Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{{100, 101, 102, 103, 104}} " } ], [ { "aoVal": "B", "content": "\\textbf{{1000.3, 999.56, 1000.49, 1000, 998.32}} " } ], [ { "aoVal": "C", "content": "\\textbf{{1, 1, 1, 1, 1}} " } ], [ { "aoVal": "D", "content": "\\textbf{{5, 10, 15, 20, 25}} " } ], [ { "aoVal": "E", "content": "\\textbf{{1, 2, 3, 4, 5}} " } ] ]

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

[ "\\textbf{variance is used to measure the spread of data. A, B, C, E are densely distributed.We can verify by calculating the variances out. (A) 2.5, (B) 0.598, (C) 0, (D) 50, (E) 2} " ]

[]

[ [ { "aoVal": "A", "content": "$$1007090$$ " } ], [ { "aoVal": "B", "content": "$$1019090$$ " } ], [ { "aoVal": "C", "content": "$$1028090$$ " } ], [ { "aoVal": "D", "content": "$$1037090$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$\\dfrac{1}{2}(2019\\times2018-2018\\times2017+2017\\times2016-2016\\times2015+\\cdots$$$$+5\\times4-4\\times3+3\\times2-2\\times1)$$ $$= \\frac{1}{2}\\left (2 \\times 2018+2 \\times 2016+ \\cdots +2 \\times 4+2 \\times 2\\right )$$ $=\\left (2018+2016+\\cdots+4+2\\right )=1009\\times1010=1019090$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{P(E) = 0.12} " } ], [ { "aoVal": "B", "content": "\\textbf{P(E) = 0.4} " } ], [ { "aoVal": "C", "content": "\\textbf{P(D or E) = 0.28} " } ], [ { "aoVal": "D", "content": "\\textbf{P(D or E) = 0.72} " } ], [ { "aoVal": "E", "content": "\\textbf{P(D or E) = 0.9} " } ] ]

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

[ "\\textbf{Since D and E are independent, P(D ∩ E) = P(D)*P(E). So P(E) = 0.3.} \\textbf{P(D ∪ E) = P(D) + P(E) -- P(D ∩ E) = 0.6 +0.3 -0.18 = 0.72} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "17, 80 " } ], [ { "aoVal": "B", "content": "18, 70 " } ], [ { "aoVal": "C", "content": "9,~~94 " } ], [ { "aoVal": "D", "content": "17, 70 " } ], [ { "aoVal": "E", "content": "20, 80 " } ] ]

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

[ "13+4=17 , 87-17=70 " ]

[]

[ [ { "aoVal": "A", "content": "$$295$$ " } ], [ { "aoVal": "B", "content": "$$520$$ " } ], [ { "aoVal": "C", "content": "$$570$$ " } ], [ { "aoVal": "D", "content": "$$580$$ " } ], [ { "aoVal": "E", "content": "$$620$$ " } ] ]

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

[ "$2^{200} \\times 4^{1000} \\times 8^{40}=16^{200 \\div 4} \\times 16^{1000 \\div 2} \\times 2^{120}$$=16^{50} \\times 16^{500} \\times 16^{30}=16^{580}$. " ]

[]

[ [ { "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" ]

[ "$$4$\\times$100-3=397$$. " ]

[ "其它" ]

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

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

[ "Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\\frac{n}{3}=\\frac{2 s}{5}$, which simplifies to $n=\\frac{6 s}{5}$. Since we are trying to find the value of $\\frac{\\frac{n}{3}+\\frac{2 s}{5}}{n+s}$, we can just substitute $\\frac{6 s}{5}$ for $n$ into the equation. We then get a value of $\\frac{\\frac{6 s}{5}\\cdot\\frac13+\\frac{2 s}{5}}{\\frac{6 s}{5}+s}=\\frac{\\frac{6 s+6 s}{15}}{\\frac{11 s}{5}}=\\frac{\\frac{4 s}{5}}{\\frac{11 s}{5}}=\\frac{4}{11}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac23$$ " } ], [ { "aoVal": "B", "content": "$$\\frac34$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$\\frac43$$ " } ], [ { "aoVal": "E", "content": "$$\\frac32$$ " } ] ]

