math-eval/TAL-SCQ5K huggingface dataset:
Problem: If $$x+2 = y$$ and $$y+4x = 12$$, then $$x=$$． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: A
Analysis: If $$x+2=y$$, then $$x=y-2$$.  We can get $$y+4(y-2)=12$$ implying that $$y=4$$ and hence $$x=y-2=2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Problem: \textbf{Scientists estimate that the distribution of the life span of the Galápagos Islands giant tortoise is approximately normal with mean 100 years and standard deviation 15 years. Based on the estimate, which of the following is closest to the age of a Galápagos Islands giant tortoise at the 90th percentile of the distribution?} 
Options:
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} 
Answer: C
Analysis: \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} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The sum of three numbers is $96$. The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers? (2022 AMC 10A Problems, Question \#3) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: E
Analysis: 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 $\textbar42-47\textbar=(\mathbf{E}) 5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $7$ less than $32$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$28$$ 
Answer: C
Analysis: $$32-7=25$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: Which of the following expression is written correctly? 
Options:
aoVal: A
content: $3 \times x+4$ 
aoVal: B
content: $x \times y+z$ 
aoVal: C
content: $6a+b$ 
aoVal: D
content: $3a+b \times c$ 
Answer: C
Analysis: Only $ \text C$ is in accordance with the rules of writing expression. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: What should be the last number in the series be?  $$1$$, $$2$$, $$5$$, $$10$$, $$17$$, $$26$$, ． 
Options:
aoVal: A
content: $$35$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$47$$ 
aoVal: D
content: $$37$$ 
aoVal: E
content: $$33$$ 
Answer: D
Analysis: $$+1$$, $$+3$$, $$+5$$, $$+7$$, $$+9$$, $$+11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: Which of the following is the largest fraction?  $$\dfrac{2}{15}$$,$$\dfrac{11}{15}$$,$$\dfrac{7}{15}$$,$$\dfrac{4}{15}$$ 
Options:
aoVal: A
content: $$\dfrac{2}{15}$$ 
aoVal: B
content: $$\dfrac{11}{15}$$ 
aoVal: C
content: $$\dfrac{7}{15}$$ 
aoVal: D
content: $$\dfrac{4}{15}$$ 
Answer: B
Analysis: Same denominator, so larger numerator means larger fraction 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Source: 其它
Problem: \textbf{One of the values in a normal distribution is 43, and its z-score is 1.65. If the mean of the distribution is 40, what is the standard deviation of the distribution?} 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$-1.82$$ 
aoVal: C
content: $$0.55$$ 
aoVal: D
content: $$1.82$$ 
aoVal: E
content: $$-0.55$$ 
Answer: D
Analysis: \textbf{Z = 1.65 = $\frac{43-40}{\sigma}$ → $\sigma = \frac{3}{1.65}=1.82$} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the ones digit of $$2015^{2015}+2016^{2016}$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: The ones digit of $$2015^{2015}+ 2016^{2016}$$ is the same as that of $$5+ 6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: Joe writes an expression $\frac59\times\frac9{13}\times\frac{13}{17}\cdots $ Following the pattern, he writes the expression with $\frac{45}{49}$ as the multiplier in the middle. What is the result of the expression? 
