math-eval/TAL-SCQ5K huggingface dataset:
Problem: Find the value of $$\left\textbackslash{ \frac{2018+1}{5} \right\textbackslash} + \left\textbackslash{ \frac{2018+2}{5} \right\textbackslash} + \cdots \cdots + \left\textbackslash{ \frac{2018+2017}{5} \right\textbackslash} + \left\textbackslash{ \frac{2018+2018}{5} \right\textbackslash}$$. 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$751$$ 
aoVal: C
content: $$810$$ 
aoVal: D
content: $$1009$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: $$807$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: \textbf{According to data from the United States Elections Project, only 36 percent of eligible voters voted in the 2014 elections. For random samples of size 40, which of the following best describes the sampling distribution of pˆ, the sample proportion of people who voted in the 2014 elections?} 
Options:
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.~} 
Answer: C
Analysis: \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{$\sigma\_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.36*0.64}{40}}=0.07589$} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: How many digits after the decimal does the product of $3.222$ and $4.22$ have? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: D
Analysis: Count the number of digits in each decimals, the product of the two decimals will have that many digits after the decimal point. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals
Source: 其它
Problem: In your family, there are $$15$$ chairs, $$5$$ tables, and $$20$$ cups. What is the ratio of chairs to cups? 
Options:
aoVal: A
content: $$15:5$$ 
aoVal: B
content: $$5:15$$ 
aoVal: C
content: $$5:20$$ 
aoVal: D
content: $$3:4$$ 
Answer: D
Analysis: There are $$15$$ chairs and $$20$$ cups. So the ratio of chairs to cups is $$15:20$$. The simplest form is $$3:4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: What is the sum of the remainders of $$1234 \div5$$ and $$6789 \div 10$$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$13$$ 
Answer: D
Analysis: The remainders are $$4$$ and $$9$$. Their sum is $$4 +9= 13$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: If the 9-digit number $$2017122\square2$$ can be divisible by $$4$$, then the number in $$\square$$ can be ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: D
Analysis: We check if it is divisible by $$4$$ by looking at the last two digits.  $$72$$ is divisible by $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: A box has fewer than $50$ cookies in it. The cookies can be divided evenly between $2, 3,$ or $4$ children. However, they cannot be divided evenly between $7$ children because $6$ more cookies would be needed. How many cookies are there in the box? (2021 Math Kangaroo Problem, Level 3-4, Question \#20) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$36$$ 
aoVal: E
content: $$48$$ 
Answer: D
Analysis: 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 $D$. 
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 $80$-th percentile. 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$135$$ 
aoVal: C
content: $$197$$ 
aoVal: D
content: $$230$$ 
aoVal: E
content: $$185$$ 
Answer: C
Analysis: $np=7(0.8)=5.6 \uparrow 6$ The $80$-th percentile is $197$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Meena had $$96$$ coconuts. She sold $$\frac{1}{3}$$ of them yesterday and $$\frac{1}{2}$$ of them today. How many coconuts did she sell altogether? 
Options:
aoVal: A
content: $$63$$ 
aoVal: B
content: $$72$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$84$$ 
Answer: C
Analysis: $$\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$$  $\frac{5}{6}\times96=80$ 
Knowledge points: 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
Problem: Mariam had ＄$$4y$$. After buying some cloth at ＄$$7$$ per metre, she had ＄$$y$$ left. How many metres of cloth did she buy? 
Options:
aoVal: A
content: $$\frac{3y}{7}$$ 
aoVal: B
content: $$\frac{5y}{7}$$ 
aoVal: C
content: $21y$ 
aoVal: D
content: $$35y$$ 
Answer: A
Analysis: Amount of money she spent to buy the cloth  $$\rightarrow$$＄$$4y-$$＄$$y$$  $$=$$＄$$3y$$,  Length of cloth she bought  $$\rightarrow$$＄$$3y\div $$＄$$7/\text{m}$$  $$= \frac{3y}{7}\text{m}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: The mid points of the four sides of a rectangle are $(−3,0)$, $(2,0)$, $(5,4)$, and $(0, 4)$. What is the area of the rectangle? 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$50$$ 
aoVal: E
content: $$80$$ 
Answer: C
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Pick two consecutive positive integers whose sum is less than $$100$$. Square both of those integers and then find the difference of the square numbers. Which of the following could be the difference? ($2007$ AMC $8$ Problem, Question \#$19$) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$64$$ 
aoVal: C
content: $$79$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$131$$ 
Answer: C
Analysis: Let\textquotesingle s say that $x$ is the smaller of the two numbers. So the question is $$(x+1)+x\textless100(x+1)^{2}-x^{2}=x^{2}+2x+1-x^{2}=2x+1$$. ~$$2x+1$$ is obviously odd, so the answer could be $$\text{C}$$ or $$\text{E}$$.  $$2x+1=131$$ doesn\textquotesingle t match with $$2x+1\textless100$$, so the answer is $$79$$.  Therefore, the answer is $$\text{C}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas-> Difference of Two Squares Formula
Problem: Two integers are inserted into the list $3,3,8,11,28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers? 