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

[ "Line $l\\_1$ has the equation $y=\\frac{3x}{2}-\\frac12$ when rearranged. Substituting $1$ for $y$, we find that line $l\\_2$ will meet this line at point $(1,1)$, which is point $B$. We call $\\overline{B C}$ the base and the altitude from $A$ to the line connecting $B$ and $C, y=1$, the height. The altitude has length $\\textbar-2-1\\textbar=3$, and the area of $\\triangle A B C=3$. Since $A=\\frac{bh}{2}, b=2$. Because $l\\_3$ has positive slope, it will meet $l\\_2$ to the right of $B$, and the point that is $2$ to the right of $B$ is $(3,1)$. $l\\_3$ passes through $(-1,-2)$ and $(3,1)$, and thus has slope $\\frac{\\textbar1-(-2)\\textbar}{\\textbar3-(-1)\\textbar}=(\\mathbf{B}) \\frac{3}{4}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3998$$ " } ], [ { "aoVal": "B", "content": "$$3999$$ " } ], [ { "aoVal": "C", "content": "$$4000$$ " } ], [ { "aoVal": "D", "content": "$$4001$$ " } ] ]

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

[ "This is $$(1+999)+(2+998)+(3+997)+(4 +996) = 4000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$x+3+5=11+5$ " } ], [ { "aoVal": "B", "content": "$2x=28$ " } ], [ { "aoVal": "C", "content": "$x+3-11=11-11$ " } ], [ { "aoVal": "D", "content": "$x+3-11=0$ " } ] ]

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

[ "Equations are equivalent when you can obtain one by subtracting, adding, dividing, or multiplying the same number on the other. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$40$$ " } ] ]

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

[ "$18 \\times 10 - 30 \\times 4 - 20 \\times 2 = 20$ " ]

[ "其它" ]

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

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

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

[]

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

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

[ "$$x+2=y$$, then $$x=y-2$$. We can get $$y+4(y-2)=12$$, $$y=4$$, $$x=y-2=2$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$200001$$ " } ], [ { "aoVal": "B", "content": "$$100001$$ " } ], [ { "aoVal": "C", "content": "$$200001\\times 10^{6}$$ " } ], [ { "aoVal": "D", "content": "$$100001\\times 10^{6}$$ " } ] ]

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

[ "$$100001^{2}-100000^{2}=10000200001-10000000000=200001$$. " ]

[]

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

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "First, note that $24+1=25$, which motivates us to factor the polynomial as $\\left(x^{2}-24\\right)\\left(x^{2}-1\\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-24\\textless0\\textless x^{2}-1$. Solving this inequality, we find $1\\textless x^{2}\\textless24$. There are exactly $6$ integers $x$ that satisfy this inequality, $\\pm\\textbackslash{2,3,4\\textbackslash}$. Thus our answer is $(\\mathbf{B}) 6$. " ]

[]

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

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

[ "Total people $=2u+10=40$ $2u=40-10=30$ Girls $=1u=30\\div2=15$ Boys $=1u+10=15+10=25$ Girls $:$ Boys $\\to$ $15:25$ $\\to$ $3:5$ " ]

[]

[ [ { "aoVal": "A", "content": "$$66$$ " } ], [ { "aoVal": "B", "content": "$$77$$ " } ], [ { "aoVal": "C", "content": "$$88$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ] ]

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

[ "The largest two-digit factor is $$3^{2} \\times11 = 99$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$115$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$135$$ " } ] ]

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

[ "$np=7(0.36)=2.52 \\uparrow 3$. The $36$-th percentile is $135$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\frac{3}{2}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{5}$ " } ], [ { "aoVal": "D", "content": "$\\frac{7}{55}$ " } ] ]

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

[ "$\\frac{2}{5}(2x-3)+\\frac{4}{11}x-\\frac{6}{11}=0$ $\\frac{2}{5}(2x-3)+\\frac{2}{11}(2x-3)=0$ $(\\frac{2}{5}+\\frac{2}{11})(2x-3)=0$ $2x-3=0$ $x=\\frac{3}{2}$ " ]

[ "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" ]

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

[]

[ [ { "aoVal": "A", "content": "$$\\frac{8}{16}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{4}{20}$$ " } ] ]

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

[ "$$\\frac{16}{24}=\\frac{8\\times 2}{8\\times 3}=\\frac{2}{3}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$68$$ " } ], [ { "aoVal": "B", "content": "$$168$$ " } ], [ { "aoVal": "C", "content": "$$270$$ " } ], [ { "aoVal": "D", "content": "$$816$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division", "Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Multiplication of Whole Numbers->Multiplication of Multi-Digit Numbers and 1-Digit Numbers->Multiplication of 3-Digit and 1-Digit (with regrouping for more than once)" ]

