Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{13}{20}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{3}{4}$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{17}{20}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{6}{5}$$ " } ] ]

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

[ "First, write down the decimal \"over\" the number 1 :~$0.65=\\dfrac{0.65}{1}$ Then multiply top and bottom by 100 since there are two numbers after the decimal point :~$\\dfrac{0.65}{1}=\\dfrac{0.65\\times100}{1\\times100}=\\dfrac{65}{100}$ This makes it a correctly formed fraction. Then simplify the fraction (in this case by dividing top and bottom by 5) :~$\\dfrac{65}{100}=\\dfrac{13}{20}$ " ]

[]

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

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

[ "$$15+3-8=10$$，$$17-9=8$$，$$8+\\left( 2 \\right)=10$$，$$number$$ has to be bigger than $$2$$ so $$5$$ fits the requirement. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\textbackslash$ 2$ " } ], [ { "aoVal": "B", "content": "$\\textbackslash$ 10$ " } ], [ { "aoVal": "C", "content": "$\\textbackslash$ 12$ " } ], [ { "aoVal": "D", "content": "$\\textbackslash$ 22$ " } ] ]

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

[ "Consumer surplus is calculated as the difference between the maximum price a consumer is willing to pay for a good and the actual price they pay, which is $\\textbackslash$ 12$- $\\textbackslash$10$ = $\\textbackslash$ 2$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$55$$ " } ], [ { "aoVal": "B", "content": "$$385$$ " } ], [ { "aoVal": "C", "content": "$$1155$$ " } ], [ { "aoVal": "D", "content": "$$2310$$ " } ] ]

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

[ "Answer is $385$ " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{77}{30}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{6}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{17}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{41}{18}$$ " } ] ]

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

[ "$\\frac{5}{6}+\\frac{77}{30}=\\frac{25}{30}+\\frac{77}{30}=\\frac{102}{30}=\\frac{17}{5}$ " ]

[]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$31$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "Sum of the previous two number. " ]

[]

[ [ { "aoVal": "A", "content": "$$2\\times \\left( 0+2+2 \\right)\\times 2021$$ " } ], [ { "aoVal": "B", "content": "$$2\\times 2021-0-2-2$$ " } ], [ { "aoVal": "C", "content": "$$2021-2\\times 0\\times 2\\times 2$$ " } ], [ { "aoVal": "D", "content": "$$2+\\left( 0\\times 2\\times 2 \\right)\\times 2021$$ " } ] ]

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

[ "C " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "$$\\frac{2}{7}=0.\\overline{285714}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$$^{th}$ digit is $4$. The $2022$$$^{nd}$$ digit is $4$, so the $2021$$^{st}$ digit is $1$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$87$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$132$$ " } ], [ { "aoVal": "D", "content": "$$135$$ " } ] ]

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

[ "After adding $3$ toys, the number of toys should be divisible by $9$, and it also should be divisible by $2, 3, $ and $4$. Thus, the answer is $B$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1.4$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$140$$ " } ], [ { "aoVal": "D", "content": "$$1400$$ " } ] ]

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

[ "$$49\\div0.035=49000\\div35=1400$$, so the answer is $$\\rm{D}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{36}$ " } ], [ { "aoVal": "B", "content": "$\\frac{7}{72}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{9}$ " } ], [ { "aoVal": "D", "content": "$\\frac{5}{36}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{6}$ " } ] ]

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

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$5.5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$6.5$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$\\left (403 \\frac{3}{5}+183 \\frac{5}{11}+155 \\frac{3}{13}+118 \\frac{12}{17}\\right ) \\div$$$$ \\left ( \\frac{1009}{15}+ \\frac{1009}{33}+ \\frac{10009}{39}+ \\frac{1009}{51}\\right )$$ $$=2018\\left ( \\frac{1}{5}+ \\frac{1}{11}+ \\frac{1}{13}+ \\frac{1}{17}\\right ) \\div \\frac{1000}{3}\\left ( \\frac{1}{5}+ \\frac{1}{11}+ \\frac{1}{13}+ \\frac{1}{17}\\right )$$ $=6$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{23}{99}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{23}{90}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{25}{99}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{25}{90}$$ " } ] ]

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

[ "$$x=0.2\\dot{5}$$, $$100x=25. \\dot{5}$$, $$10x=2. \\dot{5}$$, $$90x=23$$, $$x= \\frac{23}{90}$$. " ]