Options:
aoVal: A
content: $\frac5{49}$ 
aoVal: B
content: $\frac5{89}$ 
aoVal: C
content: $\frac5{17}$ 
aoVal: D
content: $\frac1{31}$ 
aoVal: E
content: $\frac5{81}$ 
Answer: B
Analysis: The last fraction should be $\frac{85}{89}$, so the answer is $\frac5{89}.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Calculate: $\frac{2022+4567\times 7890}{4568\times 7890 - 5778}$ 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: A 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: $1+3+5+7+9+ 11 + 13+ 15+17 +19=$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$80$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$100$$ 
aoVal: D
content: $$121$$ 
Answer: C
Analysis: $1+3+5+7+9+ 11 + 13+ 15+17 +19=10^{2}=100$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: The most likely height of a single-decker bus is． 
Options:
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: $$2500\rm cm$$ 
Answer: D
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length
Problem: $$\frac{1}{2}~ (19 \times 18-18 \times 17+17 \times 16-16 \times 15+\cdots$$$$ +5\times4-4\times3 +3\times2-2\times1 )$$． 
Options:
aoVal: A
content: $$80$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$100$$ 
aoVal: D
content: $$110$$ 
Answer: B
Analysis: $$\dfrac{1}{2}(19\times18-18\times17+17\times16-16\times15+\cdots$$$$+5\times4-4\times3+3\times2-2\times1)$$  $$=\dfrac{1}{2} (2(18) + 2(16) + \cdots + 2(2))$$  $$= 18 + 16 + 14 +\cdots + 2$$  $$= 2(1 + 2 + \cdots + 9)$$  $$= 2(45)$$  $$= 90$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: Round $$398^{}\circ \rm C$$ to the nearest $$10^{}\circ $$. 
Options:
aoVal: A
content: $$380^{}\circ \rm C$$ 
aoVal: B
content: $$390^{}\circ \rm C$$ 
aoVal: C
content: $$399^{}\circ \rm C$$ 
aoVal: D
content: $$400^{}\circ \rm C$$ 
Answer: D
Analysis: Rounding, $$398^{}\circ \rm C$$ is closer to $$400^{}\circ \rm C$$ than to $$390^{}\circ \rm C$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Estimating Large Numbers
Problem: What is the median of the following list of 4040 numbers: $$ 1,2,3, \ldots, 2020,1^{2}, 2^{2}, 3^{2}, \ldots, 2020^{2}$$? (2020 AMC 10A Problems, Question \#11) 
Options:
aoVal: A
content: $$1974.5$$ 
aoVal: B
content: $$1975.5$$ 
aoVal: C
content: $$1976.5$$ 
aoVal: D
content: $$1977.5$$ 
aoVal: E
content: $$1978.5$$ 
Answer: C
Analysis: 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) $1976.5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Big Al, the ape, ate $100$ bananas from May $1$ through May $5$ . Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May $5$ ? (2005 AMC 8 Problems, Question \#12) 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$32$$ 
aoVal: E
content: $$34$$ 
Answer: D
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating
Source: 其它
Problem: For $\triangle ABC$, all its side lengths are integer. The primeter of $\triangle ABC$ with a side of length $2$ and a side length of $5$ is at least ． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
Answer: D
Analysis: 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 $11$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: One basket can contain no more than $5$ eggs. What is the smallest number of baskets needed to contain $37$ eggs? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: $37\div5=7R2$  $7+1=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Let $f(x)=a x^{2}+b x+c$, where $a, b$, and $c$ are integers. Suppose that $f(1)=0,50\textless f(7)\textless60,70\textless f(8)\textless80$, $5000 k\textless f(100)\textless5000(k+1)$ for some integer $k$. What is $k$? (2011 AMC 12A Problems, Question 20) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: C
Analysis: 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 $k=3 \rightarrow(\mathbf{C}) 3$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the sum of the tenths and the hundredths digits in the num-ber $$12345.6789$$? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$15$$ 
Answer: C
Analysis: The tenths digit is $$6$$ and the hundredths digit is $$7$$. Their sum is $$13$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Addition and Subtraction of Decimals
Problem: $$3.87+4.62+6.13+5.38=$$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$19$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$21$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$23$$ 
Answer: B
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde3.87+4.62+6.13+5.38$$  $$=(3.87+6.13)+(4.62+5.38)$$  $$=10+10$$  $$=20$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals
Problem: Find the difference between $$\frac{5}{9}$$ and $$\frac{1}{3}$$ 
Options:
aoVal: A
content: $$\frac{4}{6}$$ 
aoVal: B
content: $$\frac{2}{9}$$ 
aoVal: C
content: $$\frac{4}{9}$$ 
aoVal: D
content: $$\frac{2}{3}$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: In the arithmetic sequence $5$, $7$, $9$, $11$ $\cdots$ , the $9$\textsuperscript{th} term is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$19$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$25$$ 
Answer: B
Analysis: Observe that each number is the sum of the number of previous term and the difference between adjacent numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: A 95\% confidence interval for the difference between two population proportions is found to be (0.07, 0.19). Which of the following statement is (are) true?  I. It is unlikely that the two populations have the same proportions.  II. We are 95\% confidence that the true difference between population proportions is between 0.07 and 0.19.  III. The probability is 0.95 that the true difference between the population proportions is between 0.07 and 0.19. 