[ "Stack the two numbers as shown below ,~ lining up the unit digits Remember that the 7 in 272 stands for 7 tens(70) ,the first 2 in the 272 stands for 2 hundreds(200) First, multiply the ones~ $2\\times3=6$~, regroup the 0 tens to the tens column Write ~6 in the ones place. Then,~ Multiply and add the tens .~$3\\times7+0=21$ Write 1 in the tens place and regroup the 2 hundreds. Last, multiply and add the hundreds.~$3\\times2+2=8$ Write 8 in the hundreds place " ]

[ "其它" ]

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

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

[ "$$\\dfrac{5}{14}\\div \\dfrac{10}{21}=\\dfrac{5}{7\\times 2}\\times \\dfrac{3\\times 7}{2\\times 5}=\\dfrac{3}{4}$$． $$\\dfrac{4}{15}\\div \\dfrac{28}{45}=\\dfrac{4}{15}\\times \\dfrac{3\\times 15}{7\\times 4}=\\dfrac{3}{7}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ], [ { "aoVal": "E", "content": "$$58$$ " } ] ]

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

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$1.1$$ " } ], [ { "aoVal": "B", "content": "$$0.98$$ " } ], [ { "aoVal": "C", "content": "$$0.9$$ " } ], [ { "aoVal": "D", "content": "$$1.09$$ " } ], [ { "aoVal": "E", "content": "$$1.9$$ " } ] ]

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

[ "Nil " ]

[]

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

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

[ "$$5-5+5-5+5-5=0$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac{1}{60}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{70}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{210}$ " } ], [ { "aoVal": "D", "content": "$\\frac{1}{380}$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$64$$ " } ], [ { "aoVal": "D", "content": "$$128$$ " } ] ]

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

[ "$$64\\div2=32=2\\times16$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "B " ]

[ "其它" ]

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

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

[ "Two same digits that add up to \"4\" in the last digit, hence it must be either 2 or 7. If we try digit 2, 42+52=94, wrong. If we try digit 7, 27+57=104. correct. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7.7$$ " } ], [ { "aoVal": "B", "content": "$$9.1$$ " } ], [ { "aoVal": "C", "content": "$$9.7$$ " } ], [ { "aoVal": "D", "content": "$$10.1$$ " } ], [ { "aoVal": "E", "content": "$$9.9$$ " } ] ]

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

[ "$$1.2+8.9=10.1$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$52$$ " } ], [ { "aoVal": "E", "content": "$$54$$ " } ] ]

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

[ "$$\\left( 330+22 \\right)\\div 11=352\\div 11=32$$. Alternatively, $$\\left(330 + 22 \\right)\\div 11 = 330 \\div 11 + 22 \\div 11= 30 + 2 = 32$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$243$$ " } ], [ { "aoVal": "B", "content": "$$244$$ " } ], [ { "aoVal": "C", "content": "$$245$$ " } ], [ { "aoVal": "D", "content": "$$246$$ " } ], [ { "aoVal": "E", "content": "$$247$$ " } ] ]

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

[ "Multiplying both sides by $(s-p)(s-q)(s-r)$ yields $$ 1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q) $$ As this is a polynomial identity, and it is true for infinitely many $s$, it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial 0$)$. This means we can plug in $s=p$ to find that $\\frac{1}{A}=(p-q)(p-r)$. Similarly, we can find $\\frac{1}{B}=(q-p)(q-r)$ and $\\frac{1}{C}=(r-p)(r-q)$. Summing them up, we get that $$ \\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r $$ By Vieta\\textquotesingle s Formulas, we know that $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$. Thus the answer is $324-80=$ 244 . " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

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

[ "$24 \\times 4 \\div 2 = 48$ " ]

[ "其它" ]

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

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

[ "You can create the equation $\\frac{x+1}{y+1}=\\frac{11 x}{10 y}$. Cross multiplying and combining like terms gives $x y+11 x-10 y=0$. This can be factored into $(x-10)(y+11)=-110$. $x$ and $y$ must be positive, so $x\\textgreater0$ and $y\\textgreater0$, so $x-10\\textgreater-10$ and $y+11\\textgreater11$. Using the factors of 110 , we can get the factor pairs: $(-1,110),(-2,55)$, and $(-5,22)$. But we can\\textquotesingle t stop here because $x$ and $y$ must be relatively prime. $(-1,110)$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn\\textquotesingle t work. $(-2,55)$ gives $x=8$ and $y=44$. This doesn\\textquotesingle t work. $(-5,22)$ gives $x=5$ and $y=11$. This does work. We found one valid solution so the answer is (B)$1$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$210$$ " } ], [ { "aoVal": "D", "content": "$$400$$ " } ], [ { "aoVal": "E", "content": "$$401$$ " } ] ]

