Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "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" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

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

[ "First, solve for $p$. Start with $3^{}p+3^{4}=90$. Then, change $3^{4}$ to $81$. Subtract $81$ from both sides to get $3^{}p=9$ and see that $p$ is $2$. Now, solve for $r$. Since $2^{}r+44=76$, $2^{}r$ must equal $32$, $r=5$. $pr$ equals $2\\times5$ which equals $10$. So, the answer is $10$. " ]

[ "其它" ]

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

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

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2017$$ " } ], [ { "aoVal": "B", "content": "$$2018$$ " } ], [ { "aoVal": "C", "content": "$$2019$$ " } ], [ { "aoVal": "D", "content": "$$2020$$ " } ], [ { "aoVal": "E", "content": "$$2021$$ " } ] ]

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

[ "Grouping the first and last terms and two middle terms gives $\\left(x^{2}+5 x+4\\right)\\left(x^{2}+5 x+6\\right)+2019$, which can be simplified to $\\left(x^{2}+5 x+5\\right)^{2}-1+2019$. Noting that squares are nonnegative, and verifying that $x^{2}+5 x+5=0$ for some real $x$, the answer is 2018. " ]

[ "其它" ]

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

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

[ "$$14-6=8$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

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

[ "To evaluate $$60\\div 4+1\\times 3$$, we first do the $$\\times $$ and $$\\div $$ in the order in which they appear. Do the addition last. We get $$15+3=18$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$199\\times 99$$ " } ], [ { "aoVal": "B", "content": "$$198\\times 99$$ " } ], [ { "aoVal": "C", "content": "$$100\\times 100$$ " } ], [ { "aoVal": "D", "content": "$$100\\times 99$$ " } ] ]

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

[ "$$(100-1)\\times 99=(100\\times 99)-(1\\times 99)$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4000$$ " } ], [ { "aoVal": "B", "content": "$$1000$$ " } ], [ { "aoVal": "C", "content": "$$1545$$ " } ], [ { "aoVal": "D", "content": "$$4545$$ " } ] ]

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

[ "$$1+2+3+\\ldots\\ldots+37+38+39+40+39+38+37+\\ldots\\ldots+13+12+11$$ $$=（1+2+3+\\ldots\\ldots+37+38+39+40+39+38+37+\\ldots\\ldots+3+2+1）-（1+2+3+\\ldots\\ldots+8+9+10）$$ $$=40\\times40-55$$ $$=1545.$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "$$45$$ " } ] ]

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

[ "$54 \\div (5 + 1) = 9$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1.2$$ " } ], [ { "aoVal": "B", "content": "$$3.1$$ " } ], [ { "aoVal": "C", "content": "$$3.2$$ " } ], [ { "aoVal": "D", "content": "$$5.2$$ " } ] ]

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

[ "When $w= 2$, $$\\frac{21w}{10}-w+1= \\frac{21 \\times 2}{10}-2+1$$ $$= \\frac{42}{10}-2+1$$ $=4.2-2+1$ $= 2.2 + 1$ $= 3.2$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2^{24}$$\\textless{} $$10^{8}$$\\textless$$5^{12}$$ " } ], [ { "aoVal": "B", "content": "$$2^{24}$$\\textless$$5^{12}$$\\textless$$10^{8}$$ " } ], [ { "aoVal": "C", "content": "$$5^{12}$$\\textless$$2^{24}$$\\textless$$10^{8}$$ " } ], [ { "aoVal": "D", "content": "$$10^{8}$$\\textless$$5^{12}$$\\textless$$2^{24}$$ " } ], [ { "aoVal": "E", "content": "$$10^{8}$$\\textless$$2^{24}$$\\textless$$5^{12}$$ " } ] ]

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

[ "$$\\rm Method$$ $$1$$: Since all of the exponents are multiples of $4$, we can simplify the problem by taking the fourth root of each number. Evaluating we get $$10^{2}=100$$, $$5^{3}=125$$, and $$2^{6}=64$$. $$64\\textless100\\textless125$$. So, $$2^{24} \\textless$$ $$10^{8}$$ $$\\textless{} 5^{12}$$. $$\\rm Method$$ $$2$$: First, let us make all exponents equal to $$8$$. Then, it will be easy to order the numbers without doing any computations. $$10^{8}$$ is fine as it is. We can rewrite $$2^{24}$$ as $$(2^{3})^{8}=8^{8}$$. We can rewrite $$5^{12}$$ as $$\\left( 5^{\\frac{3}{2}}\\right)^{8}=\\left( \\sqrt{125}\\right)^{8}$$. We take the eighth root of all of these to get $$10$$, $$8$$,~$\\sqrt{125}$. Obviously, $$8\\textless10\\textless\\sqrt{ 125}$$, so $$2^{24}\\textless10^{8}\\textless5^{12}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{307}{97}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{16}{97}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{145}{16}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{372}{97}$ " } ] ]

