ], [ { "aoVal": "D", "content": "$$10^{6}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers" ]

[ "$$10^{5}+10^{6}=1100000=11\\times 10^{5}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "(31 - 7) x 3 x 24 x 60 " } ], [ { "aoVal": "B", "content": "(31 - 7) x 24 x 60 " } ], [ { "aoVal": "C", "content": "(31 -7 x 3) x 24 x 60 " } ], [ { "aoVal": "D", "content": "(31 -7 x 3) x 24 x 60 x 60 " } ], [ { "aoVal": "E", "content": "(30 -7 x 3) x 24 x 60 " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion" ]

[ "The cat slept for exactly 3 weeks for 7 x 3 days. The cat was awake for 31 - 7 x 3 days. A day has 24 hours, and an hour has 60 minutes. Therefore, the cat was awake in (31- 7 x 3) x 24 x 60 " ]

[]

[ [ { "aoVal": "A", "content": "$$300$$ " } ], [ { "aoVal": "B", "content": "$$400$$ " } ], [ { "aoVal": "C", "content": "$$500$$ " } ], [ { "aoVal": "D", "content": "$$600$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Operation Strategy in Addition and Subtraction of Rounding Whole Numbers" ]

[ "$$(894-94)-(89+111)-(95+105)$$ $$=800-200-200$$ $$=400$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "In weight, $$45$$ mice $$=3\\times (15$$ mice$$)=3\\times (2$$ cats$$)= 2\\times (3$$ cats$$)=2\\times (2$$ dogs$$)= 4$$ dogs. " ]

[]

[ [ { "aoVal": "A", "content": "$$2:30$$ " } ], [ { "aoVal": "B", "content": "$$1:5$$ " } ], [ { "aoVal": "C", "content": "$$5:1$$ " } ], [ { "aoVal": "D", "content": "$$30:2$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio" ]

[ "We need to make the units same first. $$2$$ yards equal to $$6$$ feet. Now we could remove the same unit, feet. We get $$6:30$$ and simplify it to $$1:5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10302$$ " } ], [ { "aoVal": "B", "content": "$$10202$$ " } ], [ { "aoVal": "C", "content": "$$10201$$ " } ], [ { "aoVal": "D", "content": "$$10102$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "The $$100$$ even numbers that add up to $$10100$$ are $$2$$, $$4$$, $$\\cdots $$, $$200$$. The sum we want is $$2+4+ \\cdots + 200 + 202 = 10 100 + 202$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "The ones digit is the same as the ones digit of $$6 \\times7\\times8\\times9 \\times0$$. ` " ]

[]

[ [ { "aoVal": "A", "content": "$$4017$$ " } ], [ { "aoVal": "B", "content": "$$4007$$ " } ], [ { "aoVal": "C", "content": "$$4027$$ " } ], [ { "aoVal": "D", "content": "$$3017$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas-> Difference of Two Squares Formula" ]

[ "$$=(2008+1)\\times 2009-2008\\times 2008$$ $$=2008\\times 2009+2009-2008\\times 2008$$ $$=2008\\times (2009-2008)+2009$$ $$=2008+2009$$ $$=4017$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

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

[ "Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. $$ \\begin{aligned} 7 \\equiv 07 \\&(\\bmod 100) \\textbackslash\\textbackslash{} 7^{2} \\equiv 49 \\&(\\bmod 100) \\textbackslash\\textbackslash{} 7^{3} \\equiv 43 \\&(\\bmod 100) \\textbackslash\\textbackslash{} 7^{4} \\equiv 01 \\&(\\bmod 100) \\end{aligned} $$ We can write $7^{2011}$ as $\\left(7^{4}\\right)^{502} \\times 7^{3}$. Using this, we can say: $$ 7^{2011} \\equiv\\left(7^{4}\\right)^{502} \\times 7^{3} \\equiv 7^{3} \\equiv 343 \\equiv 43 \\quad(\\bmod 100) . $$ From the above, we can conclude that the last two digits of $7^{2011}$ are $43$. Since they have asked us to find the tens digit, our answer is (D) $4$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$81$$ " } ], [ { "aoVal": "B", "content": "$$900$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables" ]