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

[ "We can easily find out she makes $2 \\cdot 20-1=39$ widgets on Day $20$ . Then, we make the sum of $1,3,5, \\ldots \\ldots, 35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+\\ldots+(19+21)$, which include $10$ pairs of $40$ . So the sum of $1,3,5, \\ldots \\ldots \\ldots 39$ is $(40 \\cdot 10)=(\\text{D}) 400$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

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

[ "The pitcher is $\\frac{3}{5}$ full, i.e. $60 \\textbackslash\\%$ full. Therefore each cup receives $\\frac{60}{6}=(\\mathbf{C}) 10$ percent of the total capacity. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ], [ { "aoVal": "E", "content": "$$29$$ " } ] ]

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

[ "The sum of the first digit and the second digit is the third digit, and so on. The horizontal line is $11+18=29$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$84$$ " } ], [ { "aoVal": "C", "content": "$$107$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ], [ { "aoVal": "E", "content": "$$214$$ " } ] ]

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

[ "We multiply the first ratio by 4 on both sides, and the second ratio by 3 to get the same number for 8 th graders, in order that we can put the two ratios together: $$ \\begin{aligned} \\&11: 6=11(4): 6(4)=44: 24 \\textbackslash\\textbackslash{} \\&8: 13=8(3): 13(3)=24: 39 \\end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $44: 24: 39$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 44+24+39=\\text { (C) } 107 $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

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

[ "Let $f$ be a linear function with slope $m$. $$ \\begin{gathered} m=\\frac{f(5)-f(2)}{\\Delta x}=\\frac{10}{5-2}=\\frac{10}{3} \\textbackslash\\textbackslash{} f(8)-f(2)=m \\Delta x=\\frac{10}{3}(8-2)=20 \\Rightarrow(C) \\end{gathered}$$ " ]

[ "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": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "First, note that $50+1=51$, which motivates us to factor the polynomial as $\\left(x^{2}-50\\right)\\left(x^{2}-1\\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-50\\textless0\\textless x^{2}-1$. Solving this inequality, we find $1\\textless x^{2}\\textless50$. There are exactly 12 integers $x$ that satisfy this inequality, $\\pm\\textbackslash{2,3,4,5,6,7\\textbackslash}$. Thus our answer is $(\\mathbf{C}) 12$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$165$$ " } ], [ { "aoVal": "B", "content": "$$175$$ " } ], [ { "aoVal": "C", "content": "$$177$$ " } ], [ { "aoVal": "D", "content": "$$180$$ " } ] ]

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

[ "Arrange the data from the least to the largest: $157$, $160$, $165$, $175$, $177$, $180$, $184$, $197$. $np=8(0.42)=3.36 \\uparrow 4$. The $42$-th percentile is $175$. " ]

[]

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

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

[ "If $$x+2=y$$, then $$x=y-2$$. We can get $$y+4(y-2)=12$$ implying that $$y=4$$ and hence $$x=y-2=2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{80 years} " } ], [ { "aoVal": "B", "content": "\\textbf{115 years} " } ], [ { "aoVal": "C", "content": "\\textbf{120 years} " } ], [ { "aoVal": "D", "content": "\\textbf{125 years} " } ], [ { "aoVal": "E", "content": "\\textbf{130 years} " } ] ]

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

[ "\\textbf{P(X\\textgreater x) = 0.9~} \\textbf{P(Z\\textgreater$$\\frac{x-100}{15}$$) = 0.9} \\textbf{$$\\frac{x-100}{15}$$ = 1.29} \\textbf{X = 119.35} " ]

[ "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" ]

[ "Let $x$ be the third number. It follows that the first number is $6 x$, and the second number is $x+40$. We have $$ 6 x+(x+40)+x=8 x+40=96, $$ from which $x=7$. Therefore, the first number is $42$ , and the second number is $47$ . Their absolute value of the difference is $\\textbar42-47\\textbar=(\\mathbf{E}) 5$. " ]

[ "其它" ]

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

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

[ "$$32-7=25$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$3 \\times x+4$ " } ], [ { "aoVal": "B", "content": "$x \\times y+z$ " } ], [ { "aoVal": "C", "content": "$6a+b$ " } ], [ { "aoVal": "D", "content": "$3a+b \\times c$ " } ] ]

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

[ "Only $ \\text C$ is in accordance with the rules of writing expression. " ]

[]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$47$$ " } ], [ { "aoVal": "D", "content": "$$37$$ " } ], [ { "aoVal": "E", "content": "$$33$$ " } ] ]

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{2}{15}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{11}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{7}{15}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{4}{15}$$ " } ] ]

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$-1.82$$ " } ], [ { "aoVal": "C", "content": "$$0.55$$ " } ], [ { "aoVal": "D", "content": "$$1.82$$ " } ], [ { "aoVal": "E", "content": "$$-0.55$$ " } ] ]

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

[ "\\textbf{Z = 1.65 = $\\frac{43-40}{\\sigma}$ → $\\sigma = \\frac{3}{1.65}=1.82$} " ]

[]

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

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

[ "The ones digit of $$2015^{2015}+ 2016^{2016}$$ is the same as that of $$5+ 6$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac5{49}$ " } ], [ { "aoVal": "B", "content": "$\\frac5{89}$ " } ], [ { "aoVal": "C", "content": "$\\frac5{17}$ " } ], [ { "aoVal": "D", "content": "$\\frac1{31}$ " } ], [ { "aoVal": "E", "content": "$\\frac5{81}$ " } ] ]

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

[ "The last fraction should be $\\frac{85}{89}$, so the answer is $\\frac5{89}.$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "A " ]

[]

[ [ { "aoVal": "A", "content": "$$80$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$121$$ " } ] ]

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

[ "$1+3+5+7+9+ 11 + 13+ 15+17 +19=10^{2}=100$ " ]

[]

[ [ { "aoVal": "A", "content": "$$0.25\\rm cm$$ " } ], [ { "aoVal": "B", "content": "$$2.5\\rm cm$$ " } ], [ { "aoVal": "C", "content": "$$25\\rm cm$$ " } ], [ { "aoVal": "D", "content": "$$250\\rm cm$$ " } ], [ { "aoVal": "E", "content": "$$2500\\rm cm$$ " } ] ]

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

[ "omitted " ]

[]

[ [ { "aoVal": "A", "content": "$$80$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$110$$ " } ] ]

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

[ "$$\\dfrac{1}{2}(19\\times18-18\\times17+17\\times16-16\\times15+\\cdots$$$$+5\\times4-4\\times3+3\\times2-2\\times1)$$ $$=\\dfrac{1}{2} (2(18) + 2(16) + \\cdots + 2(2))$$ $$= 18 + 16 + 14 +\\cdots + 2$$ $$= 2(1 + 2 + \\cdots + 9)$$ $$= 2(45)$$ $$= 90$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$380^{}\\circ \\rm C$$ " } ], [ { "aoVal": "B", "content": "$$390^{}\\circ \\rm C$$ " } ], [ { "aoVal": "C", "content": "$$399^{}\\circ \\rm C$$ " } ], [ { "aoVal": "D", "content": "$$400^{}\\circ \\rm C$$ " } ] ]

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

[ "Rounding, $$398^{}\\circ \\rm C$$ is closer to $$400^{}\\circ \\rm C$$ than to $$390^{}\\circ \\rm C$$. " ]

[ "其它" ]

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

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

[ "We want to know the $2020$-th term and the $2021$-st term to get the median. We know that $44^{2}=1936$. So, numbers $1^{2}, 2^{2}, \\ldots, 44^{2}$ are in between $1$ and $1936$. So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$-th number. Also, notice that $45^{2}=2025$, which is larger than $2021$. Then the $2020$-th term will be $1936+40=1976$, and similarly the $2021$-th term will be $1977$. Solving for the median of the two numbers, we get (C) $1976.5$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ], [ { "aoVal": "E", "content": "$$34$$ " } ] ]

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

[ "There are $5$ days from May $1$ to May $5$ . If we set the first day as $n$, the second day can be expressed as $n+6$, the third as $n+12$, and so on, for five days. The sum $n+(n+6)+(n+12)+(n+18)+(n+24)$ is equal to $100$ , as stated in the problem. We can write a very simple equation, that is: $5 n+60=100$. Now all we do is just solve. $5 n=40$, so Big Al eats 8 bananas on the first day. The fifth day, $n+24$, is then 32 , which is your answer. " ]