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

[ "$=3+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{81}{16}}}$ $=3+\\dfrac{1}{1+\\dfrac{16}{81}}$ $=3+\\dfrac{1}{\\dfrac{97}{81}}$ $=3+\\dfrac{81}{97}$ $=\\dfrac{372}{97}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$xy+91$ " } ], [ { "aoVal": "B", "content": "$5=3$ " } ], [ { "aoVal": "C", "content": "$x+1=y+4$ " } ] ]

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

[ "both sides are algebraic expressions. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{7}{12};2$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{3}; \\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{12};2$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}; \\frac{1}{9}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{12}; \\frac{1}{9}$$ " } ] ]

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

[ "$$\\frac{3}{4}+ \\frac{1}{4}\\times \\frac{2}{3}- \\frac{1}{3}= \\frac{3}{4}+\\frac{1}{6}- \\frac{1}{3}=\\frac{9}{12}+\\frac{2}{12}-\\frac{4}{12}=\\frac{7}{12}$$; $$\\frac{2}{47}\\times 15 \\times \\frac{47}{36}\\times \\frac{2}{15}\\div \\frac{1}{18}= \\frac{4}{36}\\times 18=2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$180$$ " } ], [ { "aoVal": "B", "content": "$$200$$ " } ], [ { "aoVal": "C", "content": "$$300$$ " } ], [ { "aoVal": "D", "content": "$$480$$ " } ] ]

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

[ "$$80\\div40\\textbackslash\\%=200$$． so choose $$\\text{B}$$． " ]

[ "其它" ]

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

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

[ "$30\\div3=10$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2523$$ " } ], [ { "aoVal": "B", "content": "$$2623$$ " } ], [ { "aoVal": "C", "content": "$$2723$$ " } ], [ { "aoVal": "D", "content": "$$2823$$ " } ] ]

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

[ "The sequence increases in this order: $$300$$, $$400$$, $$500$$, $$600$$, $$\\dots$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$49$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ], [ { "aoVal": "E", "content": "$$90$$ " } ] ]

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

[ "We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$ \\begin{gathered} 15 c-15=g \\textbackslash\\textbackslash{} 12 c+6=g \\end{gathered} $$ Therefore, $15 c-15=12 c+6$. This implies that $c=7$. We therefore have $g=90$. So, our answer is (E) $90$ " ]

[ "其它" ]

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

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

[ "$$6\\times \\dfrac{11}{12}=\\dfrac{11}{2}=5\\dfrac{1}{2}$$． $$\\dfrac{11}{24}\\times 16=\\dfrac{22}{3}=7\\dfrac{1}{3}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{2}{7}$ " } ], [ { "aoVal": "B", "content": "$\\frac{5}{42}$ " } ], [ { "aoVal": "C", "content": "$\\frac{11}{14}$ " } ], [ { "aoVal": "D", "content": "$\\frac{5}{7}$ " } ], [ { "aoVal": "E", "content": "$\\frac{6}{7}$ " } ] ]

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

[ "We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put the remaining point, so we have $8 \\cdot 4$ choices. The total amount of choices is $\\left(\\begin{array}{l}8 \\textbackslash\\textbackslash{} 3\\end{array}\\right)=8 \\cdot 7$. Thus, our answer is $\\frac{8+8 \\cdot 4}{8 \\cdot 7}=\\frac{1+4}{7}=$ (D) $\\frac{5}{7}$. " ]

[]

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

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

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde1\\times12\\times23\\times34\\times45\\times \\cdots \\times78\\times89$$ $$=1\\times6\\times23\\times34\\times9\\times \\cdots \\times78\\times89\\times2\\times5$$ ∴$$1\\times6\\times3\\times4\\times9\\times6\\times7\\times8\\times9$$ has the last digit of $2$ ∴ the last two digits are $20.$ " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{9}{23}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{7}{23}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{46}$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

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

[ "$$3.75\\times 20\\textbackslash\\%\\times \\frac{3}{23}-4\\times \\frac{7}{23}+\\frac{9}{23}\\times 3\\frac{1}{4}$$ $$=\\frac{15}{4}\\times \\frac{1}{5}\\times \\frac{3}{23}-\\frac{4\\times 7}{23}+\\frac{9}{23}\\times \\frac{13}{4}$$ $$=\\frac{3}{4}\\times \\frac{3}{23}-\\frac{28}{23}+\\frac{9\\times 13}{4\\times 23}$$ $$=\\frac{9}{92}-\\frac{28\\times 4}{23\\times 4}+\\frac{117}{92}$$ $$=\\frac{9-112+117}{92}$$ $$=\\frac{14}{92}$$ $$=\\frac{7}{46}$$． " ]