Options:
aoVal: A
content: I 
aoVal: B
content: II 
aoVal: C
content: I, II 
aoVal: D
content: I, III 
aoVal: E
content: II, III 
Answer: C
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The following are the weights (in pounds) of seven people: $100, 115, 135, 140, 180, 197, 230$. Find the $36$-th percentile. 
Options:
aoVal: A
content: $$180$$ 
aoVal: B
content: $$115$$ 
aoVal: C
content: $$135$$ 
aoVal: D
content: $$140$$ 
Answer: D
Analysis: $np=7(0.36)=2.52 \uparrow 3$ The $36$-th percentile is $135$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$\frac{4}{9}\div \frac{1}{4}=$$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$\frac{1}{9}$$ 
aoVal: B
content: $$\frac{16}{9}$$ 
aoVal: C
content: $$\frac{2}{9}$$ 
aoVal: D
content: $$\frac{8}{9}$$ 
Answer: B
Analysis: $$\frac{4}{9}\div \frac{1}{4}=\frac{4}{9}\times \frac{4}{1}=\frac{16}{9}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: If $x\textgreater0$, which of the following is equivalent to $\sqrt[3]{x^{4}}$ ?  $ $ $I. x+x^{\frac{1}{3}}$  $ $ $II. \left(x^{\frac{1}{3}}\right)^{4}$  $ $ $III. x^{2}\left(x^{-\frac{2}{3}}\right)$ 
Options:
aoVal: A
content: None 
aoVal: B
content: $I$ and $II$ only 
aoVal: C
content: $II$ and $III$ only 
aoVal: D
content: $I$, $II$, and $III$ 
Answer: C
Analysis: $\sqrt[3]{x^{4}}$ can be written as $x^{}\frac{4}{3}$, which is equivalent to II and III. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the $$30$$th term of the following arithmetic sequence $$1, 4, 7, 10, \cdots$$? 
Options:
aoVal: A
content: $$82$$ 
aoVal: B
content: $$85$$ 
aoVal: C
content: $$88$$ 
aoVal: D
content: $$91$$ 
Answer: C
Analysis: $$1+(30-1)\times 3=88$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: How many digits are there in the correct answer to the calculation $$123123123123\div 123$$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$10$$ 
Answer: D
Analysis: The correct answer to the calculation $$123123123123\div123=1001001001$$.  This has $$10$$ digits. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: A computer makes $$4\times {{10}^{9}}$$ operations per second. How many operations does it make in $$5\times {{10}^{2}}$$ seconds? 
Options:
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: $$2\times {{10}^{19}}$$ 
Answer: C
Analysis: $$4\times {{10}^{9}}\times 5\times {{10}^{2}}$$  $$=20\times {{10}^{11}}$$  $$=2\times {{10}^{12}}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Problem: Kitty writes down a sequence of five integers. The rule she uses is, "after the first two terms, each term is the sum of the two previous terms." Her sequence is ---, ---, ---, ~$$18$$, $$29$$.  What is her first term? ． 
Options:
aoVal: A
content: $$0 $$ 
aoVal: B
content: $$ 3 $$ 
aoVal: C
content: $$ 4 $$ 
aoVal: D
content: $$ 5 $$ 
aoVal: E
content: $$ 7$$ 
Answer: C
Analysis: 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 $$a=4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean\textquotesingle s average speed, in miles per hour, until they meet? (2021 AMC 10A Problems, Question \#6) 
Options:
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: $$2$$ 
Answer: A
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: After simplying the following expressions, which one is different from others? 