[ "其它" ]

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

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

[ "We know from the triangle inequality that the last side, $s$, fulfills $s+2\\textgreater5$. Therefore, $P\\textgreater5+5$. The least integer value of $P$ is $11$ " ]

[ "其它" ]

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

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

[ "$37\\div5=7R2$ $7+1=8$ " ]

[ "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" ]

[ "From $f(1)=0$, we know that $a+b+c=0$. From the first inequality, we get $50\\textless49 a+7 b+c\\textless60$. Subtracting $a+b+c=0$ from this gives us $50\\textless48 a+6 b\\textless60$, and thus $\\frac{25}{3}\\textless8 a+b\\textless10$. Since $8 a+b$ must be an integer, it follows that $8 a+b=9$. Similarly, from the second inequality, we get $70\\textless64 a+8 b+c\\textless80$. Again subtracting $a+b+c=0$ from this gives us $70\\textless63 a+7 b\\textless80$, or $10\\textless9 a+b\\textless\\frac{80}{7}$. It follows from this that $9 a+b=11$. We now have a system of three equations: $a+b+c=0,8 a+b=9$, and $9 a+b=11$. Solving gives us $(a, b, c)=(2,-7,5)$ and from this we find that $f(100)=2(100)^{2}-7(100)+5=19305$. Since $15000\\textless19305\\textless20000 \\rightarrow 5000(3)\\textless19305\\textless5000(4)$, we find that $k=3 \\rightarrow(\\mathbf{C}) 3$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

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

[ "The tenths digit is $$6$$ and the hundredths digit is $$7$$. Their sum is $$13$$. " ]

[]

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

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

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde3.87+4.62+6.13+5.38$$ $$=(3.87+6.13)+(4.62+5.38)$$ $$=10+10$$ $$=20$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{4}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}$$ " } ] ]

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

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$21$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

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

[ "Observe that each number is the sum of the number of previous term and the difference between adjacent numbers. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "I " } ], [ { "aoVal": "B", "content": "II " } ], [ { "aoVal": "C", "content": "I, II " } ], [ { "aoVal": "D", "content": "I, III " } ], [ { "aoVal": "E", "content": "II, III " } ] ]

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

[ "Because 0 is not in the interval (0.07, 0.19). It is unlikely to be the true difference between the proportions. III is just plain wrong! We cannot make a probability statement about an interval we have already constructed. All we can say is that the process used to generate this interval has a 0.95 chance of producing an interval that does contain the true population proportion. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$180$$ " } ], [ { "aoVal": "B", "content": "$$115$$ " } ], [ { "aoVal": "C", "content": "$$135$$ " } ], [ { "aoVal": "D", "content": "$$140$$ " } ] ]

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

[ "$np=7(0.36)=2.52 \\uparrow 3$ The $36$-th percentile is $135$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{16}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8}{9}$$ " } ] ]

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

[ "$$\\frac{4}{9}\\div \\frac{1}{4}=\\frac{4}{9}\\times \\frac{4}{1}=\\frac{16}{9}$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "None " } ], [ { "aoVal": "B", "content": "$I$ and $II$ only " } ], [ { "aoVal": "C", "content": "$II$ and $III$ only " } ], [ { "aoVal": "D", "content": "$I$, $II$, and $III$ " } ] ]

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

[ "$\\sqrt[3]{x^{4}}$ can be written as $x^{}\\frac{4}{3}$, which is equivalent to II and III. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$82$$ " } ], [ { "aoVal": "B", "content": "$$85$$ " } ], [ { "aoVal": "C", "content": "$$88$$ " } ], [ { "aoVal": "D", "content": "$$91$$ " } ] ]

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

[ "$$1+(30-1)\\times 3=88$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "The correct answer to the calculation $$123123123123\\div123=1001001001$$. This has $$10$$ digits. " ]

[]

[ [ { "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" ]

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

[]

[ [ { "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": "$\\frac{12}{13}$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$\\frac{13}{12}$ " } ], [ { "aoVal": "D", "content": "$\\frac{24}{13}$ " } ], [ { "aoVal": "E", "content": "$$2$$ " } ] ]