[ "From 100-199: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191 (10 numbers) From 200 - 299: 202, 212, 222, 232, 242, 252, 262, 272, 282, 292 (10 numbers) From 300 to 399: 10 numbers From 400 to 499: 10 numbers $$\\cdots $$ From 900 to 999: 10 numbers. Hence, there are 10 x 9 = 90 three-digit palindromes. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$m=-3, b=4$ " } ], [ { "aoVal": "B", "content": "$m=-3, b=5$ " } ], [ { "aoVal": "C", "content": "$m=\\frac{3}{4}, b=\\frac{5}{4}$ " } ], [ { "aoVal": "D", "content": "$m=-\\frac{3}{4}, b=\\frac{5}{4}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable" ]

[ "$3x+4y=5$, $y=-\\frac{3}{4}x+\\frac{5}{4}$, Its slope is $-\\frac{3}{4}$ and intercept is $\\frac{5}{4}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$A$ " } ], [ { "aoVal": "B", "content": "$B$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion" ]

[ "$16$ ounce equals $1$ pound " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Time" ]

[ "The number of hours in $$10$$ days is $$240$$; $$240$$ minutes is $$240\\div60 =4$$ hours. " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$114$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio" ]

[ "Divide through by $$2$$ then multiply by $$7$$ to get $$4:14=2:7=14:49$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3505$$ " } ], [ { "aoVal": "B", "content": "$$3815$$ " } ], [ { "aoVal": "C", "content": "$$3515$$ " } ], [ { "aoVal": "D", "content": "$$3805$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "omitted " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly" ]

[ "If $a◆b$~ represents$(a\\times b)+b$ , $2◆3=(2\\times3)+3=9$ . So the answer is $\\rm B$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$22$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{6}{7}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{7}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions" ]

[ "$$\\frac{1}{2}\\times \\frac{22}{7}\\div \\frac{11}{5}=\\frac{1}{2}\\times \\frac{22}{7}\\times \\frac{5}{11}=\\frac{5}{7}$$. " ]

[]

[ [ { "aoVal": "A", "content": "　prime " } ], [ { "aoVal": "B", "content": "　composite " } ], [ { "aoVal": "C", "content": "　odd " } ], [ { "aoVal": "D", "content": "　even " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "The product of a whole number and an even number must be even. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{16}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{8}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{4}$ " } ], [ { "aoVal": "D", "content": "$\\frac{3}{8}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{2}$ " } ] ]

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

[ "First, we calculate that there are a total of $4 \\cdot 4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\\frac{2}{16}=$ (B) $\\frac{1}{8}$ probability that the number formed by the two spinners is a perfect square. " ]

[ "其它" ]

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

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

[ "Convert both sides into powers with the same bases: $4^{-4}=4^{-x}$ $x=4$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "46\\% " } ], [ { "aoVal": "B", "content": "32\\% " } ], [ { "aoVal": "C", "content": "11\\% " } ], [ { "aoVal": "D", "content": "43\\% " } ], [ { "aoVal": "E", "content": "3.5\\% " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables" ]

[ "The correct answer is(d). Because ethnic group categories are assumed to be mutually exclusive, P(Asian or Latino)=P(Asian)+P(Latino)=32\\%+11\\%=43\\% " ]

[]

[ [ { "aoVal": "A", "content": "$$0 $$ " } ], [ { "aoVal": "B", "content": "$$ 3 $$ " } ], [ { "aoVal": "C", "content": "$$ 4 $$ " } ], [ { "aoVal": "D", "content": "$$ 5 $$ " } ], [ { "aoVal": "E", "content": "$$ 7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences" ]

[ "Let the first six terms of Kitty\\textquotesingle s sequence be $$a$$, $$b$$, $$c$$, $$18$$ and $$29$$ respectively. Then $$c+ 18= 29$$, so $$c= 11$$. Hence $$b+11= 18$$, so $$b=7$$. Therefore, $$a+7=11$$, so $$a=4$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

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

[ "$$3\\times7=21$$, start from the $$3^{}\\rm{rd}$$ $$14\\times7=98$$, end at the $$14^{}\\rm{th}$$ Thus, the number of term remain: $$14-2=12$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12.5$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation" ]

[ "The pitcher is half full, i.e. $50 \\textbackslash\\%$ full. Therefore each cup receives $\\frac{50}{4}=(\\mathbf{C}) 12.5$ percent of the total capacity. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0.077$$ " } ], [ { "aoVal": "B", "content": "$$0.123$$ " } ], [ { "aoVal": "C", "content": "$$0.134$$ " } ], [ { "aoVal": "D", "content": "$$0.618$$ " } ], [ { "aoVal": "E", "content": "$$0.923$$ " } ] ]