Options:
aoVal: A
content: $$56$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$58$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$61$$ 
Answer: D
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The $1986^{}\text{th}$ digit at the right of the decimal point in the decimal expression of $\dfrac{1}{7}$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
Answer: D
Analysis: $$\frac{1}{7}=0.\overline{142857}$$, it is a decimal which repeats in cycles of $6$. Every $6$\textsuperscript{th}~digit is $7$. The $1986$\textsuperscript{th} digit is $7$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation
Problem: There are four more girls than boys in Mr. Tse\textquotesingle s class of 28 students. What is the ratio of number of girls to the number of boys in her class? 
Options:
aoVal: A
content: $4:3$ 
aoVal: B
content: $3:2$ 
aoVal: C
content: $7:4$ 
aoVal: D
content: $2:1$ 
Answer: A
Analysis: We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$. Since the total number of students is equal to 28 , we get $x+x-4=28$. Solving this equation, we get $x=16$. There are $16-4=12$ boys in our class, and our answer is $16: 12=$ (B) $4: 3$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Cassandra is helping her mother to pack $$75$$ cupcakes. The boxes that her mother prepares can only fit $$7$$ cupcakes. She must ensure the box is full before she can use the next box. How many boxes she can fill up? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$68$$ 
aoVal: E
content: $$70$$ 
Answer: C
Analysis: $$75\div7=10R5$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers
Source: 其它
Problem: If $*abcd*=a\times d+b\times c$, then $*2543*=$~\uline{~~~~~~~~~~}~. ($2004$ Math League.com contest problem, $8$\textsuperscript{th} Grade, Question \#$33$) 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$120$$ 
Answer: C
Analysis: $*2543*=2\times3+5\times4=6+20=26$.  So the answer is $\rm C$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: Louis had $8$ sticks. He broke three of them into two pieces. How many sticks does he have now? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$13$$ 
Answer: C
Analysis: $8 + 3 = 11$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $3^{}p+3^{4}=90$, $2^{}r+44=76$, and $5^{3}+6^{}s=1421$, what is the product of $p$, $r$, and $s$? (2013 AMC 8 Problem, Question \#15) 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$70$$ 
aoVal: E
content: $$90$$ 
Answer: B
Analysis: Start with $3^{}p+3^{4}=90$. Then, change $3^{4}$ to $81$. Subtract from $81$ from both sides to get $3^{}p=9$ and see that $p$ is $2$. Now, solve for $r$. Since $2^{}r+44=76$, $2^{}r$ must equal $32$, so $r=5$. Now, solve for $s$. $5^{3}+6^{}s=1421$ can be simplified to $125+6^{}s=1421$, which simplifies further to $6^{}s=1296$. Therefore, $s=4$. $prs$ equals $2\times5\times4$ which equals $40$. So, the answer is $40$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers
Source: 其它
Problem: Hannah bought $$5$$ whole pizzas: $$1$$ for herself and $$1$$ for each of her $$4$$ students.  Hannah sliced her pizza into $$5$$ equal parts and ate only $$3$$ slices.  Student $$A$$ sliced his pizza into $$4$$ equal parts, but ate only $$3$$ slices.  Student $$B$$ sliced his pizza into $$8$$ equal parts, but ate only $$7$$ slices.  Student $$C$$ sliced her pizza into $$3$$ equal parts, but ate only $$2$$ slices.  Student $$D$$ sliced her pizza into $$6$$ equal parts, but ate only $$3$$ slices.  Who ate less pizza than Hannah? 