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

[ "Let $2 d$ miles be the distance from the trailhead to the fire tower, where $d\\textgreater0$. When Chantal meets Jean, the two have traveled for $$ \\frac{d}{4}+\\frac{d}{2}+\\frac{d}{3}=d\\left(\\frac{1}{4}+\\frac{1}{2}+\\frac{1}{3}\\right)=d\\left(\\frac{3}{12}+\\frac{6}{12}+\\frac{4}{12}\\right)=\\frac{13}{12} d $$ hours. At that point, Jean has traveled for $d$ miles, so his average speed is $\\frac{d}{\\frac{13}{12} d}=(\\mathbf{A}) \\frac{12}{13}$ miles per hour. " ]

[]

[ [ { "aoVal": "A", "content": "$(x+y)-(x-y)$ " } ], [ { "aoVal": "B", "content": "$(2x+3y)+(-2x-y)$ " } ], [ { "aoVal": "C", "content": "$(3x-2y)+(-3x+4y)$ " } ], [ { "aoVal": "D", "content": "$(x-y)+(-x+y)$ " } ] ]

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

[ "$(x+y)-(x-y)=2y$ $(2x+3y)+(-2x-y)=2x+3y-2x-y=2y$ $(3x-2y)+(-3x+4y)=3x-2y-3x+4y=2y$ $(x-y)+(-x+y)=x-y-x+y=0$ " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{33}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{8}$$ " } ] ]

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

[ "$0.33 = 33\\div100 = \\dfrac{33}{100}$, so choice $\\text{B}$ is correct. " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

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

[ "$$9\\times 9+9\\times 8+9\\times 7+9\\times 6=9\\times~ \\left( {9+8+7+6} \\right) $$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{$0.32 \\pm 1.960(\\frac{(0.32)(0.68)}{\\sqrt{1,005}})$} " } ], [ { "aoVal": "B", "content": "\\textbf{$0.32 \\pm 1.645(\\frac{(0.32)(0.68)}{\\sqrt{1,005}})$} " } ], [ { "aoVal": "C", "content": "\\textbf{$0.32 \\pm 2.575\\sqrt{(\\frac{(0.32)(0.68)}{1,005})}$} " } ], [ { "aoVal": "D", "content": "\\textbf{$0.32 \\pm 1.960\\sqrt{(\\frac{(0.32)(0.68)}{1,005})}$} " } ], [ { "aoVal": "E", "content": "\\textbf{$0.32 \\pm 1.645\\sqrt{(\\frac{(0.32)(0.68)}{1,005})}$} " } ] ]

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

[ "\\textbf{$\\hat{p} \\pm z^{*}\\sigma\\_{\\hat{p}}=\\hat{p} \\pm Z\\_{\\frac{\\alpha}{2}}\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}= 0.32 \\pm Z\\_{0.05}\\sqrt{\\frac{(0.32)(0.68)}{1005}}$} " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{2022}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{2021}$ " } ], [ { "aoVal": "C", "content": "$$2022$$ " } ], [ { "aoVal": "D", "content": "$$2021$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$3- \\frac{2}{1}=1$$ $$5- \\frac{4}{1}=1$$ $$\\cdots \\cdots$$ $$2022- \\frac{2021}{1}=1$$ $$\\frac{2022}{1}=2022$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$126$$ " } ], [ { "aoVal": "B", "content": "$$127$$ " } ], [ { "aoVal": "C", "content": "$$128$$ " } ], [ { "aoVal": "D", "content": "$$129$$ " } ], [ { "aoVal": "E", "content": "$$130$$ " } ] ]

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

[ "C " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$225$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "$23\\times45\\div23=45$ " ]

[]

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

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

[ "$$6+6+6-6-6=6$$． " ]

[]

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

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems" ]

[ "In weight, $$45$$ mice $$=3\\times (15$$ mice$$)=3\\times (2$$ cats$$)= 2\\times (3$$ cats$$)=2\\times (2$$ dogs$$)= 4$$ dogs. On Planet Pythagoras, the people use a different money system to us. two pog is worth six pings. " ]

[]

[ [ { "aoVal": "A", "content": "$$10\\times 99$$ " } ], [ { "aoVal": "B", "content": "$$10\\div 99$$ " } ], [ { "aoVal": "C", "content": "$$9\\times 99$$ " } ], [ { "aoVal": "D", "content": "$$99\\times 99$$ " } ] ]

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

[ "The sum of ten $$99$$\\textquotesingle s is the same as $$10\\times 99$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$52$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ] ]

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

[ "$$\\left( 100-98 \\right)+\\left( 96-94 \\right)+\\ldots +\\left( 8-6 \\right)+\\left( 4-2 \\right)+0=2\\times 25+0=50$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$4.035$$ " } ], [ { "aoVal": "B", "content": "$$4.14$$ " } ], [ { "aoVal": "C", "content": "$$4.35$$ " } ], [ { "aoVal": "D", "content": "$$4.7$$ " } ], [ { "aoVal": "E", "content": "$$4.72$$ " } ] ]

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

[ "$$4\\frac{7}{20}=4\\frac{35}{100}=4.35$$ " ]

[ "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" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

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

[ "$*2543*=2\\times3+5\\times4=6+20=26$. So the answer is $\\rm C$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0.111$$ " } ], [ { "aoVal": "B", "content": "$$0.331$$ " } ], [ { "aoVal": "C", "content": "$$0.558$$ " } ], [ { "aoVal": "D", "content": "$$0.669$$ " } ], [ { "aoVal": "E", "content": "\\textbf{Not enough information is given to determine the probability} " } ] ]

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

[ "\\textbf{The sample in the report is about veterinarians. The population cannot be a household in the United States.} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$62$$ " } ], [ { "aoVal": "B", "content": "$$63$$ " } ], [ { "aoVal": "C", "content": "$$64$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ], [ { "aoVal": "E", "content": "$$66$$ " } ] ]

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

[ "You can see that since the ratio of real building\\textquotesingle s heights to the model building\\textquotesingle s height is $1: 35$. If the height of the tower is $2250$ feet, to find the height of the model, we divide by $35$ . That gives us $64.28$ which rounds to $64$ . Therefore, to the nearest whole number, the duplicate is (C) $64$ feet. " ]

[]

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

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

[ "$$25$$ dimes $$=250$$￠$$=125\\times 2$$￠$$=125\\times 2$$ pennies. " ]

[]

[ [ { "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 $$4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2022$$ " } ], [ { "aoVal": "B", "content": "$$202$$ " } ], [ { "aoVal": "C", "content": "$$22$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$2026$ " } ] ]

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

[ "$2+0+2+2=6$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{13}{3}$$，$$2$$． " } ], [ { "aoVal": "B", "content": "$$\\dfrac{14}{3}$$，$$2$$． " } ], [ { "aoVal": "C", "content": "$$\\dfrac{14}{3}$$，$$\\dfrac{3}{2}$$． " } ], [ { "aoVal": "D", "content": "$$\\dfrac{13}{3}$$，$$\\dfrac{3}{2}$$． " } ] ]

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

[ "$$\\dfrac{8}{9}\\div \\dfrac{4}{7}\\div \\dfrac{1}{3}=\\dfrac{4\\times 2}{3\\times 3}\\times \\dfrac{7}{4}\\times \\dfrac{3}{1}=\\dfrac{2\\times 7}{3}=\\dfrac{14}{3}$$． $$\\dfrac{7}{9}\\div \\dfrac{1}{3}\\div 1\\dfrac{5}{9}=\\dfrac{7}{9}\\times \\dfrac{3}{1}\\times \\dfrac{9}{14}==\\dfrac{7}{9}\\times \\dfrac{3}{1}\\times \\dfrac{9}{2\\times 7}=\\dfrac{3}{2}$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "75 is divisible by 1, 3, 5,15, 25 and 75. 30 is divisible by 1, 2, 3, 5\\textsubscript{Z} 6,10,15 and 30. 15 is the highest common divisor for both 75 and 30. $\\textasciitilde$ To get the least number of squares, the length of the squares must be the largest possible. Hence the length must be 15 and there must be 5 x 2 = \\textbf{10} squares. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2223$$ " } ], [ { "aoVal": "B", "content": "$$2022$$ " } ], [ { "aoVal": "C", "content": "$$1921$$ " } ], [ { "aoVal": "D", "content": "$$1821$$ " } ], [ { "aoVal": "E", "content": "$$1793$$ " } ] ]

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

[ "$522-321=201$, so the sum will decrease by $201$. $2022-201=1821$. " ]

[]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$66$$ " } ], [ { "aoVal": "C", "content": "$$78$$ " } ], [ { "aoVal": "D", "content": "$$132$$ " } ] ]

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

[ "($13$ \\&~$22$) $\\times$ ($3$ \\&~$6$) = $22\\times6=132$ " ]