[]

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

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Extracting Common Factors from Whole Numbers" ]

[ "$$5+10+15+20+25=1\\times 5+2\\times 5+\\cdots+5\\times 5$$ $$=\\left(1+2+3+4+5\\right)\\times 5$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$405$$ " } ], [ { "aoVal": "B", "content": "$$407$$ " } ], [ { "aoVal": "C", "content": "$$409$$ " } ], [ { "aoVal": "D", "content": "$$411$$ " } ], [ { "aoVal": "E", "content": "$$413$$ " } ] ]

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

[ "D " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{P(B) = 0.37 , if A and B are mutually exclusive.} " } ], [ { "aoVal": "B", "content": "\\textbf{P(B) = 0.561, if A and B are independent.} " } ], [ { "aoVal": "C", "content": "\\textbf{P(B) cannot be determined if A and B are neither mutually exclusive nor independent.} " } ], [ { "aoVal": "D", "content": "\\textbf{P(A and B) = 0.191 , if A and B are independent.} " } ], [ { "aoVal": "E", "content": "\\textbf{P(A\\textbar B) = 0.34 , if A and B are mutually exclusive.} " } ] ]

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

[ "\\textbf{P(A) = 0.34, P(A ∪ B) = 0.71 =~ P(A) + P(B) -- P(A ∩ B)} \\textbf{(A) If A and B are mutually exclusive, P(A ∩ B) = 0. Then, P(A ∪ B) = P(A) + P(B) which is 0.71= 0.34+P(B). So P(B) = 0.37.} \\textbf{(B) If A and B is independent, P(A ∩ B) = P(A)*P(B). Then, P(A ∪ B) =~ P(A) + P(B) --~ P(A)*P(B) which 0.71 = 0.34 + P(B) - 0.34*P(B). So P(B) = 0.561} \\textbf{(C) P(A ∪ B) = P(A) + P(B) -- P(A ∩ B). We have to know it is mutually exclusive or independent in order to know P(A ∩ B).} \\textbf{(D) If A and B is independent, P(A ∩ B) = P(A)*P(B) =0.34*0.561=0.191} \\textbf{(E) P(A \\textbar{} B)=P(A∩B)P(B) = 0} " ]

[]

[ [ { "aoVal": "A", "content": "$$20:9$$ " } ], [ { "aoVal": "B", "content": "$$5:3$$ " } ], [ { "aoVal": "C", "content": "$$40:21$$ " } ], [ { "aoVal": "D", "content": "$$5:8$$ " } ] ]

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

[ "$$a:b=20:24$$, $$b:c=24:9$$, $$a:c=20:9$$ " ]

[ "其它" ]

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

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

[ "There were $$4 \\times 4 = 16$$ monkeys in total. After the second whistle, there were $$16 \\div 8 = 2$$ monkeys in each row. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4444$$ " } ], [ { "aoVal": "B", "content": "$$4494$$ " } ], [ { "aoVal": "C", "content": "$$4944$$ " } ], [ { "aoVal": "D", "content": "$$9444$$ " } ], [ { "aoVal": "E", "content": "$$9944$$ " } ] ]

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

[ "C " ]

[ "其它" ]

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

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

[ "Like terms are terms that have the same variables. For each variable, the number of times it is multiplied by itself is also the same. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2021}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2019}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2019}{2021}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2019}{3}$$ " } ] ]

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

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{1}{3}\\times \\frac{3}{5}\\times \\frac{5}{7}\\times \\cdots \\cdots \\times \\frac{2019}{2021}$$ $$=\\frac{1}{2021}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$\\frac12$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ], [ { "aoVal": "E", "content": "It cannot be determined. " } ] ]

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

[ "Their fractions are just reciprocal of the other one. Thus, the product of the two fractions should be $1.$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ], [ { "aoVal": "E", "content": "$$64$$ " } ] ]

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

[ "C " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{0.15} " } ], [ { "aoVal": "B", "content": "\\textbf{0.50; either he makes it or he doesn't} " } ], [ { "aoVal": "C", "content": "\\textbf{0.80} " } ], [ { "aoVal": "D", "content": "\\textbf{1.2} " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

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

[ "\\textbf{160/200=0.8} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$140$$ " } ], [ { "aoVal": "B", "content": "$$160$$ " } ], [ { "aoVal": "C", "content": "$$180$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ], [ { "aoVal": "E", "content": "None of the answer above " } ] ]

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

[ "Grouping: $$5+15+17+23=60$$ $$60\\times3=180$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2016$$ " } ], [ { "aoVal": "B", "content": "$$2107$$ " } ], [ { "aoVal": "C", "content": "$$2018$$ " } ], [ { "aoVal": "D", "content": "$$2109$$ " } ], [ { "aoVal": "E", "content": "$$2020$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas->1²+2²+3²+......+n²=1/6n(n+1)(n+2)" ]