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

[ "\\textbf{A: receive a merit B: hour \\textless{} 90} \\textbf{P(B) = P(hour \\textless{} 90) = P(Z\\textless(90-80)/7) = 0.9236} \\textbf{P(A) = 0.2} \\textbf{P(hour \\textgreater{} x) = 0.2} \\textbf{P(hour ≤ x) = 0.8} \\textbf{P(Z ≤ (x-80)/7) =0.8} \\textbf{P(A∩B) = P(x\\textless hour\\textless90) = P($$\\frac{x-80}{7} \\textless{} Z \\textless{} \\frac{90-80}{7}$$) = P($$Z \\textless{} \\frac{90-80}{7}$$) - P($$Z \\textless{} \\frac{x-80}{7}$$) = 0.9236-0.8 = 0.1236} \\textbf{P(A\\textbar B) = $$\\frac{P(A∩B)}{P(B)}$$ = $$\\frac{0.1236}{0.9236}$$=0.134} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "A " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "$$1\\textasciitilde\\text{kg}$$ of strawberries costs $$2a+b$$, so $$3\\textasciitilde\\text{kg}$$ of strawberries costs $$3\\times (2a+b)=6a+3b$$. So, the answer is C. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1974.5$$ " } ], [ { "aoVal": "B", "content": "$$1975.5$$ " } ], [ { "aoVal": "C", "content": "$$1976.5$$ " } ], [ { "aoVal": "D", "content": "$$1977.5$$ " } ], [ { "aoVal": "E", "content": "$$1978.5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables" ]

[ "We want to know the $2021^{th}$ term and the $2022^{th}$ term to get the median. We know that $44^{2}=1936\\textless2021$, and $45^{2}=2025\\textgreater2021$. So, the number $1^{2}$, $2^{2}$, $3^{2}$, $\\cdots $, $44^{2}$ are between $1$ to $1936$. $1936+44=1980$, which mean that $1936$ is the $1980^{th}$ number. Thus, the $2021^{th}$ term will be $1936+41=1977$, and similarly the $2021^{th}$ term will be $1978$. So, the answer is $1977.5$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions" ]

[ "The sum of the ratios is 10 . Since the sum of the angles of a triangle is $180^{\\circ}$, the ratio can be scaled up to $54: 54: 72(3 \\cdot 18: 3 \\cdot 18: 4 \\cdot 18)$. The numbers in the ratio $54: 54: 72$ represent the angles of the triangle. The question asks for the largest, so the answer is " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0.0918$$ " } ], [ { "aoVal": "B", "content": "$$0.4082$$ " } ], [ { "aoVal": "C", "content": "$$0.9082$$ " } ], [ { "aoVal": "D", "content": "$$-0.0918$$ " } ], [ { "aoVal": "E", "content": "$$0$$ " } ] ]

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

[ "P(X\\textless3.0) = $P(z\\textless\\frac{3-3.4}{0.3}=-1.33)$ = 0.0918 or normalcdf(-100, 3, 3.4, 0.3) = 0.0912 " ]

[ "其它" ]

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

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

[ "A " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{2020}{2021}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2021}{2022}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2022}{2023}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2023}{2024}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating" ]

[ "Sugar water theory. 1 gram of sugar added each time, and the sugar water gets sweeter. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ], [ { "aoVal": "E", "content": "$$33$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities" ]

[ "We know from the triangle inequality that the last side, $s$, fulfills $s\\textless12+13$. Adding $12+13$ to both sides of the inequality, we get $s+12+13\\textless50$, and because $s+12+13$ is the perimeter of our triangle, (B) 50 is our answer. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$1$ to $5$, $\\frac{0.5}{1}$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{5}$$, $1:5$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{5}$, $5$ to $10$ " } ], [ { "aoVal": "D", "content": "$1:5$, $\\frac{5}{10}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas" ]

[ "$$A$$, $1$ to $5$ $$=\\frac{1}{5}$$, $$\\frac{0.5}{1}=\\frac{5}{10}=\\frac{1}{2}$$, so wrong. $$B$$, $$\\frac{1}{5}=1:5$$, so true. $$C$$, $5$ to $10$ $$=\\frac{5}{10}=\\frac{1}{2}$$, so wrong. $$D$$, $$1:5=\\frac{1}{5}$$, $$\\frac{5}{10}=\\frac{1}{2}$$, so wrong. " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$200$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$2000$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "$$10000\\div 200=50$$; $$50\\times \\underline{200}=10000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$120.221$$ " } ], [ { "aoVal": "B", "content": "$$4.401$$ " } ], [ { "aoVal": "C", "content": "$$2424.390$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals" ]

[ "$$2424.390=2424.39$$ " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable" ]

[ "$6x-3y=2$, $y=2x-\\frac{2}{3}$, Its $y$-intercept is $-\\frac{2}{3}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition" ]

[ "Nil " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$360$$ " } ], [ { "aoVal": "B", "content": "$$400$$ " } ], [ { "aoVal": "C", "content": "$$420$$ " } ], [ { "aoVal": "D", "content": "$$440$$ " } ], [ { "aoVal": "E", "content": "$$480$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations" ]

[ "$$ x+\\frac{x^{3}}{y^{2}}+\\frac{y^{3}}{x^{2}}+y=x+\\frac{x^{3}}{y^{2}}+y+\\frac{y^{3}}{x^{2}}=\\frac{x^{3}}{x^{2}}+\\frac{y^{3}}{x^{2}}+\\frac{y^{3}}{y^{2}}+\\frac{x^{3}}{y^{2}} $$ Continuing to combine $$ \\frac{x^{3}+y^{3}}{x^{2}}+\\frac{x^{3}+y^{3}}{y^{2}}=\\frac{\\left(x^{2}+y^{2}\\right)\\left(x^{3}+y^{3}\\right)}{x^{2} y^{2}}=\\frac{\\left(x^{2}+y^{2}\\right)(x+y)\\left(x^{2}-x y+y^{2}\\right)}{x^{2} y^{2}} $$ From the givens, it can be concluded that $x^{2} y^{2}=4$. Also, $$ (x+y)^{2}=x^{2}+2 x y+y^{2}=16 $$ This means that $x^{2}+y^{2}=20$. Substituting this information into $\\frac{\\left(x^{2}+y^{2}\\right)(x+y)\\left(x^{2}-x y+y^{2}\\right)}{x^{2} y^{2}}$, we have $\\frac{(20)(4)(22)}{4}=20 \\cdot 22=$ 440. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "multiply by $4$ " } ], [ { "aoVal": "B", "content": "divide by $3$ " } ], [ { "aoVal": "C", "content": "add by $8$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions" ]

[ "$\\frac{3}{4}=\\frac{9}{12}$ $12-4=8$ " ]

[]

[ [ { "aoVal": "A", "content": "$$+++-$$ " } ], [ { "aoVal": "B", "content": "$$++++$$ " } ], [ { "aoVal": "C", "content": "$$++-\\/-$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities" ]

[ "$$6+6+6-6-6=6$$． " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution" ]

[ "We can write their relationships as the equations below: $A+B+P+P=12$ $A+P=5$ $B+P+P=A+P+5$ So, $B+P+P=5+5=10$, $A=12-10=2$, $P=5-2=3$, $B=12-2-3-3=4$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers" ]

[ "n/a. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$249$$ " } ], [ { "aoVal": "B", "content": "$$250$$ " } ], [ { "aoVal": "C", "content": "$$251$$ " } ], [ { "aoVal": "D", "content": "$$252$$ " } ], [ { "aoVal": "E", "content": "$$253$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Addition and Subtraction of Equations" ]

[ "Add the two equations. $$ 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c=14 . $$ Now, this can be rearranged and factored. $$ \\begin{aligned} \\&\\left(a^{2}-2 a b+b^{2}\\right)+\\left(a^{2}-2 a c+c^{2}\\right)+\\left(b^{2}-2 b c+c^{2}\\right)=14 \\textbackslash\\textbackslash{} \\&(a-b)^{2}+(a-c)^{2}+(b-c)^{2}=14 \\end{aligned} $$. $a, b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14=9+4+1$ $(a-c)^{2}=9 \\Rightarrow a-c=3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b=1$ or 2 , or whether $b-c=1$ or 2 . We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c)=(a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$. $a^{2}-(a-1)^{2}-(a-3)^{2}+a(a-1)=2011$, after some algebra, simplifies to $7 a=2021$. 2021 is not divisible by 7 , so $a$ is not an integer. The other case gives $a^{2}-(a-2)^{2}-(a-3)^{2}+a(a-2)=2011$, which simplifies to $8 a=2024$. Thus, $a=253$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns" ]