Options:
aoVal: A
content: Student $$A$$ 
aoVal: B
content: Student $$B$$ 
aoVal: C
content: Student $$C$$ 
aoVal: D
content: Student $$D$$ 
aoVal: E
content: Nobody 
Answer: D
Analysis: All of them eat more than $$\frac{1}{2}$$, except student $$D$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: When Koko the Koala is awake, he eats $84$ grams of leaves per hour. He was awake for $2$ hours yesterday and $10$ hours today. How many grams of leaves did he eat in total in the two days? (adapted from 2014 Math Kangaroo Problem, Level 3-4, Question \#4) 
Options:
aoVal: A
content: $$840$$ 
aoVal: B
content: $$168$$ 
aoVal: C
content: $$672$$ 
aoVal: D
content: $$1008$$ 
Answer: D
Analysis: $84 \times (10 + 2) = 84 \times 10 + 84 \times 2 = 840 + 168 = 1008$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Source: 其它
Problem: Which of the following is not an equivalent ratio of $$4:12$$? 
Options:
aoVal: A
content: $$1:3$$ 
aoVal: B
content: $$2:6$$ 
aoVal: C
content: $$8:36$$ 
aoVal: D
content: $$32:96$$ 
Answer: C
Analysis: Option A: $$1:3$$ $\to$ $4:12$  Option B: $2:6$ $\to$ $$1:3$$ $\to$ $4:12$  Option C: $8:36$ $\to$ $4:18$  Option D: $32:96$ $\to$ $$1:3$$ $\to$ $4:12$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions
Problem: In a math competition, each participant has a unique $5$-digit registration number of the form $\overline{BBCAC}$, where $0 \leq A \textless{} B\textless{} C \leq 9$ and $B$ is the average of $A$ and $C$. What is the maximum number of participant that can join this competition? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$25$$ 
Answer: D
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: For $\triangle ABC$, all its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $14$ and a side length of $8$ is at least ． 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: $$29$$ 
Answer: E
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+8\textgreater14$. Therefore, $P\textgreater14+14$. The least integer value of $P$ is $29$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: There are two positive integers, $$x$$ and $$y$$. $$x$$ equals to $$3^{2}$$, and $$y$$ is the base of $$5^{3}$$. What is the product of $$x$$ and $$y$$? 
Options:
aoVal: A
content: $15$ 
aoVal: B
content: $45$ 
aoVal: C
content: $90$ 
aoVal: D
content: $1125$ 
Answer: B
Analysis: $$x=3^{2}=9$$  $$5$$ cubed is $$5^{3}$$. And the base is $$5$$.  So, $$x\cdot y=9\times 5=45$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers
Source: 其它
Problem: SASMO 2016 P2 Q5  What number does () stands for? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: banana = 32-24= 8  A + 8 = 12  A = 4 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns->Encryption and Decryption
Source: 其它
Problem: For $\triangle ABC$, all its side lengths are integers. The perimeter of $\triangle ABC$ with a side of length $14$ and a side length of $8$ is at least ． 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: $$29$$ 
Answer: E
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+8\textgreater14$. Therefore, $P\textgreater14+14$. The least integer value of $P$ is $29$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $2018$ is an interesting number. This is because when we add the first digit and the last digit, we will get the reverse of the middle two digits. Which of the options below is also an interesting number? 
Options:
aoVal: A
content: $$3014$$ 
aoVal: B
content: $$3129$$ 
aoVal: C
content: $$4319$$ 
aoVal: D
content: $$2017$$ 
aoVal: E
content: $$4913$$ 
Answer: C
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: Given: $a-b\textgreater a$ and $a+b\textless b$. Which of the following must be true?~\uline{~~~~~~~~~~}~  $I$. $a b$ is negative  $II$. $a+b$ is negative  $III$. $a-b$ is negative 
Options:
aoVal: A
content: $I$ only 
aoVal: B
content: $I$ and $III$ only 
aoVal: C
content: $II$ only 
aoVal: D
content: $I$, $II$, $III$ 
Answer: C
Analysis: $a-b\textgreater a$ then $b\textless0$  $a+b\textless b$ then $a\textless0$.  So both $a$ and $b$ are negative. $I$ is false, since the product will be positive. $II$ is true since the sum of two negative numbers is negative. $III$ is false because if $a=-1$ and $b=-2$ then $a-b$ is positive. Only $II$ is true. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: What is the remainder when $2^{2023}+2023$ is divided by $2^{20}+1$? (Adapted From 2020 AMC 10B Problems, Question \#22) 
Options:
aoVal: A
content: $$2015$$ 
aoVal: B
content: $$2^{5}$$ 
aoVal: C
content: $$2023$$ 
aoVal: D