Options:
aoVal: A
content: $(x+y)-(x-y)$ 
aoVal: B
content: $(2x+3y)+(-2x-y)$ 
aoVal: C
content: $(3x-2y)+(-3x+4y)$ 
aoVal: D
content: $(x-y)+(-x+y)$ 
Answer: D
Analysis: $(x+y)-(x-y)=2y$  $(2x+3y)+(-2x-y)=2x+3y-2x-y=2y$  $(3x-2y)+(-3x+4y)=3x-2y-3x+4y=2y$  $(x-y)+(-x+y)=x-y-x+y=0$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: $ 0.33 =$． 
Options:
aoVal: A
content: $$\frac{3}{10}$$ 
aoVal: B
content: $$\frac{33}{100}$$ 
aoVal: C
content: $$\frac{1}{3}$$ 
aoVal: D
content: $$\frac{3}{8}$$ 
Answer: B
Analysis: $0.33 = 33\div100 = \dfrac{33}{100}$, so choice $\text{B}$ is correct. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: $$81+72+63+54=9\times $$． 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$36$$ 
Answer: C
Analysis: $$9\times 9+9\times 8+9\times 7+9\times 6=9\times~ \left( {9+8+7+6} \right) $$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: \textbf{From a random sample of 1,005 adults in the United States, it was found that 32 percent own an e-reader. Which of the following is the appropriate 90 percent confidence interval to estimate the proportion of all adults in the United States who own an e-reader?~} 
Options:
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{$0.32 \pm 1.645\sqrt{(\frac{(0.32)(0.68)}{1,005})}$} 
Answer: E
Analysis: \textbf{$\hat{p} \pm z^{*}\sigma\_{\hat{p}}=\hat{p} \pm Z\_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.32 \pm Z\_{0.05}\sqrt{\frac{(0.32)(0.68)}{1005}}$} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: ~$\dfrac{2022}{2021-\dfrac{2020}{2019-\dfrac{2018}{5-\dfrac{4}{3-\dfrac{2}{1}}}}}$$=$． 
Options:
aoVal: A
content: $\dfrac{1}{2022}$ 
aoVal: B
content: $\dfrac{1}{2021}$ 
aoVal: C
content: $$2022$$ 
aoVal: D
content: $$2021$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: $$3- \frac{2}{1}=1$$  $$5- \frac{4}{1}=1$$  $$\cdots \cdots$$  $$2022- \frac{2021}{1}=1$$  $$\frac{2022}{1}=2022$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions
Problem: It is known that the number $2017$ is placed at $M^{\text{th}}$ row, $N^{\text{th}}$ entry. Find the value of $M+N$. 
Options:
aoVal: A
content: $$126$$ 
aoVal: B
content: $$127$$ 
aoVal: C
content: $$128$$ 
aoVal: D
content: $$129$$ 
aoVal: E
content: $$130$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns of Number Tables->Triangle Number Table
Source: 其它
Problem: Definite @ as the operation of choosing the larger number between two numbers. Definite \&~as the operation of choosing the smaller number between two numbers. Find the result of $(37$\&$23)\times(25$@$45)\div(23$@$15)$. 
Options:
aoVal: A
content: $$23$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$225$$ 
aoVal: E
content: $$100$$ 
Answer: C
Analysis: $23\times45\div23=45$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition
Source: 其它
Problem: Fill "$$+$$" or~"$$-$$" in the blanks to make the equation true.~\uline{~~~~~~~~~~}~  $$6$$~~~~ $$6$$~~~~ $$6$$~~~~ $$6$$~~~~ $$6=6$$ 
Options:
aoVal: A
content: $$+++-$$ 
aoVal: B
content: $$++++$$ 
aoVal: C
content: $$++-\/-$$ 
Answer: C
Analysis: $$6+6+6-6-6=6$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Problem: If $$2$$ dogs weigh as much as $$3$$ cats, and $$2$$ cats weigh as much as $$15$$ mice, how many dogs weigh as much as $$45$$ mice? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: In weight, $$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. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems