Datasets:

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math-eval

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TAL-SCQ5K

like 57

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[ [ { "aoVal": "A", "content": "It will rain $$75\\textbackslash\\% $$ of the time tomorrow in New York City. " } ], [ { "aoVal": "B", "content": "It will rain in $$75\\textbackslash\\%$$ of the regions in New York City tomorrow. " } ], [ { "aoVal": "C", "content": "It will definitely rain tomorrow in New York City. " } ], [ { "aoVal": "D", "content": "The probability of raining in New York City tomorrow is high. " } ] ]

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

[ "The statement that \"there is a $$75\\textbackslash\\% $$ chance that it will rain tomorrow in New York City\"~shows that it is more likely to rain tomorrow in New York city. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

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

[ "$$3\\times 2\\times 1=6$$. " ]

[]

[ [ { "aoVal": "A", "content": "southwest　 " } ], [ { "aoVal": "B", "content": "southeast　 " } ], [ { "aoVal": "C", "content": "northwest　 " } ], [ { "aoVal": "D", "content": "northeast　 " } ] ]

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

[ "Since $$225^{}\\circ =180^{}\\circ +45^{}\\circ $$, the bird turns $$45^{}\\circ $$ past south. That\\textquotesingle s southwest. " ]

[]

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

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$500$$ " } ], [ { "aoVal": "B", "content": "$$540$$ " } ], [ { "aoVal": "C", "content": "$$545$$ " } ], [ { "aoVal": "D", "content": "$$550$$ " } ] ]

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

[ "$$71+82+93+104+195=70+80+90+100+200+5=545$$ " ]

[]

[ [ { "aoVal": "A", "content": "March　 " } ], [ { "aoVal": "B", "content": "April　 " } ], [ { "aoVal": "C", "content": "June　 " } ], [ { "aoVal": "D", "content": "November　 " } ] ]

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

[ "Together, July \\& August have $$62$$ days. That\\textquotesingle s twice the $$31$$ days in March. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$40$$ " } ] ]

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

[ "Let $f$ be a linear function with slope $m$. $$ \\begin{gathered} m=\\frac{f(3)-f(2)}{\\Delta x}=\\frac{5}{3-2}=5\\textbackslash\\textbackslash{} f(8)-f(2)=m \\Delta x=5(8-2)=30 \\Rightarrow(D) \\end{gathered}$$ " ]

[]

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

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

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

[]

[ [ { "aoVal": "A", "content": "$$\\dfrac{9}{23}\\textless{} \\dfrac{7}{21}\\textless\\dfrac{5}{19}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{5}{19}\\textless{} \\dfrac{7}{21}\\textless{} \\dfrac{9}{23}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{9}{23}\\textless{} \\dfrac{5}{19}\\textless{} \\dfrac{7}{21}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{5}{19}\\textless{} \\dfrac{9}{23}\\textless{} \\dfrac{7}{21}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{7}{21}\\textless{} \\dfrac{5}{19}\\textless{} \\dfrac{9}{23}$$ " } ] ]

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

[ "$$\\rm Method$$ $$1$$: The value of $$\\dfrac{7}{21}$$ is $$\\dfrac{1}{3}$$. Now we give all the fractions a common denominator. $$\\dfrac{5}{19} \\Rightarrow \\dfrac{345}{1311}$$, $$\\dfrac{1}{3} \\Rightarrow \\dfrac{437}{1311}$$, $$\\dfrac{9}{23} \\Rightarrow \\dfrac{513}{1311}$$. Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, $$\\frac{5}{19}\\textless{} \\frac{7}{21}\\textless{} \\frac{9}{23}$$. $$\\rm Method$$ $$2$$: Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus, $$1- \\dfrac{5}{19}= \\dfrac{14}{19}$$, $$1- \\dfrac{7}{21}= \\dfrac{14}{21}$$, $$1- \\dfrac{9}{23}= \\dfrac{14}{23}$$. All three fraction have common numerator $$14$$. Now the order of the fractions is obvious . $$\\dfrac{14}{19}\\textgreater\\dfrac{14}{21}\\textgreater\\dfrac{14}{23}\\Rightarrow\\dfrac{5}{19}\\textless\\dfrac{7}{21}\\textless\\dfrac{9}{23}$$. Therefore, $$\\dfrac{5}{19}\\textless\\dfrac{7}{21}\\textless\\dfrac{9}{23}$$. $$\\rm Method$$ $$3$$: Change $$\\frac7{21}$$ into $$\\frac13$$, $$\\dfrac{1}{3} \\times \\dfrac{5}{5}=\\dfrac{5}{15}$$, $$\\dfrac{5}{15}\\textgreater\\dfrac{5}{19}$$, $$\\dfrac{7}{21}\\textgreater\\dfrac{5}{19}$$, and $$\\dfrac{1}{3} \\times \\dfrac{9}{9}=\\dfrac{9}{27}$$, $$\\dfrac{9}{27}\\textless\\dfrac{9}{23}$$, $$\\dfrac{7}{21} \\textless{} \\dfrac{9}{23}$$. Therefore, $$\\dfrac{5}{19}\\textless\\dfrac{7}{21}\\textless\\dfrac{9}{23}$$. $$\\rm Method$$ $$4$$: When $$\\dfrac{x}{y}\\textless1$$ and $$z\\textgreater0$$, $$\\dfrac{x+z}{y+z}\\textgreater\\dfrac{x}{y}$$. Hence, $$\\dfrac{5}{19}\\textless\\dfrac{7}{21}\\textless\\dfrac{9}{23}$$. " ]