[ "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->Ratios and Proportions" ]

[ "$1083\\div 1000\\times 12=12.996$ " ]

[ "其它" ]

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

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

[ "We know from the triangle inequality that the last side, $s$, fulfills $s+7\\textgreater12$. $P=s+7+12\\textgreater12+12$. Therefore, $P\\textgreater24+1=25$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$13$$ " } ] ]

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

[ "$7+4-6+7-3=9$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ], [ { "aoVal": "E", "content": "$$27$$ " } ] ]

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

[ "$3+7+2+8+1$ $=(3+7)+(2+8)+1$ $=10+10+1$ $=21$ " ]

[]

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

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

[ "$$3+4=7$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$1002$$ " } ], [ { "aoVal": "C", "content": "$$1003$$ " } ], [ { "aoVal": "D", "content": "$$2005$$ " } ], [ { "aoVal": "E", "content": "$$2006$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Fast Calculation in Fractions->Reducing Fractions by cancelling out successively" ]

[ "By telescoping, it\\textquotesingle s easy to see the sum becomes $$\\frac {2006}2=1003$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1:4$$ " } ], [ { "aoVal": "B", "content": "$$15:20$$ " } ], [ { "aoVal": "C", "content": "$$4:3$$ " } ], [ { "aoVal": "D", "content": "$$3:4$$ " } ] ]

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

[ "There are $$15$$ chairs and $$20$$ cups. So the ratio of chairs to cups is $$15:20$$. The simplest form is $$3:4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3.2\\times0.16$$ " } ], [ { "aoVal": "B", "content": "$$0.32\\times0.16$$ " } ], [ { "aoVal": "C", "content": "$$32\\times0.016$$ " } ], [ { "aoVal": "D", "content": "$$0.032\\times160$$ " } ] ]

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

[ "$$3.2\\times0.16=0.512$$, $$0.32\\times0.16=0.0512$$, $$32\\times0.016=0.512$$, $$0.032\\times160=5.12$$. Therefore, expression $$\\rm D$$ has the maximum value. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$29$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$39$$ " } ] ]

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

[ "$(71-7) \\div 2 = 32$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$190$$ " } ], [ { "aoVal": "B", "content": "$$200$$ " } ], [ { "aoVal": "C", "content": "$$210$$ " } ], [ { "aoVal": "D", "content": "$$220$$ " } ] ]

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

[ "C " ]

[ "其它" ]

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

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

[ "If these $10$ sentences are true, the nose will shorten $10$ cm. There are $(13-11+10)\\div(5+1)=2$ sentences which are false. Thus, $10-2=8$ sentences are true. " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{2}{5}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{3}{8}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{2}{9}$ " } ], [ { "aoVal": "E", "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": "$4; 20$ " } ], [ { "aoVal": "B", "content": "$16; 32$ " } ], [ { "aoVal": "C", "content": "$4; 32$ " } ], [ { "aoVal": "D", "content": "$4; 40$ " } ] ]

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

[ "When the denominator and numerator are multiplied by the same number, the value of the fraction remains equal. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$79$$ " } ], [ { "aoVal": "E", "content": "$$89$$ " } ] ]

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

[ "We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8 th graders, in order that we can put the two ratios together: $$ \\begin{aligned} \\&5: 3=5(8): 3(8)=40: 24 \\textbackslash\\textbackslash{} \\&8: 5=8(5): 5(5)=40: 25 \\end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40: 25: 24$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 40+25+24=\\text { (E) } 89 $$ " ]

[]

[ [ { "aoVal": "A", "content": "$$357$$ " } ], [ { "aoVal": "B", "content": "$$2357$$ " } ], [ { "aoVal": "C", "content": "$$2 \\times5 \\times7$$ " } ], [ { "aoVal": "D", "content": "$$3 \\times5 \\times7$$ " } ] ]

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

[ "The largest divisor of $$2\\times3\\times5\\times7$$ that is less than~ is $$3\\times5\\times7$$, which we get by dropping the smallest factor. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{h}$ " } ], [ { "aoVal": "C", "content": "$3x=5y$ " } ], [ { "aoVal": "D", "content": "$xyzabc$ " } ] ]