[ "We can first assume that all games ended in a $win-lose$ scenario. Number of games played $=3+2+1=6$ Maximum total score $=6\\times3=18$ Everytime a $win-lose$ scenario changes to a $draw-draw$ scenario, the total score decreases by $3-2=1$ Difference in score $=18-16=2$ Number of matches that ended in a draw $=2\\div1=2$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$255$$ " } ], [ { "aoVal": "B", "content": "$$256$$ " } ], [ { "aoVal": "C", "content": "$$257$$ " } ], [ { "aoVal": "D", "content": "$$258$$ " } ], [ { "aoVal": "E", "content": "$$259$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations" ]

[ "The sum of the first $m$ odd integers is given by $m^{2}$. The sum of the first $n$ even integers is given by $n(n+1)$. Thus, $m^{2}=n^{2}+n+212$. Since we want to solve for $n$, rearrange as a quadratic equation: $n^{2}+n+\\left(212-m^{2}\\right)=0$. Use the quadratic formula: $n=\\frac{-1+\\sqrt{1-4\\left(212-m^{2}\\right)}}{2}$. Since $n$ is clearly an integer, $1-4\\left(212-m^{2}\\right)=4 m^{2}-847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer. Let $x=\\sqrt{4 m^{2}-847}$; note that this means $n=\\frac{-1+x}{2}$. It can be rewritten as $x^{2}=4 m^{2}-847$, so $4 m^{2}-x^{2}=847$. Factoring the left side by using the difference of squares, we get $(2 m+x)(2 m-x)=847=7 \\cdot 11^{2}$. Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2 m+x)-(2 m-x)=2 m+x-2 m+x=2 x$. We have three pairs of factors, $847 \\cdot 1,121 \\cdot 7$, and $77 \\cdot 11$. The differences between these factors are 846,114 , and 66 - those are all possible values for $2 x$. Thus the possibilities for $x$ are $423$, $57$, and $33$. Now plug in these values into the equation $n=\\frac{-1+x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $255$. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{18}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions" ]

[ "The sum of the five given fractions is $$\\frac{1}{2}+\\frac{2}{3}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{9}+\\frac{1}{18}=\\frac{18+24+9+6+4+2}{36}$$. $$\\frac{63}{36}= \\frac{7}{4}=1 \\frac{3}{4}$$. So the fraction which is not used is $$1 \\frac{3}{4}-1 \\frac{1}{2}=\\frac{1}{4}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division with Remainders" ]

[ "$75\\div8=9R3$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9968$$ " } ], [ { "aoVal": "B", "content": "$$9982$$ " } ], [ { "aoVal": "C", "content": "$$9989$$ " } ], [ { "aoVal": "D", "content": "$$9996$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers" ]

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$\\rm 80 mm$$ " } ], [ { "aoVal": "B", "content": "$$\\rm 800 mm$$ " } ], [ { "aoVal": "C", "content": "$$\\rm 8000 mm$$ " } ], [ { "aoVal": "D", "content": "$$\\rm 80000 mm$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length" ]

[ "0.08km=80m；80m=80000mm " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$58$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

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

[ "D " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "$$2014\\times400 = 805600$$; the hundreds digit is $$6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Number Machine" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde f\\left( f\\left( f\\left( f\\left( f\\left( 17 \\right) \\right) \\right) \\right) \\right)$$ $$=f\\left( f\\left( f\\left( f\\left( 18 \\right) \\right) \\right) \\right)$$ $$=f\\left( f\\left( f\\left( 9 \\right) \\right) \\right)$$ $$=f\\left( f\\left( 10 \\right) \\right)$$ $$=f\\left( 5 \\right)$$ $$=6$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$0.0006$$ " } ], [ { "aoVal": "B", "content": "$$0.006$$ " } ], [ { "aoVal": "C", "content": "$$0.06$$ " } ], [ { "aoVal": "D", "content": "$$0.6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals" ]

[ "$$0.1\\times 0.2\\times 0.3=(0.1\\times 0.2)\\times 0.3=0.02\\times 0.3=0.006$$. Therefore, the answer is $$\\rm B$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$a$$ " } ], [ { "aoVal": "C", "content": "$$a+b=a+b$$ " } ], [ { "aoVal": "D", "content": "$$b-3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "Equations are not expressions. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{17}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering" ]

[ "$$Omitted.$$ " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{2}{5}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{2}{9}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{5}{12}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions" ]