[]

[ [ { "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 " ]

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

[ "其它" ]

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

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

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

[]

[ [ { "aoVal": "A", "content": "$5:30$ " } ], [ { "aoVal": "B", "content": "$1:6$ " } ], [ { "aoVal": "C", "content": "$6:1$ " } ], [ { "aoVal": "D", "content": "$10:1$ " } ] ]

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

[ "We need to make units same first. $5$ minutes equal to $300$ seconds. Now we could remove the same unit, second. We get $300:30$ and simplify it to $10:1$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

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

[ "Let $f$ be a linear function with slope $m$. $$ \\begin{gathered} m=\\frac{f(6)-f(2)}{\\Delta x}=\\frac{12}{6-2}=3 \\textbackslash\\textbackslash{} f(12)-f(2)=m \\Delta x=3(12-2)=30 \\Rightarrow(D) \\end{gathered}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$401$$ " } ], [ { "aoVal": "B", "content": "$$403$$ " } ], [ { "aoVal": "C", "content": "$$405$$ " } ], [ { "aoVal": "D", "content": "$$407$$ " } ], [ { "aoVal": "E", "content": "$$409$$ " } ] ]

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

[ "Let the five consecutive integers be $$n-2$$, $$n-1$$, $$n$$, $$n + 1$$ and $$n + 2$$. These have a sum of $$5n$$. Hence $$5n=2015$$ and therefore $$n=403$$. Therefore, the smallest integer is $$403-2=401$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$59$$ " } ], [ { "aoVal": "B", "content": "$$65$$ " } ], [ { "aoVal": "C", "content": "$$74$$ " } ], [ { "aoVal": "D", "content": "$$92$$ " } ], [ { "aoVal": "E", "content": "$$95$$ " } ] ]

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

[ "\\textbf{Z = $$\\frac{40.6-38.5}{1.25}$$=1.68 → percentile = 0.9535} " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "$$(2+4+6+8+10)\\div (10+8+6+4+2)=30\\div30=1$$. " ]

[]

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

[ "The problem can be regarded as an equation $8$ is the number of chocolates Jack had, $1, 2, 3$ is the number of chocolates Jack gave out, and $2$ is the number of chocolates his mother gave Jack. So the remaining chocolate is $8-1-2-3+2=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$4\\times {{10}^{11}}$$ " } ], [ { "aoVal": "B", "content": "$$2\\times {{10}^{11}}$$ " } ], [ { "aoVal": "C", "content": "$$2\\times {{10}^{12}}$$ " } ], [ { "aoVal": "D", "content": "$$20\\times {{10}^{18}}$$ " } ], [ { "aoVal": "E", "content": "$$2\\times {{10}^{19}}$$ " } ] ]

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

[ "$$4\\times {{10}^{9}}\\times 5\\times {{10}^{2}}$$ $$=20\\times {{10}^{9}}+2$$ $$=2\\times {{10}^{1}}\\times {{10}^{11}}$$ $$=2\\times {{10}^{1+11}}$$ $$=2\\times {{10}^{12}}$$ " ]

[]

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

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

[ "$$9+9+9+9 +9+9+9=$$ seven $$9$$\\textquotesingle s $$=9\\times7$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$1+2+3+4$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$1\\times 2\\times 3\\times 4$$ " } ] ]

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

[ "$$3+6+9+12=30=3\\times 10=3\\times (1+2+3+4)$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{3}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac{5}{3}$ " } ], [ { "aoVal": "C", "content": "$\\frac{7}{4}$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ], [ { "aoVal": "E", "content": "$\\frac{13}{4}$ " } ] ]

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

[ "B " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

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

[ "E " ]

[]

[ [ { "aoVal": "A", "content": "$$111$$ " } ], [ { "aoVal": "B", "content": "$$1101$$ " } ], [ { "aoVal": "C", "content": "$$1011$$ " } ], [ { "aoVal": "D", "content": "$$10101$$ " } ] ]

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

[ "$$100\\times 100+10\\times 10+1\\times 1=10000+100+1=10101$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$144$$ in the blue box represents $$9\\times 16$$. " } ], [ { "aoVal": "B", "content": "$$32$$ in the red box represents $$2\\times 16$$. " } ], [ { "aoVal": "C", "content": "The product is $$32+144 =176$$. " } ], [ { "aoVal": "D", "content": "There is no regrouping in the calculation. " } ] ]

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

[ "$$32$$ in the red box represents $$20 \\times 16 = 320$$, and the product should be $$320 +144 = 464$$. " ]

[]

[ [ { "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 $$\\leqslant 15.01$$ " } ], [ { "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{} " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Unknowns Using the Given Operations" ]

[ "$3\\square 5=2\\square x \\implies 15+3+5=2x+2+x$, hence $x=7$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$49$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "$B-A=(2-1)+(4-3)+(6-5)+\\cdots +(100-99)=50$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Mean = 5.4; standard deviation = 2.8 " } ], [ { "aoVal": "B", "content": "Mean = 5.4; standard deviation = 2.95 " } ], [ { "aoVal": "C", "content": "Mean = 16.2; standard deviation = 2.90 " } ], [ { "aoVal": "D", "content": "Mean = 16.2; standard deviation = 8.4 " } ], [ { "aoVal": "E", "content": "Mean = 21.6standard deviation = 11.2 " } ] ]

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

[ "The mean is a measure of center. While it can be affected by adding new values ,the purchase of second deck would merely add more cards with the same values .As a result ,the mean would not change .Eliminate (C), (D),and (E). The standard deviation is a measure of sprees. As above, unique values will cause the standard deviation to change. However, since the new decks are identic -call to the old, the standard deviation will not change. Eliminate (B). " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$2:3:4$ " } ], [ { "aoVal": "B", "content": "$8:6:15$ " } ], [ { "aoVal": "C", "content": "$6:3:9$ " } ], [ { "aoVal": "D", "content": "$16:6:15$ " } ] ]

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

[ "$A:B=4:3$ $B:C=2:5$ $A:B:C=8:6:15$ " ]

[]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ] ]

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

[ "Lollipop: $$7$$ Toffee: $$7\\times5=35$$ Chocolate bar: $$7+5=12$$ Total: $$7+35+12=54$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{4}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{8}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{16}$ " } ], [ { "aoVal": "D", "content": "$\\frac{1}{32}$ " } ] ]

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

[ "$\\frac{1}{2}-\\frac{1}{2}\\times \\frac{1}{2}-\\frac{1}{2}\\times \\frac{1}{4}=\\frac{1}{8}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3.5$$ " } ], [ { "aoVal": "B", "content": "$$3.3$$ " } ], [ { "aoVal": "C", "content": "$$2.9$$ " } ], [ { "aoVal": "D", "content": "$$3.7$$ " } ], [ { "aoVal": "E", "content": "$$3.1$$ " } ] ]

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

[ "\\textbf{Z = 1.28} \\textbf{$$\\frac{x-2.05}{1}$$ = 1.28} \\textbf{x= 3.33} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$12$ dollars " } ], [ { "aoVal": "B", "content": "$24$ dollars " } ], [ { "aoVal": "C", "content": "$$36$$ dolllars " } ], [ { "aoVal": "D", "content": "$$40$$ dollars " } ], [ { "aoVal": "E", "content": "$$45$$ dollars " } ] ]

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

[ "$48 \\div (3 + 1) = 12$ " ]

[]

[ [ { "aoVal": "A", "content": "$130 " } ], [ { "aoVal": "B", "content": "$$260$$ " } ], [ { "aoVal": "C", "content": "$$300$$ " } ], [ { "aoVal": "D", "content": "$$360$$ " } ] ]

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

[ "Sum $$=$$ Middle Term $$\\times$$ Number of Terms, $$20 \\times 13 = 260$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$123$$ " } ], [ { "aoVal": "B", "content": "$$132$$ " } ], [ { "aoVal": "C", "content": "$$129$$ " } ], [ { "aoVal": "D", "content": "$$-123$$ " } ] ]

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

[ "$$\\begin{eqnarray}\\&\\&123\\times (-129+130)\\textbackslash\\textbackslash{} \\&=\\&123\\times (130-129)\\textbackslash\\textbackslash{} \\&=\\&123\\times1\\textbackslash\\textbackslash{} \\&=\\&123.\\end{eqnarray}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

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

[ "$(3$◆$4)=(3\\times 3)-4=5$ , and $(3$◆$4)$◆$5=5$◆$5=(5\\times5)-5=20$ . " ]

[ "其它" ]

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

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

[ "$\\dfrac{2\\left( x+y\\right)}{x}=\\dfrac{14}{3}\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} \\dfrac{x+y}{x}$~$=\\dfrac{7}{3}\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} 1+\\dfrac{y}{x}=1+\\dfrac{4}{3}$ $\\Rightarrow\\textbackslash{} \\textbackslash{} \\dfrac{y}{x}=\\dfrac{4}{3}\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} \\dfrac{y\\times y}{x\\times y}=\\dfrac{4}{3}$ The area is~$x\\times y=27\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} \\dfrac{y\\times y}{27}=\\dfrac{4}{3}\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} y\\times y=36\\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} y=6$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "$$\\frac{6}{7}=0.\\overline{857142}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$$^{th}$ digit goes back to $2$. The $2022$$$^{nd}$$ digit is $2$, so the $2021$$^{st}$ digit is $4$. " ]

[ "其它" ]

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

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

[ "Note that $x = 6-y \\Rightarrow x+y = 6$. Then, $$ \\frac{{{x}^{2}}}{x-y}+\\frac{{{y}^{2}}}{y-x}=\\frac{{{x}^{2}}-{{y}^{2}}}{x-y}=\\frac{(x+y)(x-y)}{x-y}=x+y = 6$$. " ]

[]

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

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

[ "$2◆3=(2\\times3)+3=9$. So the answer is $\\rm B$. " ]

[]

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

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

[ "We can instead calculate the probability that their product is odd, and subtract this from $$1$$. In order to get an odd product, we have to draw an odd number from each box. We have a $$\\frac23$$ probability of drawing an odd mumber from one box, so there is a $$\\left(\\frac{2}{3}\\right)^{2}= \\frac{4}{9}$$ probability of having an odd product. Thus, there is a $$1- \\frac{4}{9}= \\frac{5}{9}$$ probability of having an even product. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Heuristics Skills-> Equivalent Substitution" ]

[ "$$8$$ watermelons = $$12$$ pears $$8\\times3$$ watermelons = $24$ watermelons $$12\\times3$$ pears = $36$ pears " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\pi mv^{2}$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$\\pi mv^{2}r$ " } ], [ { "aoVal": "D", "content": "$2\\pi mv^{2}r$ " } ] ]

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

[ "Since the centripetal force always points along a raduis toward the center of the od the circle, and the velocity of the object is always tangent to the circle (and thus perpendicular to the radius), the work done by the centripetal force is zero. Alternatively, since the object\\textquotesingle s speed remains constant, the Work-Energy Theorem tells you that no work is being performed. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "Wrong: $$90 - A = 55$$ -\\/-\\textgreater{} $$A=90-55=35$$ Correct: $$75 - 35 = 40$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$54$$ " } ], [ { "aoVal": "B", "content": "$$78$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

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

[ "$$4\\times 21+ 6 = 90$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$16\\times 2$$ " } ], [ { "aoVal": "B", "content": "$$12\\times 3$$ " } ], [ { "aoVal": "C", "content": "$$7\\times 5$$ " } ], [ { "aoVal": "D", "content": "$$38\\times 1$$ " } ] ]

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

[ "$$4\\times 9=36$$. $$\\text{A}$$: $$16\\times 2=32$$; $$\\text{B}$$: $$12\\times 3=36$$; $$\\text{C}$$: $$7\\times 5=35$$; $$\\text{D}$$: $$38\\times 1=38$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$65$$ " } ], [ { "aoVal": "D", "content": "$$98$$ " } ] ]

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

[ "If a number is divided by $$100$$, the remainder is the number\\textquotesingle s last $$2$$ digits. " ]

[]

[ [ { "aoVal": "A", "content": "$$197$$ " } ], [ { "aoVal": "B", "content": "$$371$$ " } ], [ { "aoVal": "C", "content": "$$826$$ " } ], [ { "aoVal": "D", "content": "$$1455$$ " } ] ]

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

[ "$$1000-174=826$$ $$826-629=197$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ a.m. " } ], [ { "aoVal": "B", "content": "$$11$$ a.m. " } ], [ { "aoVal": "C", "content": "$$12$$ p.m. " } ], [ { "aoVal": "D", "content": "$$1$$ p.m. " } ] ]

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

[ "$$(350-80)$$$\\div$$$(40+50)=3$$hr $$8+2+3=13$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$202$$ " } ], [ { "aoVal": "B", "content": "$$223$$ " } ], [ { "aoVal": "C", "content": "$$224$$ " } ], [ { "aoVal": "D", "content": "$$225$$ " } ], [ { "aoVal": "E", "content": "$$234$$ " } ] ]

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

[ "To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223+1+1=$ (D) $225$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "($x$,$y$)=($2$,$-1$) " } ], [ { "aoVal": "B", "content": "($x$,$y$)=($2$,$1$) " } ], [ { "aoVal": "C", "content": "($x$,$y$)=($-2$,$1$) " } ], [ { "aoVal": "D", "content": "($x$,$y$)=($-2$,$-1$) " } ] ]

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

[ "$x=2$ $8-y=7$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$MSL$ " } ], [ { "aoVal": "B", "content": "$SML$ " } ], [ { "aoVal": "C", "content": "$LSM$ " } ], [ { "aoVal": "D", "content": "$SLM$ " } ] ]

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

[ "Suppose the small size costs $$$1$$ and the large size has $10$ grams of cake. The medium size then costs $$$1.40$$ and has $7.5$ grams of cake. The small size has $5$ grams of cake and the large size costs $$$2.24$$. The small, medium, and large size cost respectively, $0.200$, $0.187$, $0.224$ dollars per gram. The sizes from best to worst buy are $MSL$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "$(100-98)+(96-94)+(92-90)+\\cdots +(8-6)+(4-2)=2\\times25=50$ " ]

[ "其它" ]

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

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

[ "$\\dfrac{x}{y}=\\dfrac{y}{x-z}=\\dfrac{x+y}{z}\\textbackslash{} \\textbackslash{} \\textbackslash{} \\textbackslash{} \\Rightarrow\\textbackslash{} \\textbackslash{} \\textbackslash{} \\dfrac{x}{y}=\\dfrac{x+y+(x+y)}{y+\\left( x-z\\right)+z}=\\dfrac{2x+2y}{x+y}=\\dfrac{2\\left( x+y\\right)}{x+y}=2.$ " ]

[]

[ [ { "aoVal": "A", "content": "$$145$$ " } ], [ { "aoVal": "B", "content": "$$245$$ " } ], [ { "aoVal": "C", "content": "$$625$$ " } ], [ { "aoVal": "D", "content": "$$395$$ " } ] ]

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

[ "$280+445=725$. $725-330=395$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$3\\dfrac{3}{4}$ " } ], [ { "aoVal": "B", "content": "$\\frac{42}{12}$ " } ], [ { "aoVal": "C", "content": "$\\frac{10}{3}$ " } ], [ { "aoVal": "D", "content": "$3\\dfrac{2}{3}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions->Using Common Factors to Simplify Fractions" ]

[ "Factors of $8$ are $1, 2, 4, 8$ and factors of $12$ are $1, 2, 3, 4, 6, 12$. The highest commom factor of $8$ and $12$ is $4$. Dividing $8$ and $12$ by $4$ respectively, we get $\\frac{8 \\div 4}{12 \\div 4}$=$\\frac{2}{3}$; Write the whole and the simplified fraction together, $3\\dfrac{2}{3}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5055$$ " } ], [ { "aoVal": "B", "content": "$$5505$$ " } ], [ { "aoVal": "C", "content": "$$5550$$ " } ], [ { "aoVal": "D", "content": "$$55550$$ " } ] ]

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

[ "$$5\\textasciitilde\\text{m}+5\\textasciitilde\\text{cm}+5\\textasciitilde\\text{mm}=5000\\textasciitilde\\text{mm}+50\\textasciitilde\\text{mm}+5\\textasciitilde\\text{mm}=5055\\textasciitilde\\text{mm}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Only those people who took this particular survey " } ], [ { "aoVal": "B", "content": "All future customers of this automobile manufacturer " } ], [ { "aoVal": "C", "content": "All people who have purchased from that manufacturer in the previous five years " } ], [ { "aoVal": "D", "content": "Only those people who have purchased a vehicle in the previous five years " } ], [ { "aoVal": "E", "content": "All vehicle owners in the USA " } ] ]

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

[ "A safe generalization will broaden the survey to a population that closely resembles the group of people surveyed. Since the sample was taken from purchasers of a single auto manufacturer over the past five years, the safest generalization is a population with these same qualities. " ]

[ "其它" ]

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

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

[ "For the system to have infinitely many solutions, the second equation must be a constant multiple of the first. Comparing the coefficients of $x$, the second equation must be $\\frac{2}{5}$ times the first equation. Therefore, $q+r=\\left(\\frac{2}{5}\\right)\\times (7)+\\left(\\frac{2}{5}\\right)\\times(8)=6$. " ]

[ "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, $P\\textgreater24+1=25$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1700$$ " } ], [ { "aoVal": "B", "content": "$$1716$$ " } ], [ { "aoVal": "C", "content": "$$1717$$ " } ] ]

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

[ "$$\\left( 1+100 \\right)\\times 34\\div 2=1717$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$77$$ " } ], [ { "aoVal": "B", "content": "$$86$$ " } ], [ { "aoVal": "C", "content": "$$92$$ " } ], [ { "aoVal": "D", "content": "$$98$$ " } ], [ { "aoVal": "E", "content": "$$104$$ " } ] ]

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

[ "NA " ]

[]

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

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

[ "$\\frac{2}{3}=\\frac{6}{9}$ " ]

[]

[ [ { "aoVal": "A", "content": "$$100:1000$$ " } ], [ { "aoVal": "B", "content": "$$1000:100$$ " } ], [ { "aoVal": "C", "content": "$$1:100$$ " } ], [ { "aoVal": "D", "content": "$$100:1$$ " } ] ]

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

[ "$$1\\rm m=100\\rm cm$$ and $$1\\rm km=1000m$$; the correct ratio is $$100:1000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

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

[ "Pay attention to look at both sides of the formula. One side is addition, and the other side is subtraction. " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$105$$ " } ], [ { "aoVal": "D", "content": "$$210$$ " } ], [ { "aoVal": "E", "content": "$$315$$ " } ] ]

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

[ "$$6$$ * $$5 = (6+5+4+3+2+1) \\times 5 = 21\\times 5 = 105$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0.86$$ " } ], [ { "aoVal": "B", "content": "$$1.56$$ " } ], [ { "aoVal": "C", "content": "$$0.48$$ " } ], [ { "aoVal": "D", "content": "None of the above " } ], [ { "aoVal": "E", "content": "\\textbf{Can't tell from the information given} " } ] ]

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

[ "$$\\beta\\_1 =r \\frac{S\\_y}{S\\_x} = 0.86 * \\frac{3.8}{2.1} = 1.56$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "0 dollar " } ], [ { "aoVal": "B", "content": "50,000 dollars " } ], [ { "aoVal": "C", "content": "400,000 dollars " } ], [ { "aoVal": "D", "content": "650,000 dollars " } ], [ { "aoVal": "E", "content": "1,000,000 dollars " } ] ]

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

[ "Accounting profit = total revenue - explicit cost " ]

[ "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 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$95$$ " } ], [ { "aoVal": "B", "content": "$$85.5$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$95.5$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "\\textbf{$$\\hat{y}$$= 10 + .9*95=95.5. The predicted final exam score is 95.5.} " ]

[ "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" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$35.2$ " } ], [ { "aoVal": "B", "content": "$3.52$ " } ], [ { "aoVal": "C", "content": "$37.4$ " } ], [ { "aoVal": "D", "content": "$3.74$ " } ], [ { "aoVal": "E", "content": "$3.47$ " } ] ]

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

[ "$$19.2\\div6\\times1.1$$ $$=3.2\\times1.1$$ $$=3.52$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "Let $$x$$ be the number of £$$10$$ books and $$y$$ be the number of £$$15$$ books $$10x+15y=90$$, which has two positive integer solutions only. $$\\begin{cases} x=6 \\textbackslash\\textbackslash{} y=2 \\end{cases}$$, $$\\begin{cases} x=3 \\textbackslash\\textbackslash{} y=4 \\end{cases}$$ $$6+2=8$$ " ]

[ "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 $y=x^{2}-6x+a^{2}$ is $\\left(3, a^{2}-9\\right)$. If $\\left(2, a^{2}-1\\right)$ will be on the line $y=x+a^{2}-a-3$, $a^{2} -9=3+a^{2}-a-3$. Solve for $a$, there is one solution, $a=9$. " ]

[ "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 $y=x^{2}+a^{2}$ is $\\left(0, a^{2}\\right)$. The $y$-intercept of the line $y=x+a$ is $(0, a)$. We want to find the values (if any) such that $a=a^{2}$. Solving for $a$, the only values that satisfy this are 0 and 1 , so the answer is $(\\text{C}) 2$ " ]

[]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$99$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ] ]

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

[ "$$(100-99)+(98-97)+ \\cdots +(2-1) = 1+1+ \\cdots +1 = 50$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$+24$$ " } ], [ { "aoVal": "B", "content": "$$\\times 4$$ " } ], [ { "aoVal": "C", "content": "$$\\times 24$$ " } ], [ { "aoVal": "D", "content": "$$+4$$ " } ] ]

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

[ "$$8+24=32=8\\times 4$$ So, $$3\\times 4=12$$ " ]

[]

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

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

[ "Solution 1 Rearranging the terms, we get $$(1-2)+(3-4)+(5-6)+\\cdots (2017-2018)+2019$$, and our answer is $$-1009+2019=1010$$. Solution 2 We can rewrite the given expression as $$1+(3-2)+(5-4)+\\cdots +(2017-2016)+(2019-2018)=1+1+1+\\cdots +1$$. The number of $$1$$s is the same as the number of terms in $$1$$, $$3$$, $$5$$, $$7\\cdots $$, $$2017$$, $$2019$$. Thus the answer is $$1010$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14811850$$ " } ], [ { "aoVal": "B", "content": "$$14811851$$ " } ], [ { "aoVal": "C", "content": "$$14811852$$ " } ], [ { "aoVal": "D", "content": "$$14811853$$ " } ], [ { "aoVal": "E", "content": "$$14811854$$ " } ] ]

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

[ "$$14811852$$ It should be noted that the units digit of the product of $$3333$$ and $$4444$$ will be the same as the units digit of the product of $$3$$ and $$4$$, namely $$2$$: this succinctly identifies the correct option as $$14811852$$. Alternatively, the calculation $$3333\\times 4444$$ will have an answer that is $$3\\times 2$$ times greater than $$1111\\times 2222$$. Now we can work out $$2468642\\times 6$$ exactly, which is indeed $$14811852$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "The distribution is skewed to the left. " } ], [ { "aoVal": "B", "content": "The median temperature could be 98.2$^{\\circ}F$ " } ], [ { "aoVal": "C", "content": "The median temperature could be 98.4$^{\\circ}F$ " } ], [ { "aoVal": "D", "content": "The minimum temperature is exactly 97.0$^{\\circ}F$ " } ], [ { "aoVal": "E", "content": "The distribution is normal. " } ] ]

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

[ "The correct answer is (b).Since there are 28 patients, the median will be between the 14th and 15th data points, which puts it in the bar centered at 98.25°$F$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$\\frac{\\pi}{2}$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$\\sqrt{1+\\pi}$ " } ], [ { "aoVal": "E", "content": "$1+\\sqrt{\\pi}$ " } ] ]

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

[ "Since points on the graph make the equation true, substitute $\\sqrt{\\pi}$ in to the equation and then solve to find $a$ and $b$. $$ \\begin{aligned} \\&y^{2}+\\sqrt{\\pi}^{4}=2 \\sqrt{\\pi}^{2} y+1 \\textbackslash\\textbackslash{} \\&y^{2}+\\pi^{2}=2 \\pi y+1 \\textbackslash\\textbackslash{} \\&y^{2}-2 \\pi y+\\pi^{2}=1 \\textbackslash\\textbackslash{} \\&(y-\\pi)^{2}=1 \\textbackslash\\textbackslash{} \\&y-\\pi=\\pm 1 \\textbackslash\\textbackslash{} \\&y=\\pi+1 \\textbackslash\\textbackslash{} \\&y=\\pi-1 \\end{aligned} $$ There are only two solutions to the equation $(y-\\pi)^{2}=1$, so one of them is the value of $a$ and the other is $b$. The order does not matter because of the absolute value sign. $$ \\textbar(\\pi+1)-(\\pi-1)\\textbar=2 $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$38$$ " } ], [ { "aoVal": "C", "content": "$$41$$ " } ], [ { "aoVal": "D", "content": "$$44$$ " } ], [ { "aoVal": "E", "content": "$$47$$ " } ] ]

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

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$2⊕4=4⊕2$$ " } ], [ { "aoVal": "B", "content": "$$3⊕6=6⊕3$$ " } ], [ { "aoVal": "C", "content": "$$4⊕8=8⊕4$$ " } ], [ { "aoVal": "D", "content": "$$1008⊕2016=2016⊕1008$$ " } ] ]