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

[ "equation is not algebraic expression " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$3\\times3\\times3\\times3=81$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$793$$ " } ], [ { "aoVal": "B", "content": "$$737$$ " } ], [ { "aoVal": "C", "content": "$$757$$ " } ], [ { "aoVal": "D", "content": "$$743$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "NA " ]

[ "其它" ]

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

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

[ "We can write the two digit number in the form of $10 a+b$; reverse of $10 a+b$ is $10 b+a$. The sum of those numbers is: $$ \\begin{gathered} (10 a+b)+(10 b+a)=132 \\textbackslash\\textbackslash{} 11 a+11 b=132 \\textbackslash\\textbackslash{} a+b=12 \\end{gathered} $$ We can use brute force to find order pairs $(a, b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$. Thus our ordered pairs are $(3,9) ;(4,8) ;(5,7) ;(6,6) ;(7,5) ;(8,4) ;(9,3)$ or $(\\mathbf{B}) 7$ ordered pairs. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

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

[ "Let $$x$$ be the number of cars. $$2x+4x=48$$ $$6x=48$$ $$x=8$$ " ]

[ "其它" ]

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

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

[ "$3 + 7 + 1 = 11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1007$$ " } ], [ { "aoVal": "B", "content": "$$1008$$ " } ], [ { "aoVal": "C", "content": "$$1009$$ " } ], [ { "aoVal": "D", "content": "$$1010$$ " } ], [ { "aoVal": "E", "content": "$$1011$$ " } ] ]

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

[ "$1+3+5+\\ldots+2017+2019-2 -4-6-\\ldots-2016-2018$ $=1+(3-2)+(5-4)+\\cdots +(2017-2016)+(2019-2018)$ $=1+1+1+\\cdots +1+1$ $=1010$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ], [ { "aoVal": "E", "content": "$$720$$ " } ] ]

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

[ "${6\\choose 2} = \\frac{6!}{2!(6-2)!} =15$ " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ cents " } ], [ { "aoVal": "B", "content": "$$30$$ cents " } ], [ { "aoVal": "C", "content": "$$40$$ cents " } ], [ { "aoVal": "D", "content": "$$50$$ cents " } ], [ { "aoVal": "E", "content": "$$60$$ cents " } ] ]

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

[ "$3.20$ dollars = $320$ cents $(320-40\\times6)\\div2=40$ cents " ]

[]

[ [ { "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": "$$0.6247$$ " } ], [ { "aoVal": "B", "content": "$$4115$$ " } ], [ { "aoVal": "C", "content": "$$3650$$ " } ], [ { "aoVal": "D", "content": "$$3123$$ " } ], [ { "aoVal": "E", "content": "$$3227$$ " } ] ]

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

[ "\\textbf{P(63 \\textless{} X \\textless{} 75) calculator: normalcdf(63, 75, 72, 6) = 0.6247~} \\textbf{0.6247*5000 = 3123.3} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$-3\\textless x\\textless-1$$ " } ], [ { "aoVal": "B", "content": "$$x\\textgreater-1$$ or $$x\\textless-3$$ " } ], [ { "aoVal": "C", "content": "$$1\\textless x\\textless3$$ " } ], [ { "aoVal": "D", "content": "$$x\\textgreater3$$ or $$x\\textless1$$ " } ] ]

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

[ "$$\\frac{2\\textbar x-2\\textbar+1}{3}\\textless{}1$$ $$2\\left\\textbar x-2\\right\\textbar+1\\textless3$$ $$2\\left\\textbar x-2\\right\\textbar\\textless2$$ $$1\\left\\textbar x-2\\right\\textbar\\textless1$$ $$-1\\textless x-2\\textless1$$ $$1\\textless x\\textless3$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$991$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$9000$$ " } ] ]

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

[ "$$900+90+9+1=999+1=1000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$189$$ minutes " } ], [ { "aoVal": "B", "content": "$$172$$ minutes " } ], [ { "aoVal": "C", "content": "$$216$$ minutes " } ], [ { "aoVal": "D", "content": "$$206$$ minutes " } ], [ { "aoVal": "E", "content": "None of the above. " } ] ]

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

[ "3: 55pm - 1: 47om + 15 minutes = 145 minutes. " ]

[]

[ [ { "aoVal": "A", "content": "$$33$$ " } ], [ { "aoVal": "B", "content": "$$44$$ " } ], [ { "aoVal": "C", "content": "$$66$$ " } ], [ { "aoVal": "D", "content": "$$77$$ " } ] ]