[ "$$\\frac{1}{2+\\dfrac{1}{2+\\dfrac{1}{2}}}=\\frac{1}{2+ \\dfrac{1}{ \\dfrac{5}{2}}}= \\frac{1}{2+ \\dfrac{2}{5}}= \\frac{1}{ \\dfrac{12}{5}}= \\frac{5}{12}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "E " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

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

[ "\\textbf{This follows a geometric distribution.} $$\\mu = \\frac{1}{p} = \\frac{1}{0.10} = 10$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2454$$ " } ], [ { "aoVal": "B", "content": "$$2404$$ " } ], [ { "aoVal": "C", "content": "$$2444$$ " } ], [ { "aoVal": "D", "content": "$$2464$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division" ]

[ "omitted " ]

[]

[ [ { "aoVal": "A", "content": "$$600$$ " } ], [ { "aoVal": "B", "content": "$$900$$ " } ], [ { "aoVal": "C", "content": "$$1200$$ " } ], [ { "aoVal": "D", "content": "$$1800$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences->Sum of Terms in Arithmetic Sequences" ]

[ "$3\\times（1+24）\\times24\\div 2=3\\times300=900$. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "$$6+3=9$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{A and B only} " } ], [ { "aoVal": "B", "content": "\\textbf{A and C only} " } ], [ { "aoVal": "C", "content": "\\textbf{A, B, and C} " } ], [ { "aoVal": "D", "content": "\\textbf{B and C only} " } ], [ { "aoVal": "E", "content": "\\textbf{None are independent} " } ] ]

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

[ "\\textbf{P(B\\textbar A)\\ne P(B) → A and B not independent} \\textbf{P(B\\textbar C) =P(B) → B and C independent} \\textbf{P(A\\textbar C)=P(A∩B)/P(C)=0.05/0.1=0.5\\ne P(A)→ A and C not independent} " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form" ]

[ "$5+4=9$ " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities" ]

[ "When add the same number to both sides of the inequality, the equation is still true " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "If $$2$$ hoots $$=1$$ holler, then $$(10\\times 1)$$ hollers $$=(10\\times 2)$$ hoots. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$5$, $4$ and $2$ " } ], [ { "aoVal": "B", "content": "$6$, $2$ and $7$ " } ], [ { "aoVal": "C", "content": "$7$, $5$ and $8$ " } ], [ { "aoVal": "D", "content": "$8$, $7$ and $11$ " } ], [ { "aoVal": "E", "content": "$9$, $7$ and $12$ " } ] ]

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

[ "$6 \\times 7 - 5 = 21 + 8 \\times 2$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

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

[ "the new price is $$70\\textbackslash\\%$$ of the original price, so the new price is $$80\\times 70\\textbackslash\\%=56$$； $$80-56=24$$． so choose $$\\text{B}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$5\\dfrac{1}{8}$ " } ], [ { "aoVal": "B", "content": "$6\\dfrac{1}{4}$ " } ], [ { "aoVal": "C", "content": "$7\\dfrac{1}{2}$ " } ], [ { "aoVal": "D", "content": "$8\\dfrac{3}{4}$ " } ], [ { "aoVal": "E", "content": "$9\\dfrac{7}{8}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions" ]

[ "Assuming excesses of the other ingredients, the chocolate can make~$\\dfrac{5}{2}\\cdot5=12.5$~servings, the sugar can make~$\\dfrac{2}{1/4}\\cdot5=40$~servings, the water can make unlimited servings, and the milk can make~$\\dfrac{7}{4}\\cdot5=8.75$~servings. Limited by the amount of milk, Jordan can make at most~$\\boxed{\\left( D\\right)\\textbackslash{} 8\\dfrac{3}{4}}$~servings. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences" ]

[ "$1,3,5,7,9,11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

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

[ "D " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{773}{778}\\textgreater\\frac{884}{889}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{773}{778}\\textless{}\\frac{884}{889}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{773}{778}=\\frac{884}{889}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions" ]

[ "$$\\frac{773}{778}\\textless{}\\frac{(773+111)}{(778+111)}=\\frac{884}{889}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$200$$ " } ], [ { "aoVal": "B", "content": "$$300$$ " } ], [ { "aoVal": "C", "content": "$$600$$ " } ], [ { "aoVal": "D", "content": "$$919$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction " ]

[ "$$99+101 + 99+101 + 99+101 = 200 + 200 + 200 = 600$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "$(28 + 4) \\div 2 = 16$ " ]