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

[ "method $$1$$: （$$1$$）$$\\frac{2}{2^{2}}+ \\frac{1}{4}= \\frac{2}{4}+ \\frac{1}{4}= \\frac{3}{4}$$， $$\\frac{2}{4^{2}}+ \\frac{1}{2}= \\frac{2}{16}+ \\frac{1}{2}= \\frac{10}{16}= \\frac{5}{8}$$. Rejected. （$$2$$）$$\\frac{2}{3^{2}}+ \\frac{1}{6}= \\frac{2}{9}+ \\frac{1}{6}= \\frac{4}{18}+ \\frac{3}{18}= \\frac{7}{18}$$,~ $$\\frac{2}{6^{2}}+ \\frac{1}{3}= \\frac{2}{36}+\\frac{1}{3}=\\frac{1}{18}+\\frac{1}{3}=\\frac{1}{18} +\\frac{6}{18}=\\frac{7}{18}$$. ($$3$$）$$\\frac{2}{4^{2}}+ \\frac{1}{8}= \\frac{2}{16}+ \\frac{1}{8}= \\frac{1}{4}$$, $$\\frac{2}{8^{2}}+ \\frac{1}{4}= \\frac{2}{64}+ \\frac{1}{4}= \\frac{1}{32}+ \\frac{8}{32}= \\frac{9}{32} \\neq \\frac{1}{4}.$$ ($$4$$)$$\\frac{2}{1008^{2}}+ \\frac{1}{2016}= \\frac{1}{252 \\times 2016}+ \\frac{1}{2016}= \\frac{253}{252 \\times 2016}.$$ $$\\frac{2}{2016^{2}}+ \\frac{1}{1008}= \\frac{1}{2016 \\times 1008}+ \\frac{1}{1008}= \\frac{2017}{2016 \\times 1008}.$$ $$1008⊕2016\\neq 2016⊕1008$$. method $$2$$: $$4$$ is twice of $$2$$, $$6$$ is twice of $$3$$, $$8$$ is twice of $$4 $$, $$2016$$ is twice of $$1008$$. Specifically, the question is asking when is $$k⊕2k=2k⊕k$$? $$\\frac{2}{k^{2}}+ \\frac{1}{2k}= \\frac{2}{\\left(2k\\right)^{2}}+ \\frac{1}{k} \\Rightarrow \\frac{2}{k^{2}}+ \\frac{1}{2k}= \\frac{1}{2k^{2}}+ \\frac{1}{k}$$ $$\\Rightarrow \\frac{4}{2k^{2}}+ \\frac{1}{2k}= \\frac{1}{2k^{2}}+ \\frac{2}{2k} \\Rightarrow \\frac{3}{2k^{2}}= \\frac{1}{2k}$$ $$\\Rightarrow \\frac{3}{k}=1 \\Rightarrow k=3$$ Hence, ($$2$$) is the only possible answer. " ]

[]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$52$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

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

[ "$$10+8\\times6-4\\div2=10+48-2=56$$. " ]

[ "其它" ]

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

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

[ "$$Omitted.$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{and I get anything other than 2500 heads, then something is wrong with the way I flip coins.} " } ], [ { "aoVal": "B", "content": "\\textbf{the proportion of heads will be close to 0.5} " } ], [ { "aoVal": "C", "content": "\\textbf{a run of 10 heads in a row will increase the probability of getting a run of 10 tails in a row.} " } ], [ { "aoVal": "D", "content": "\\textbf{the proportion of heads in these tosses is a parameter} " } ], [ { "aoVal": "E", "content": "\\textbf{the proportion of heads will be close to 50.} " } ] ]

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

[ "\\textbf{Since this is a fair coin, the probability to get a head is always 0.5.} " ]

[ "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 median of this dataset is $2$, and the $25$-th percentile is also $2$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "$(100-98)+(96-94)+(92-90)+\\cdots +(8-6)+(4-2)=2\\times25=50$ " ]

[]

[ [ { "aoVal": "A", "content": "$$497$$ " } ], [ { "aoVal": "B", "content": "$$498$$ " } ], [ { "aoVal": "C", "content": "$$499$$ " } ], [ { "aoVal": "D", "content": "$$500$$ " } ] ]

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

[ "Regrouping,$$ (98+102) + (99+101) + 100 = 500$$. " ]

[ "其它" ]