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

[ "$$55+55+55 = 44+11+44+11+55 = 44+44+(11+11+55)= 44+44+77$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "66.8\\% " } ], [ { "aoVal": "B", "content": "47.2\\% " } ], [ { "aoVal": "C", "content": "33.4\\% " } ], [ { "aoVal": "D", "content": "15.9\\% " } ], [ { "aoVal": "E", "content": "2.28\\% " } ] ]

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

[ "$P(X\\textgreater60) = P(Z\\textgreater\\frac{60-55.6}{2.2}) = p(Z\\textgreater2) =1-0.9772 = 0.0228$ " ]

[ "其它" ]

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

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

[ "We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$. Since the total number of students is equal to 28 , we get $x+x-4=28$. Solving this equation, we get $x=16$. There are $16-4=12$ boys in our class, and our answer is $16: 12=$ (B) $4: 3$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$625\\rm L$$ " } ], [ { "aoVal": "B", "content": "$$375\\rm L$$ " } ], [ { "aoVal": "C", "content": "$$525\\rm L$$ " } ], [ { "aoVal": "D", "content": "$$270\\rm L$$ " } ], [ { "aoVal": "E", "content": "$$675\\rm L$$ " } ] ]

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

[ "$$1050\\times \\frac {70-25}{70} = 675$$ " ]

[]

[ [ { "aoVal": "A", "content": "$\\left (10^{4}+1\\right )^{2}$ " } ], [ { "aoVal": "B", "content": "$\\left (10^{8}+1\\right )^{2}$ " } ], [ { "aoVal": "C", "content": "$\\left (10^{5}\\right )^{2}$ " } ], [ { "aoVal": "D", "content": "$\\left (10^{8}\\right )^{2}$ " } ], [ { "aoVal": "E", "content": "$\\left (10^{4}\\right )^{2}+1$ " } ] ]

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

[ "As we know, $1$, $4$, $9$, $16$ are square numbers. $10^{8}=\\left (10^{4}\\right )^{2}$, so, the next one is $\\left (10^{4}+1\\right )^{2}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$x\\leq-4$ " } ], [ { "aoVal": "B", "content": "$x\\geq-4$ " } ], [ { "aoVal": "C", "content": "$x\\leq-2$ " } ], [ { "aoVal": "D", "content": "$x\\geq-2$ " } ] ]

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

[ "$-10x\\leq40$ $x\\leq-4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

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

[ "Three pencils are three dollars. Notebook is two dollars. A three dollar pencil case and a one dollar free at last. So the last answer is $3+2+3-1=7$ " ]

[]

[ [ { "aoVal": "A", "content": "$$7y + 1$$ " } ], [ { "aoVal": "B", "content": "$$7y + 1 -x$$ " } ], [ { "aoVal": "C", "content": "$$7y-x$$ " } ], [ { "aoVal": "D", "content": "$$7y$$ " } ] ]

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

[ "$$7y+1-x-1=7y-x$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5150$$ " } ], [ { "aoVal": "B", "content": "$$10100$$ " } ], [ { "aoVal": "C", "content": "$$11050$$ " } ], [ { "aoVal": "D", "content": "$$20100$$ " } ] ]

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

[ "($$1$$ to $$200$$)$$=$$($$1$$ to $$100$$)$$+[(100+1)+(100+2)+\\cdots +(100+100)]=$$($$1$$ to $$100$$)$$+[(100\\times 100)+$$($$1$$ to $$100$$)$$]=5050+[10000+5050]=20100$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "2 metres and 18 centimetres " } ], [ { "aoVal": "B", "content": "2 kilometres and 18 centimetres " } ], [ { "aoVal": "C", "content": "20 metres and 18 centimetres " } ], [ { "aoVal": "D", "content": "201 metres and 8 centimetres " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "1m = 100cm; 1km=1000m A. 200 cm + 18 cm = 218 cm B. 200 000cm + 18cm = 200 018cm \\textbf{C. 2000cm + 18cm = 2018 cm} D. 20100cm + 8cm = 20108cm. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{50}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{200}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{400}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{800}$$ " } ] ]

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

[ "As $$1=1^{3}$$ and $$1000 = 10^{3}$$, there are $$10$$ cubes from $$1$$ to $$1000$$. So the fraction of the integers from $$1$$ to $$1000$$ inclusive which are cubes is $$\\frac{10}{1000}= \\frac{1}{100}$$. " ]