[ "其它" ]

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

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

[ "Rearranging, we find $3 x+y=-2 x+6 y$, or $5 x=5 y \\Longrightarrow x=y$. Substituting, we can convert the second equation into $$\\frac{x+3 x}{3 x-x}=\\frac{4 x}{2 x}=\\text { (D) } 2$$. " ]

[ "其它" ]

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

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

[ "Dividing a term by $2$ is the same as multiplying it by half ($\\frac{1}{2}$)! " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$235$$ " } ], [ { "aoVal": "B", "content": "$$241$$ " } ], [ { "aoVal": "C", "content": "$$299$$ " } ], [ { "aoVal": "D", "content": "$$274$$ " } ], [ { "aoVal": "E", "content": "$$171$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences" ]

[ "By the formula~$d=\\dfrac{a\\_{m}-a\\_{n}}{m-n}\\textbackslash{} \\Rightarrow\\textbackslash{} d=\\dfrac{a\\_{55}-a\\_{16}}{55-16}=\\dfrac{157-40}{39}=3.$ By the formula~$a\\_{n}=a\\_{m}+\\left( n-m\\right)d\\textbackslash{} \\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} a\\_{81}=a\\_{55}+\\left( 81-55\\right)\\times3=235.$ " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution" ]

[ "We can write their relationships as the equations below: $$F+P+M+M+M+M=24$$ $$F+M=11$$ $$P+M+M=F+M-1$$ $ $ $$P+M+M=F+M-1$$, so $P+M+M=10$ $$F+P+M+M+M+M=24$$, so $11+10+M=24$, $M=3$, $P=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$9.25$$ " } ], [ { "aoVal": "B", "content": "$$92.5$$ " } ], [ { "aoVal": "C", "content": "$$925$$ " } ] ]

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

[ "$$9.25\\times 0.8+9\\frac{1}{4}\\times 0.2$$ $$=9.25\\times 0.8+9.25\\times 0.2$$ $$=9.25\\times (0.8+0.2)$$ $$=9.25\\times 1$$ $$=9.25$$ So, $$\\text{A}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$a-b=2c$$ " } ], [ { "aoVal": "B", "content": "$$z$$ " } ], [ { "aoVal": "C", "content": "$$a+b-5$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions" ]

[ "Equations are not expressions. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$140$$ " } ], [ { "aoVal": "B", "content": "$$175$$ " } ], [ { "aoVal": "C", "content": "$$210$$ " } ], [ { "aoVal": "D", "content": "$$245$$ " } ], [ { "aoVal": "E", "content": "$$280$$ " } ] ]

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

[ "Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives. Let $d$ be the distance still needed to travel after $1$ hour. We have that $\\frac{d}{50}+1.5=\\frac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early. Simplifying gives $7 d+525=10 d$, or $d=175$. Now, we must add an extra $35$ miles traveled in the first hour, giving a total of (C) $210$ miles. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$300$$ " } ], [ { "aoVal": "B", "content": "$$383$$ " } ], [ { "aoVal": "C", "content": "$$384$$ " } ], [ { "aoVal": "D", "content": "$$385$$ " } ], [ { "aoVal": "E", "content": "$$400$$ " } ] ]

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

[ "\\textbf{$n \\geq (\\frac{Z\\_{\\alpha/2}}{ME})^{2} {p(1-p)}$} \\textbf{Since p is unknown, then use p = ½. $n \\geq (\\frac{Z\\_{\\alpha/2}}{2ME})^{2}$} \\textbf{$Z\\_{\\alpha/2}=1.96$ The ME is how far off from the true proportion you are willing to be, in this case, 0.05} \\textbf{$n \\geq (\\frac{1.96}{2*0.05})^{2} = 384.16$ round up} " ]

[]

[ [ { "aoVal": "A", "content": "$$99$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$101$$ " } ], [ { "aoVal": "D", "content": "$$102$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction " ]

[ "$$(101+100+\\cdots +3+2)-(100+99+\\cdots +2+1)=(101-1)=100$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\sqrt{36}$$ " } ], [ { "aoVal": "B", "content": "$$\\sqrt{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{144}$$ " } ], [ { "aoVal": "D", "content": "$$\\sqrt{169}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers" ]

[ "$$\\sqrt{9+16+144}=\\sqrt{169}=13=3+4+6=\\sqrt{9}+\\sqrt{16}+\\sqrt{36}$$. " ]