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

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

[ "Solution 1 Using the formula for the volume of a cylinder, we get Alex,~$\\pi108$, and Felicia,~$\\pi216$. We can quickly notice that~~cancels out on both sides, and that Alex\\textquotesingle s volume is~$\\dfrac{1}{2}$~~of Felicia\\textquotesingle s leaving~$\\dfrac{1}{2}=\\boxed{1:2}$~as the answer. Solution 2 Using the formula for the volume of a cylinder, we get that the volume of Alex\\textquotesingle s can is~$3^{2}\\cdot12\\cdot\\pi$, and that the volume of Felicia\\textquotesingle s can is~$6^{2}\\cdot6\\cdot\\pi$. Now, we divide the volume of Alex\\textquotesingle s can by the volume of Felicia\\textquotesingle s can, so we get~$\\dfrac{1}{2}$, which is$\\boxed{\\left( B\\right)\\textbackslash{} 1:2}.$ Solution 3 The ratio of the numbers is~$\\dfrac{1}{2}$. Looking closely at the formula~$r^{2}*h*\\pi$, we see that the~$r*h*\\pi$~will cancel, meaning that the ratio of them will be~$\\dfrac{1\\left( 2\\right)}{2\\left( 2\\right)}=\\boxed{\\left( B\\right)\\textbackslash{} 1:2}$ Solution 4 The second can is 2 size in each of 2 dimensions, and~$\\dfrac{1}{2}$~size in 1 dimension.~$\\dfrac{2^{2}}{2}=\\boxed{\\left( B\\right)\\textbackslash{} 1:2}$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$-9009$$ " } ], [ { "aoVal": "B", "content": "$$-8008$$ " } ], [ { "aoVal": "C", "content": "$$-7007$$ " } ], [ { "aoVal": "D", "content": "$$-6006$$ " } ], [ { "aoVal": "E", "content": "$$-5005$$ " } ] ]

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

[ "$f(x)$ must have four roots, three of which are roots of $g(x)$. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that $$ f(x)=g(x)(x-r) $$ where $r \\in \\mathbb{C}$ is the fourth root of $f(x)$. Substituting $g(x)$ and expanding, we find that $$ \\begin{aligned} f(x) \\&=\\left(x^{3}+a x^{2}+x+10\\right)(x-r) \\textbackslash\\textbackslash{} \\&=x^{4}+(a-r) x^{3}+(1-a r) x^{2}+(10-r) x-10 r \\end{aligned} $$ Comparing coefficients with $f(x)$, we see that $$ \\begin{aligned} a-r \\&=1 \\textbackslash\\textbackslash{} 1-a r \\&=b \\textbackslash\\textbackslash{} 10-r \\&=100 \\textbackslash\\textbackslash{} -10 r \\&=c . \\end{aligned} $$ Let\\textquotesingle s solve for $a, b, c$, and $r$. Since $10-r=100, r=-90$, so $c=(-10)(-90)=900$. Since $a-r=1, a=-89$. Then, since $b=1-a r, b=-8009$. Thus, we know that $$ f(x)=x^{4}+x^{3}-8009 x^{2}+100 x+900 . $$ Taking $f(1)$, we find that $$ \\begin{aligned} f(1) \\&=1^{4}+1^{3}-8009(1)^{2}+100(1)+900 \\textbackslash\\textbackslash{} \\&=1+1-8009+100+900 \\textbackslash\\textbackslash{} \\&=(\\mathbf{C})-7007 . \\end{aligned} $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$751$$ " } ], [ { "aoVal": "C", "content": "$$810$$ " } ], [ { "aoVal": "D", "content": "$$1009$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$807$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{The sampling distribution is skewed to the left, with mean 0.36 and standard deviation 0.076.} " } ], [ { "aoVal": "B", "content": "\\textbf{The sampling distribution is skewed to the right, with mean 0.64 and standard deviation 0.006.~} " } ], [ { "aoVal": "C", "content": "\\textbf{The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.076.~} " } ], [ { "aoVal": "D", "content": "\\textbf{The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.006.~} " } ], [ { "aoVal": "E", "content": "\\textbf{The sampling distribution is approximately normal,with mean 0.64 and standard deviation 0.076.~} " } ] ]

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

[ "\\textbf{For a large n, the sampling distribution of $\\hat{p}$ is approximately normal distribution with $E(\\hat{p})=p$} \\textbf{$\\sigma\\_{\\hat{p}}=\\sqrt{\\frac{p(1-p)}{n}}$.~} \\textbf{$E(\\hat{p})=0.36$} \\textbf{$\\sigma\\_{\\hat{p}}=\\sqrt{\\frac{p(1-p)}{n}} = \\sqrt{\\frac{0.36*0.64}{40}}=0.07589$} " ]

[ "其它" ]

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

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

[ "Count the number of digits in each decimals, the product of the two decimals will have that many digits after the decimal point. " ]

[]

[ [ { "aoVal": "A", "content": "$$15:5$$ " } ], [ { "aoVal": "B", "content": "$$5:15$$ " } ], [ { "aoVal": "C", "content": "$$5:20$$ " } ], [ { "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": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

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

[ "The remainders are $$4$$ and $$9$$. Their sum is $$4 +9= 13$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "We check if it is divisible by $$4$$ by looking at the last two digits. $$72$$ is divisible by $$4$$. " ]

[ "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" ]

[ "The cookies can be divided evenly between $2, 3,$ or $4$ children, which means the number of cookies can be divisible by $2, 3,$ and $4$ at the same time. The cookies cannot be divided evenly between $7$ children because $6$ more cookies would be needed, which means the number of cookies divided by $7$ with $1$ remaining. Thus, the right answer is $D$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$135$$ " } ], [ { "aoVal": "C", "content": "$$197$$ " } ], [ { "aoVal": "D", "content": "$$230$$ " } ], [ { "aoVal": "E", "content": "$$185$$ " } ] ]

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

[ "$np=7(0.8)=5.6 \\uparrow 6$ The $80$-th percentile is $197$. " ]

[]

[ [ { "aoVal": "A", "content": "$$63$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ] ]

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

[ "$$\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}$$ $\\frac{5}{6}\\times96=80$ " ]