[]

[ [ { "aoVal": "A", "content": "Alice " } ], [ { "aoVal": "B", "content": "Bob " } ], [ { "aoVal": "C", "content": "Alan " } ], [ { "aoVal": "D", "content": "Tom " } ], [ { "aoVal": "E", "content": "Susan " } ] ]

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

[ "Alice：$12+8=20$ Bob：$7+9=16$ Alan：$14+5=19$~ Tom：$6+8=14$ Susan: $6+15=21$ $21\\textgreater20\\textgreater19\\textgreater16\\textgreater14$ " ]

[ "其它" ]

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

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

[ "Subtract a negative number is the same as add a postive number with the same absolute value. $a-(-b)=a+b$ " ]

[]

[ [ { "aoVal": "A", "content": "$7.35\\text{p.m.}$ " } ], [ { "aoVal": "B", "content": "$8.35\\text{p.m.}$ " } ], [ { "aoVal": "C", "content": "$8.55\\text{p.m.}$ " } ], [ { "aoVal": "D", "content": "$9.45\\text{p.m.}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules", "Overseas In-curriculum->Knowledge Point->Measurement->Time->Time Calculation" ]

[ "$225$ minutes $=3$ hours $45$ minutes. He started playing at $7.35\\text{p.m.}$ " ]

[ "其它" ]

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

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

[ "$$36 \\div 5 = 7R1$$, so the answer is $$5$$. " ]

[]

[ [ { "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->Division of Whole Numbers->Division with Remainders" ]

[ "$160\\div50=3R10$, $3+1=4$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$30^{\\circ}$ " } ], [ { "aoVal": "B", "content": "$45^{\\circ}$ " } ], [ { "aoVal": "C", "content": "$60^{\\circ}$ " } ], [ { "aoVal": "D", "content": "$75^{\\circ}$ " } ], [ { "aoVal": "E", "content": "$90^{\\circ}$ " } ] ]

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

[ "$\\dfrac{6}{3+4+5+6}\\times360^{\\circ}-\\dfrac{3}{3+4+5+6}\\times360^{\\circ}=\\dfrac{360}{18}\\left( 6-3\\right)=20\\times3=60^{\\circ}$ " ]

[]

[ [ { "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->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers->Questions Involving Addition and Subtraction" ]

[ "$5-3+4+3=9$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$81$$ " } ] ]

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

[ "$2 \\times \\_3P\\_1 \\times \\_2C\\_2=6$. " ]

[ "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 . " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes " } ], [ { "aoVal": "B", "content": "No " } ] ]

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

[ "NA " ]

[ "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" ]

[ "$y=x+1$ and $y=-x+1$ have $y$-intercepts at $(0,1)$ and slopes of 1 and $-1$, respectively. Since the product of these slopes is $-1$, the two lines are perpendicular. From $y=5$, we see that $(-4,5)$ and $(4,5)$ are the other two intersection points, and they are 8 units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4 \\sqrt{2}$ each and the area is $\\frac{(4 \\sqrt{2})^{2}}{2}=(\\mathbf{E}) 16$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes " } ], [ { "aoVal": "B", "content": "No " } ] ]

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

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$63$$，$$54$$ " } ], [ { "aoVal": "B", "content": "$$63$$，$$55$$ " } ], [ { "aoVal": "C", "content": "$$62$$，$$54$$ " } ], [ { "aoVal": "D", "content": "$$62$$，$$55$$ " } ] ]

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

[ "$$(\\textbackslash{} \\textbackslash{} \\textbackslash{} \\textbackslash{} )\\div 9=6 R (\\textbackslash{} \\textbackslash{} \\textbackslash{} \\textbackslash{} )$$ Largest remainder: $$8$$, $$9\\times 6+8=62$$ Smallest remainder: $$1$$, $$9\\times 6+1=55$$ " ]

[]

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

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

[ "$100-15-27-36-8=14$ " ]

[ "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->Basic Concepts of Equation->Linear Equations with Multiple Variables" ]

[ "Function Form is written as: $y=$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "infinitely many " } ] ]