content: $$2048$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: Let $x=2^{20}$.  We are now looking for the remainder of $\frac{8x^{101}+2023}{x+1}$. By Polynomial Remainder Theorem, the remainder is $8\times (-1)^{101} + 2023 = 2015$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The last four digits in Andy\textquotesingle s ID card are $2025$. What is the the difference between the largest and the smallest digit in $2025$?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$0$$ 
Answer: D
Analysis: The largest digit is $5$, and the smallest digit is $0$.  $5-0=5$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: Let $a$ and $b$ be relatively prime positive integers with $a\textgreater b\textgreater0$ and $\frac{a^{3}-b^{3}}{(a-b)^{3}}=\frac{73}{3}$. What is $a-b$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: C
Analysis: Slightly expanding, we have that $\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$. Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that $0=70 a^{2}-149 a b+70 b^{2}$. Dividing everything by $b^{2}$, we get that $$ 0=70\left(\frac{a}{b}\right)^{2}-149\left(\frac{a}{b}\right)+70 \text {. } $$ Applying the quadratic formula and following the restriction that $a\textgreater b\textgreater0$ $$ \frac{a}{b}=\frac{10}{7} \text {. } $$ Hence, $7 a=10 b$. Since they are relatively prime, $a=10, b=7$. $$ 10-7= 3 \text {. } $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Which number has to be subtracted from $17$ in order to obtain $-33$? (Adapted from 2017 Math Kangaroo Problem, Level 7-8, Question \#3) 
Options:
aoVal: A
content: $$-50$$ 
aoVal: B
content: $$-16$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$40$$ 
aoVal: E
content: $$50$$ 
Answer: E
Analysis: $17-50=-33$, so the answer is $E$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Source: 其它
Problem: Fill in the blank:~\uline{~~~~~~~~~~}~is $$3$$ tens $$7$$ ones less than $$4$$ tens $$6$$ ones. 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$73$$ 
aoVal: D
content: $$83$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: $$3$$ tens $$7$$ ones: $$37$$  $$4$$ tens $$6$$ ones: $$46$$  less than: $$46-37=9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: There were $$16$$ monkeys in total in the animal school. After the whistle, they arranged themselves into $$8$$ rows. How many monkeys were there in each row after the whistle? (Adapted from 2005 Math Kangaroo Problem, Level 3-4, Question \#4) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: B
Analysis: There were $$16 \div 8 = 2$$ monkeys in each row. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Source: 其它
Problem: Find all values of $x$ such that $\textbar3 x+12\textbar\textless9$ and $\textbar x+2\textbar\textless\textbar-3 x-6\textbar$.~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $x\textless-2$ 
aoVal: B
content: $-7\textless x\textless-1$ 
aoVal: C
content: $-7\textless x\textless-2$ 
aoVal: D
content: $-7\textless x\textless-1 ; x \neq-2$ 
Answer: D
Analysis: Based on the first inequality, we know that $-9\textless3 x+12\textless9$ because $3 x+2$ must be less than 9 units away from zero. We can subtract 12 from all three parts of the inequality to arrive at $-21\textless3 x\textless-3 \rightarrow-7\textless x\textless-1$. From the second inequality we can rewrite $\textbar-3 x-6\textbar$ as $\textbar3 x+6\textbar=3\textbar x+2\textbar$ because they must be equal. The second inequality must be true for all numbers except for when $\textbar x+2\textbar=0$, or when $x=-2$. Thus the answer includes all numbers from $-7$ to $-1$ with the exception of $-2$. The answer is $\mathbf{D}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: For $\triangle ABC$, all its side lengths are integers. The perimeter of $\triangle ABC$ with a side of length $3$ and a side length of $4$ is at least ． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
Answer: C
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+3\textgreater4$. Therefore, $P\textgreater4+4$. The least integer value of $P$ is $9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3 x+y}{x-3 y}=-2$. What is the value of $\frac{x+3 y}{3 x-y}$? 
Options:
aoVal: A
content: $$-3$$ 
aoVal: B
content: $$-1$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$3$$ 
Answer: D
Analysis: Rearranging, we find $3 x+y=-2 x+6 y$, or $5 x=5 y \Longrightarrow x=y$. Substituting, we can convert the second equation into $\frac{x+3 x}{3 x-x}=\frac{4 x}{2 x}= 2$ More step-by-step explanation:  $$ \begin{aligned} \&\frac{3 x+y}{x-3 y}=-2 \textbackslash\textbackslash{} \&3 x+y=-2(x-3 y) \textbackslash\textbackslash{} \&3 x+y=-2 x+6 y \textbackslash\textbackslash{} \&5 x=5 y \textbackslash\textbackslash{} \&x=y \textbackslash\textbackslash{} \&\frac{x+3 y}{3 x-y}=\frac{1+3(1)}{3(1)-1}=\frac{4}{2}=2 \end{aligned} $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
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