[ "Directly substitute $$n$$ with $$18$$ into the formula to get $$1^{2}+2^{2}+3^{2}+\\cdots +18^{2}=\\frac{18\\times 19\\times 37}{6}=2109$$. So the answer is $$\\text{D}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$49$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$66$$ " } ], [ { "aoVal": "D", "content": "$$74$$ " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

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

[ "Before page $50$, the digit $5$ appears $5$ times. From page $50$ to $59$, the digit $5$ appears $11$ times. Then, the digit $5$ appears the $17$\\textsuperscript{th}~time in $65$. Thus, the maximum page number is $64$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$98$$ " } ], [ { "aoVal": "B", "content": "$$2.5$$ " } ], [ { "aoVal": "C", "content": "$$-2.5$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "\\textbf{residual =$$\\hat{y}$$-y = 98-95.5 = 2.5} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$110$ " } ], [ { "aoVal": "B", "content": "$$104$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

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

[ "$1.3\\times 0.8=1.04$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

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

[ "$$(2-1)+(4 -3)+(6-5)+(8-7)+(10-9)=5$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2010$$ " } ], [ { "aoVal": "B", "content": "$$2008$$ " } ], [ { "aoVal": "C", "content": "$$1610$$ " } ], [ { "aoVal": "D", "content": "$$2015$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$400 \\times 5 + 1+2+3+4+5 = 2000 + 15 = 2015$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "two " } ], [ { "aoVal": "B", "content": "three " } ], [ { "aoVal": "C", "content": "four " } ], [ { "aoVal": "D", "content": "five " } ] ]

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

[ "$-4.3$, $-9.97$, and $$-\\frac{20}{21}$$ are negative numbers. " ]

[ "其它" ]

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

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

[ "let $a$ be legs, $b$ be the base. $a: 1,2,3,4,5,6,7,8,9,10,11$ $b:21,19,17,15,13,11,9,7,5,3,1$ Since $2a\\textgreater b$, $a\\textgreater5$, there are $6$ possible $a$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$x=60-34\\times 10\\textbackslash\\%-34$, let $x$ be the amount of tips " } ], [ { "aoVal": "B", "content": "$x=60-34$, let $x$ be the amount of tips " } ], [ { "aoVal": "C", "content": "$x \\leq 60-34\\times 10\\textbackslash\\%-34$, let $x$ be the amount of tips " } ], [ { "aoVal": "D", "content": "$x\\leq 60-34$, let $x$ be the amount of tips " } ] ]

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

[ "Robin needs to pay the food and taxes, the rest he can decide how much he wants to pay for tips. Therefore, the tip should be represented using inequality. " ]

[]

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

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

[ "$$\\frac{5}{37}=0.\\overline{135}$$, it is a decimal which repeats in cycles of $3$ digits. $2021\\div 3=673$$R2$, so the $2021$$^{st}$ digit is $3$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$29$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "$(65 + 7) \\div 2 = 36$ " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{99}{50}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{99}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{99}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{99}{200}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{50}{99}$$ " } ] ]

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

[ "$${{a}\\_{n}}=\\frac{{{\\left( n+1 \\right)}^{2}}}{\\left( n+1+1 \\right)\\left( n+1-1 \\right)}=\\frac{{{\\left( n+1 \\right)}^{2}}}{n\\left( n+2 \\right)}$$. $$=\\frac{2\\times 2}{(2+1)\\times (2-1)}\\times \\frac{3\\times 3}{(3+1)\\times (3-1)}\\times \\frac{4\\times 4}{(4+1)\\times (4-1)}\\times \\cdots \\times \\frac{98\\times 98}{(98+1)\\times (98-1)}\\times \\frac{99\\times 99}{(99+1)\\times (99-1)}$$ $$=\\frac{2\\times 2}{3\\times 1}\\times \\frac{3\\times 3}{4\\times 2}\\times \\frac{4\\times 4}{5\\times 3}\\times \\frac{5\\times 5}{6\\times 4}\\times \\cdots \\times \\frac{98\\times 98}{99\\times 97}\\times \\frac{99\\times 99}{100\\times 98}$$ $$=\\frac{2}{1}\\times \\frac{2}{3}\\times \\frac{3}{2}\\times \\frac{3}{4}\\times \\frac{4}{3}\\times \\frac{4}{5}\\times \\cdots \\times \\frac{98}{97}\\times \\frac{98}{99}\\times \\frac{99}{98}\\times \\frac{99}{100}$$ $$=\\frac{2}{1}\\times \\frac{99}{100}$$ $$=\\frac{99}{50}$$ " ]

[ "其它" ]

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

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

[ "Let the smallest number be $x$. It follows that the largest number is $4 x$. Since $x, 15$, and $4 x$ are equally spaced on a number line, we have $$ \\begin{aligned} 4 x-15 \\& =15-x \\textbackslash\\textbackslash{} 5 x \\& =30 \\textbackslash\\textbackslash{} x \\& =(C) 6. \\end{aligned} $$ " ]

[]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

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

[ "$$=\\left( 100-99 \\right)+\\left( 98-97 \\right)+\\left( 96-95 \\right)+\\cdots +\\left( 4-3 \\right)+\\left( 2-1 \\right)$$ $$=1+1+1+\\cdots +1+1=50$$． " ]

[]

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

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

[ "$$\\frac{12}{3}=\\frac{4}{1}=\\frac{20}{5}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14.25 $$ " } ], [ { "aoVal": "B", "content": "$$28.5 $$ " } ], [ { "aoVal": "C", "content": "$$42.75 $$ " } ], [ { "aoVal": "D", "content": "$$57 $$ " } ], [ { "aoVal": "E", "content": "$$71.25$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Extracting Common Factors from Decimals" ]

[ "It is a general rule that $$x\\textbackslash\\%$$ of $$y$$ equals $$y\\textbackslash\\%$$ of $$x$$. This is because $$x\\textbackslash\\%$$ of $$y= \\frac{x}{100} \\times y= \\frac{xy}{100}$$ and $$y \\textbackslash\\% $$ of $$x= \\frac{y}{100} \\times x= \\frac{yx}{100}$$. So $$40\\textbackslash\\%$$ of $$28.5$$ plus $$28.5\\textbackslash\\%$$ of $$60 = 40\\textbackslash\\%$$ of $$28.5$$ plus $$60\\textbackslash\\%$$ of $$28.5 = 100\\textbackslash\\%$$ of $$28.5 = 28.5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$77$$ " } ] ]

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

[ "The $25$\\textsuperscript{th} number: $5+(25-1)\\times3=77$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$203$$ " } ], [ { "aoVal": "B", "content": "$$181$$ " } ], [ { "aoVal": "C", "content": "$$194$$ " } ], [ { "aoVal": "D", "content": "$$212$$ " } ], [ { "aoVal": "E", "content": "$$185$$ " } ] ]

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

[ "$77+9\\times (14-1)=194$ " ]

[ "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": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ] ]

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

[ "$$94 - 49 = 45$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0.99$$ " } ], [ { "aoVal": "B", "content": "$$0.994$$ " } ], [ { "aoVal": "C", "content": "$$0.995$$ " } ], [ { "aoVal": "D", "content": "$$1.00$$ " } ] ]

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

[ "Since the thousandth\\textquotesingle s digit is $$4$$, round $$0.9949$$ down to $$0.99$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$54$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

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

[ "$$B$$-$$A$$-$$D$$-$$C$$ has the rule: divide the input number by $$3$$; add $$3$$ to the input number;~multiply the input number by $$3$$; subtract $$3$$ from the input. So the inverse operation gives $$((60+3)\\div3-3)\\times3=54$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$36\\frac{5}{14}$$ " } ], [ { "aoVal": "B", "content": "$$25\\frac{5}{14}$$ " } ], [ { "aoVal": "C", "content": "$$36\\frac{1}{3}$$ " } ], [ { "aoVal": "D", "content": "$$25\\frac{1}{3}$$ " } ] ]

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

[ "$$\\begin{eqnarray}\\&\\&1+3\\frac{1}{6}+5\\frac{1}{12}+7\\frac{1}{20}+9\\frac{1}{30}+11\\frac{1}{42}\\textbackslash\\textbackslash{} \\&=\\&1+3+5+7+9+11+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}+\\frac{1}{42}\\textbackslash\\textbackslash{} \\&=\\&\\left[ (1+9)+(3+7)+(5+11) \\right]+\\left( \\frac{1}{2}-\\frac{1}{3} \\right)+\\left( \\frac{1}{3}-\\frac{1}{4} \\right)+\\left( \\frac{1}{4}-\\frac{1}{5} \\right)+\\left( \\frac{1}{5}-\\frac{1}{6} \\right)+\\left( \\frac{1}{6}-\\frac{1}{7} \\right)\\textbackslash\\textbackslash{} \\&=\\&36+\\left( \\frac{1}{2}-\\frac{1}{7} \\right)\\textbackslash\\textbackslash{} \\&=\\&36\\frac{5}{14}\\end{eqnarray}$$ " ]

[ "其它" ]

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

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

[ "The quality of each candy: $(105 - 75) \\div 2 = 15$ g The amount of the apples at first: $75~ \\div 15 + 5 = 10$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ], [ { "aoVal": "E", "content": "$$55$$ " } ] ]

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

[ "E " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

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

[ "The second to last digit is always $$2$$. For example $$5^{3}$$ is $$125$$. When multiplying this by $$5$$, we see that is it inevitable that the second to last digit remains $$2$$. Try repeated multiplication of $$5$$ on a calculator. " ]

[]

[ [ { "aoVal": "A", "content": "$$6724$$ " } ], [ { "aoVal": "B", "content": "$$6402$$ " } ], [ { "aoVal": "C", "content": "$$6416$$ " } ], [ { "aoVal": "D", "content": "$$6714$$ " } ] ]

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

[ "$$(8x)^{2}+2\\cdot(8x)\\cdot2+2^{2}=8^{2}x^{2}+32x+4=8^{2}10^{2}+32\\times10+4=6724$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$0$$ " } ] ]

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

[ "A " ]

[]

[ [ { "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 $$6 \\times7\\times8\\times9 \\times0$$. ` " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$63$$ " } ], [ { "aoVal": "E", "content": "$$81$$ " } ] ]

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

[ "We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$ \\begin{gathered} 8 c-1=g \\textbackslash\\textbackslash{} 6 c+15=g \\end{gathered} $$ Therefore, $8 c-1=6 c+15$. This implies that $c=8$. We therefore have $g=63$. So, our answer is (D) $63$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$2 x-4\\textless-5$ " } ], [ { "aoVal": "B", "content": "$1-x \\geq 2$ " } ], [ { "aoVal": "C", "content": "$2+x=1$ " } ], [ { "aoVal": "D", "content": "$\\frac{2}{x}\\textgreater x$ " } ] ]

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

[ "$A$ gives $-6\\textless-5$. True. $B$ gives $2 \\geq 2$. True. $C$ gives 1=1. True. $D$ gives $-2\\textgreater-1$. False. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{2}{15}$ " } ], [ { "aoVal": "B", "content": "$\\frac{4}{11}$ " } ], [ { "aoVal": "C", "content": "$\\frac{11}{30}$ " } ], [ { "aoVal": "D", "content": "$\\frac{3}{8}$ " } ], [ { "aoVal": "E", "content": "$\\frac{11}{15}$ " } ] ]

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

[ "Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\\frac{n}{3}=\\frac{2 s}{5}$, which simplifies to $n=\\frac{6 s}{5}$. Since we are trying to find the value of $\\frac{\\frac{n}{3}+\\frac{2 s}{5}}{n+s}$, we can just substitute $\\frac{6 s}{5}$ for $n$ into the equation. We then get a value of $\\frac{\\frac{6 s}{5}+\\frac{2 s}{5}}{\\frac{6 s}{5}+s}=\\frac{\\frac{6 s+6 s}{15}}{\\frac{11 s}{5}}=\\frac{\\frac{4 s}{5}}{\\frac{11 s}{5}}=\\left(\\right.$ B) $\\frac{4}{11}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

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

[ "Set up the proportion $\\frac{12 \\text { meals }}{18 \\text { people }}=\\frac{x \\text { meals }}{12 \\text { people }}$. Solving for $x$ gives us $x=8$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{2}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{2}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{2}{11}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{2}{9}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{2}{10}$$ " } ] ]

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

[ "Same numerator, so smaller denominator means larger fraction. " ]

[]

[ [ { "aoVal": "A", "content": "$$14.995\\leqslant $$ the actual length $$\\leqslant 15.005$$ " } ], [ { "aoVal": "B", "content": "$$14.9\\leqslant $$ the actual length $$\\textless~15.1$$ " } ], [ { "aoVal": "C", "content": "$$14.95\\leqslant $$ the actual length $$\\textless{} 15.05$$ " } ], [ { "aoVal": "D", "content": "$$14.99\\leqslant $$ the actual length $$\\textless15.01$$ " } ], [ { "aoVal": "E", "content": "$$14.5\\textless$$ the actual length $$\\textless15.5$$ " } ] ]

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

[ "14.9500000\\ldots{} 15.0499999\\ldots{} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ], [ { "aoVal": "E", "content": "$$70$$ " } ] ]

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

[ "$30 + (30 - 5) = 55$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

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

[ "$40\\times40\\textbackslash\\%=16$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1996$$ " } ], [ { "aoVal": "B", "content": "$$2006$$ " } ], [ { "aoVal": "C", "content": "$$2016$$ " } ], [ { "aoVal": "D", "content": "$$2026$$ " } ] ]

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

[ "$$33$$ hours and $$36$$ minutes $$=33\\times60+36$$ minutes $$=2016$$ minutes. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$132$$ " } ], [ { "aoVal": "C", "content": "$$135$$ " } ], [ { "aoVal": "D", "content": "$$168$$ " } ], [ { "aoVal": "E", "content": "$$171$$ " } ] ]

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

[ "$9 \\times 11 - 3 = 96$ " ]

[ "其它" ]

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

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

[ "$$15-8=7$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$9003$$ " } ], [ { "aoVal": "B", "content": "$$9002$$ " } ], [ { "aoVal": "C", "content": "$$8003$$ " } ], [ { "aoVal": "D", "content": "$$8002$$ " } ] ]

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

[ "$$8002-2008=(8002+1)-(2008+1)=8003-2009$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$-19$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

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

[ "By the definition, we can remove the mimus sign before $$-19$$. We get $$19$$ and choose $$\\text{A}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "As long as we can find the factor $2$ and $5$, the ones digit is $0$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2^{20}-2^{6}$$ " } ], [ { "aoVal": "B", "content": "$$2^{20}-2^{7}$$ " } ], [ { "aoVal": "C", "content": "$$2^{21}-2^{6}$$ " } ], [ { "aoVal": "D", "content": "$$2^{21}-2^{7}$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$1+2+\\cdots +2^{20}=2^{21}-1$$ $$1+2+\\cdots +2^{6} = 2^{7}-1$$ Subtracting the two equations we have: $$2^{7}+2^{8}+\\cdots +2^{20} = 2^{21}-2^{7}$$ " ]

[]

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

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

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4077$$ " } ], [ { "aoVal": "B", "content": "$$4160$$ " } ], [ { "aoVal": "C", "content": "$$4167$$ " } ], [ { "aoVal": "D", "content": "$$4170$$ " } ], [ { "aoVal": "E", "content": "$$4177$$ " } ] ]

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

[ "$4077 + 93 = 4170$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2122$$ " } ], [ { "aoVal": "B", "content": "$$2993$$ " } ], [ { "aoVal": "C", "content": "$$3013$$ " } ], [ { "aoVal": "D", "content": "$$3293$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "B " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$61$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ] ]

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

[ "$$7\\times3=21$$~ $21+40=61$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

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

[ "There is no dark blue ball in the box. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Her estimate is larger than $x-y$ " } ], [ { "aoVal": "B", "content": "Her estimate is smaller than $x-y$ " } ], [ { "aoVal": "C", "content": "Her estimate equals $x-y$ " } ], [ { "aoVal": "D", "content": "Her estimate equals $y-x$ " } ], [ { "aoVal": "E", "content": "Her estimate is $0$ " } ] ]

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

[ "Let\\textquotesingle s define $z$ as the amount rounded up by and down by. The problem statement tells us that Xiaoli performed the following computation: $(x+z)-(y-z)=x+z-y+z=x-y+2 z$ We can see that $x-y+2 z$ is greater than $x-y$, and so the answer is $A$. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\dfrac{9}{10}\\textless{} \\dfrac{7}{8}\\textless\\dfrac{5}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{5}{6}\\textless{} \\dfrac{7}{8}\\textless{} \\dfrac{9}{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{9}{10}\\textless{} \\dfrac{5}{6}\\textless{} \\dfrac{7}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{5}{6}\\textless{} \\dfrac{9}{10}\\textless{} \\dfrac{7}{8}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{7}{8}\\textless{} \\dfrac{5}{6}\\textless{} \\dfrac{9}{10}$$ " } ] ]

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

[ "Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus, $$1- \\dfrac{5}{6}= \\dfrac{1}{6}$$, $$1- \\dfrac{7}{8}= \\dfrac{1}{8}$$, $$1- \\dfrac{9}{10}= \\dfrac{1}{10}$$. All three fractions have the common numerator $$1$$. Now the order of the fractions is obvious. $$\\dfrac{1}{6}\\textgreater\\dfrac{1}{8}\\textgreater\\dfrac{1}{10}\\Rightarrow\\dfrac{5}{6}\\textless\\dfrac{7}{8}\\textless\\dfrac{9}{10}$$. Therefore, $$\\dfrac{5}{6}\\textless\\dfrac{7}{8}\\textless\\dfrac{9}{10}$$. " ]

[ "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->Basic Concepts of Equation->Proportional Equations" ]

[ "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 $\\frac{x+3 x}{3 x-x}=\\frac{4 x}{2 x}= 2$ More step-by-step explanation: $$ \\begin{aligned} \\&\\frac{3 x+y}{x-3 y}=-2 \\textbackslash\\textbackslash{} \\&3 x+y=-2(x-3 y) \\textbackslash\\textbackslash{} \\&3 x+y=-2 x+6 y \\textbackslash\\textbackslash{} \\&5 x=5 y \\textbackslash\\textbackslash{} \\&x=y \\textbackslash\\textbackslash{} \\&\\frac{x+3 y}{3 x-y}=\\frac{1+3(1)}{3(1)-1}=\\frac{4}{2}=2 \\end{aligned} $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "24, 3.49 " } ], [ { "aoVal": "B", "content": "24, 14.4 " } ], [ { "aoVal": "C", "content": "4.90, 3.79 " } ], [ { "aoVal": "D", "content": "4.90, 14.4 " } ], [ { "aoVal": "E", "content": "2.4, 3.79 " } ] ]

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

[ "$$\\mu\\_X = 60 * 0.4 = 24$$ $$\\sigma\\_X = \\sqrt{60*0.4*0.6} = 3.79$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

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

[ "The sum of the previous two number. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$x=y$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{h}$ " } ], [ { "aoVal": "D", "content": "$123xyzabc$ " } ] ]

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

[ "equation is not algebraic expression " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$70$$ " } ], [ { "aoVal": "B", "content": "$$84$$ " } ], [ { "aoVal": "C", "content": "$$98$$ " } ], [ { "aoVal": "D", "content": "$$105$$ " } ], [ { "aoVal": "E", "content": "$$126$$ " } ] ]

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

[ "Let $x$ be the length of the ship. Then, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps. Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps. Since the ship and Emily both travel at some constant rate, $\\frac{210}{210-x}=\\frac{42}{x-42}$. Dividing both sides by $42$ and cross multiplying, we get $5(x-42)=210-x$, so $6 x=420$, and $x=70$. " ]

[]

[ [ { "aoVal": "A", "content": "$$+$$；$$-$$；$$+$$；$$-$$；$$+$$ " } ], [ { "aoVal": "B", "content": "$$-$$；$$+$$；$$-$$；$$+$$；$$-$$ " } ], [ { "aoVal": "C", "content": "$$+$$；$$+$$；$$+$$；$$-$$；$$-$$ " } ], [ { "aoVal": "D", "content": "I don\\textquotesingle t know o(╥﹏╥)o " } ] ]

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

[ "$$5-5+5-5+5-5=0$$． Option$$\\text{B}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{~Approximately normal with mean 11.4 minutes and standard deviation 2.6 minutes} " } ], [ { "aoVal": "B", "content": "\\textbf{~Approximately normal with mean 11.4 minutes and standard deviation $\\frac{2.6}{\\sqrt{84}}$ minute~} " } ], [ { "aoVal": "C", "content": "\\textbf{Approximately normal with mean 12.0 minutes and standard deviation 2.6 minutes} " } ], [ { "aoVal": "D", "content": "\\textbf{Binomial with mean 84(0.41) minutes and standard deviation 84(0.41)(0.59) minutes~} " } ], [ { "aoVal": "E", "content": "\\textbf{Binomial with mean 84(0.5) minutes and standard deviation 84(0.5)(0.5) minutes} " } ] ]

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

[ "\\textbf{For a sufficiently large n, the sampling distribution of $$\\bar\\_{X}$$ is approximately normal, with mean $\\mu\\_{\\bar\\_{X}}=\\mu$ and standard deviation $\\sigma\\_\\bar\\_{X}=\\sqrt{\\frac{\\sigma}{\\sqrt{n}}}$} " ]

[ "其它" ]

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

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

[ "\\uline{NA} " ]

[]

[ [ { "aoVal": "A", "content": "$$2\\times \\left( 0+2+2 \\right)$$ " } ], [ { "aoVal": "B", "content": "$$2-0-2-2$$ " } ], [ { "aoVal": "C", "content": "$$2\\times 0\\times 2\\times 2$$ " } ], [ { "aoVal": "D", "content": "$$2+0+2+2$$ " } ] ]

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

[ "A " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac 1{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac 14$$ " } ], [ { "aoVal": "C", "content": "$$\\frac 12$$ " } ], [ { "aoVal": "D", "content": "$$\\frac {3}{4}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac {1}{8}$$ " } ] ]

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

[ "We notice that a lot of terms can be canceled. In fact, every term in the numerator except for the $$1$$ and every term in the denominator except for the $$4$$ will be canceled out, so the answer is $$\\frac 1{4}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$3.10$ dollars " } ], [ { "aoVal": "B", "content": "$10.70$ dollars " } ], [ { "aoVal": "C", "content": "$9.30$ dollars " } ], [ { "aoVal": "D", "content": "$8.70$ dollars " } ], [ { "aoVal": "E", "content": "$10.50$ dollars " } ] ]

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

[ "$2.40$ dollars=$240$ cents $240\\times 3+3\\times70=930$ cents=$9.30$ dollars. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$199$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "$${{100}^{2}}-{{99}^{2}}=(100+99)\\times (100-99)=199$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "omitted " ]