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

[ "We see that the vertex of the quadratic function $y=4x^{2}-8x+a^{2}$ is $\\left(1, a^{2}-4\\right)$. If $\\left(2, a^{2}-1\\right)$ will be on the line $y=x+a^{2}-6$, $a^{2} -4=1+a^{2}-6$. Solve for $a$, there is no solution. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{A binomial distribution with p = 0.15 and n = 20} " } ], [ { "aoVal": "B", "content": "\\textbf{A binomial distribution with p = 0.2 and n = 15} " } ], [ { "aoVal": "C", "content": "\\textbf{A geometric distribution with p = 0.15} " } ], [ { "aoVal": "D", "content": "\\textbf{A geometric distribution with p = 0.2} " } ], [ { "aoVal": "E", "content": "\\textbf{A cumulative geometric distribution with p = 0.15} " } ] ]

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

[ "\\textbf{Binomial distribution: how likely it is to get x successes in n trials given that your probability of success is p} \\textbf{Geometric distribution: how likely is it that I'll have my first critical strike on the 5th attack} \\textbf{Cumulative geometric distribution: how likely is it that I'll have my first critical strike on or before the 5th attack.} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "True " } ], [ { "aoVal": "B", "content": "False " } ] ]

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

[ "For example, consider a dataset with the following values: 0, 2, 2, 3, 3. The mean of this dataset is $2$, and the $25$-th percentile is also $2$. " ]

[ "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": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$43$$ " } ], [ { "aoVal": "D", "content": "$$44$$ " } ], [ { "aoVal": "E", "content": "$$57$$ " } ] ]

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

[ "We know from the triangle inequality that the last side, $s$, fulfills $s\\textless7+15$. Adding $7+15$ to both sides of the inequality, we get $s+7+15\\textless44$, and because $s+7+15$ is the perimeter of our triangle, (D) 44 is our answer. " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

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

[ "$$12345+123450=12345\\times1+12345\\times10=12345\\times11$$ " ]

[]

[ [ { "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 $2$ and $0$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2^{8}$$ " } ], [ { "aoVal": "B", "content": "$$2^{6}$$ " } ], [ { "aoVal": "C", "content": "$$4^{8}$$ " } ], [ { "aoVal": "D", "content": "$$4^{7}$$ " } ] ]

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

[ "omitted " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$79$$ " } ], [ { "aoVal": "D", "content": "$$89$$ " } ] ]

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

[ "We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8 th graders, in order that we can put the two ratios together: $$ \\begin{aligned} \\&5: 3=5(8): 3(8)=40: 24 \\textbackslash\\textbackslash{} \\&8: 5=8(5): 5(5)=40: 25 \\end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40: 25: 24$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 40+25+24=\\text { (E) } 89 $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

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

[ "~\\uline{~~~~~~~~~~}~$\\div$ $7=$~\\uline{~~~~~~~~~~}~$R1$ $36-1=35$, $35\\div7=5$, so the answer is $D$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30.41$$ " } ], [ { "aoVal": "B", "content": "$$3.5260$$ " } ], [ { "aoVal": "C", "content": "$$42.09$$ " } ] ]

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

[ "Only $$3.5260$$\\textquotesingle s $$\"0\"$$ can be removed without causing other digits to change their places from the original number. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$33 \\frac{1}{3}$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ], [ { "aoVal": "E", "content": "$$54$$ " } ] ]

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

[ "After Gilda gives $20 \\textbackslash\\%$ of the marbles to Pedro, she has $80 \\textbackslash\\%$ of the marbles left. If she then gives $10 \\textbackslash\\%$ of what\\textquotesingle s left to Ebony, she has $(0.8 * 0.9)=72 \\textbackslash\\%$ of what she had at the beginning. Finally, she gives $25 \\textbackslash\\%$ of what\\textquotesingle s left to her brother, so she has $(0.75 * 0.72)$ (E) 54 . of what she had in the beginning left. " ]

[]

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

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

[ "$$25+35+45=(25+35)+45=60+45$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$\\frac{56}{15}$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "We want the lines to be parallel and not the same line in order to have 0 solutions. Parallel implies the slopes of the lines are equal, so we have $-\\frac{8}{6 k}=-\\frac{k}{12}$. Cross multiplying, we get $96=6 k^{2}$, so $k^{2}=$ 16 and our positive solution is then $k=4$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$0$$ " } ] ]

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

[ "$5+3+2-1=9$，$10-9=1$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$210$$ " } ], [ { "aoVal": "B", "content": "$$215$$ " } ], [ { "aoVal": "C", "content": "$$231$$ " } ], [ { "aoVal": "D", "content": "$$236$$ " } ] ]

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

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$22222$$ " } ], [ { "aoVal": "B", "content": "$2222^{2}$ " } ], [ { "aoVal": "C", "content": "$222^{22}$ " } ], [ { "aoVal": "D", "content": "$22^{222}$ " } ], [ { "aoVal": "E", "content": "$2^{2222}$ " } ] ]

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

[ "$2^{2222}\\textgreater2^{2220}=\\left (2^{10}\\right )^{222}=1024^{222}\\textgreater22^{222}$. According to this rule, $\\text{E}\\textgreater\\text{D}\\textgreater\\text{C}\\textgreater\\text{B}\\textgreater\\text{A}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "The ones digits of exponents based on $3$ follow the rule: $3, 9, 7, 1, 3, 9, 7, 1\\cdots $ $50\\div4=12R2$, which means the ones digit is $9$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{7}{11}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{57}{90}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{636}{1000}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{63}{100}$ " } ] ]

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

[ "Let $$x = 0.636363\\cdots$$ Multiply by $$100$$, which means move the decimal point two places to the right: $$100x = 63.636363\\cdots$$ $$x$$ and $$100x$$ have exactly the same decimal part, so if we subtract, it will disappear: $$100x - x = 63.636363\\cdots - 0.636363\\cdots$$ Which is: $$99x = 63$$ ,$x=\\dfrac{63}{99}=\\dfrac{7}{11}$ So there is our answer： $$0.636363\\cdots=$$$\\dfrac{7}{11}$ " ]

[ "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 (c).There are 12 values in the A and E cell out the total of 125. When we are given colwnn E, the total is 63. Of those,28 are C. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{111}{200}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{11}{20}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{5}{9}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{2}{3}$ " } ] ]

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

[ "Let $$x = 0.555\\cdots$$ Multiply by $$10$$ , which means move the decimal point one place to the right: $$10x = 5.555\\cdots$$. $$x$$ and $$10x$$ have exactly the same decimal part, so if we subtract, it will disappear: $$10x - x = 5.555\\cdots - 0.555\\cdots$$. Simplify:~ $$9x = 5$$，$x=\\dfrac{5}{9}$ So there\\textquotesingle s our answer : $$0.555\\cdots=$$$\\dfrac{5}{9}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

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

[ "C " ]

[]

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

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

[ "The ones digit of a $$4\\text{th}$$ power canbe $$0$$, $$1$$, $$5$$, or $$6$$. It can never be $$3$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$27$$ " } ] ]

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

[ "The number of values you multiply together is equal to the number of factors in a study. There are three factors, so multiply \\_\\_x\\_\\_x\\_\\_. The values that go into each slot represent the number of levels for each factor. In this case, 3x3x3=27. " ]

[ "其它" ]

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

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

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$990-9$$ " } ], [ { "aoVal": "B", "content": "$$990-90$$ " } ], [ { "aoVal": "C", "content": "$$900-99$$ " } ], [ { "aoVal": "D", "content": "$$900-9$$ " } ] ]

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

[ "$$99\\times9=(100-1)\\times9$$. This is slightly less than $$100\\times9$$, so it\\textquotesingle s $$\\text{D}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "The sum can be divisible by both $3$ and $4$, so the sum is the common multiple of $3$ and $4$. If the sum is $12$, $12-5=7$, so the sum of the other numbers should be $7$. The other two numbers can be $1$ and $6$ or $3$ and $4$. If the sum is $24$, $24-5=19$, the other two numbers can only be $10$ and $9$. If the sum is $36$, it is impossible, because $36-5=31$ and the largest sum of two numbers from $1$ to $10$ is $19$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$33x-55$$ " } ], [ { "aoVal": "B", "content": "$$-25x-32$$ " } ], [ { "aoVal": "C", "content": "$-\\frac{x}{3}-\\frac{1}{4}$ " } ] ]

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

[ "the sign in front is also part of constant or coefficient " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "$6$\\#$3=3$, $10$\\#$9=9$ $3 \\times 9=27$ " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

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

[ "$1◆(2◆3)=1◆[2+(2\\times3)]=1◆8=1+16=17$. So the answer is $\\rm C$. " ]

[]

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

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

[ "NA " ]